Stochastic Volatility Models with Closed-Form Solutions

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```					                Stochastic Volatility models
with closed-form solutions

Solomon M. Antoniou

SKEMSYS
Scientific Knowledge Engineering
and Management Systems
37 Κoliatsou Street, Corinthos 20100, Greece
solomon_antoniou@yahoo.com

Abstract
We introduce a methodology of building stochastic volatility models with closed-form
solutions. The valuation of options makes use of the option pricing PDE, derived under
SV, and Heston-Bakshi-Madan (HBM) formula, accompanied by the Fourier inversion
formula (the Gil-Pelaez formula). The probabilities P1 and P2 entering to the HBM

formula are expressed in terms of the characteristic functions f 1 and f 2 respectively and

satisfy a pair of partial differential equations. The two functions entering to the
exponentiation of the characteristic functions satisfy a system of Riccati equations. The
various quantities appearing in the Riccati equations are proved to be linear with respect
to the underlined stochastic process Y and thus by assigning simple functions and
converting Riccati to second order differential equation, we obtain closed form solutions,
expressed in terms of elementary or special functions (hypergeometric functions, Heun’s
functions) .

Keywords: stochastic volatility models, closed-form solutions, Heston model, Heston-
Bakshi-Madan formula, Gil-Pelaez inversion formula, special functions, hypergeometric
functions, Heun functions.
1. Introduction
The Black-Scholes option pricing model (Black and Scholes [6]) was
derived on the assumption of constant volatility. However this assumption does
not appear to be true in everyday market transactions, giving rise to the so-called
volatility smile (see for example Dupire [11] and [12], Bergomi [5], Fouque,
Papanicolaou and Sircar [18], Gatheral [19], Lee [34], Lewis [35], Savine [47],
Schönbucher [48]). In this paper we consider, as in Heston (Heston [23]), that
volatility σ t is not a constant quantity, but it is instead a function of an underlying
stochastic process Yt : σ t = f (Yt ) where f (⋅) is a smooth function.
The paper is organized as follows:
In Section 2 we introduce a general form of models where the stochastic volatility
is a function that depends on an underlying stochastic process. The pricing PDE of
options under SV is derived by considering a portfolio Π which consists of an
option F we want to make pricing, − ∆1 units of the underlying asset S and
− ∆ 2 units of the benchmark option G to implement the hedging. We thus derive
a generic PDE for pricing options under SV. In Section 3 the pricing function F
of a European call option is expressed in terms of the probability functions P1 and

P2 through the Heston-Bakshi-Madan (HBM) formula: F = S ⋅ P1 − K ⋅ e − r (T − t ) P2
(see Heston [23], Bakshi and Madan [4]). Upon substituting this form of F into
the generic PDE for option pricing derived in Section 2, we obtain two
independent PDEs satisfied by P1 and P2 , which in turn, the same PDEs are
satisfied by the characteristic functions f1 and f 2 . Assuming that f1 and f 2
admit an exponential form representation, the functions which enter, satisfy a
system of Riccati ODEs. We then prove that the various unspecified functions
entering the Riccati ODEs, are linear in Y, the underlined stochastic process. One
of Riccati’s equation (for each one of the functions f1 and f 2 ), is then converted
into a second order ODE. This approach has the obvious advantage: There is no

2
need of considering any pre-assigned model, like Feller’s square-root process
(Feller [14] and [15]), Uhlenbeck-Ornstein process (Uhlenbeck and Ornstein [52])
or any other stochastic process like the 3/2-process considered by Lewis (Lewis
[35]).We are thus free to consider any number of models by assigning values to
the time-dependent unspecified functions, so that the second order ODEs admits a
closed-form solution (in terms of special functions). The other Riccati equation
(provided that we have found a solution to the first Riccati equation through the
second order ODE), can easily be integrated. In Section 4 we consider a simple
model (very similar to the Heston model) which serves as a guide to our approach.
In the next four Sections (Section 5 until Section 8) we introduce a number of
models in which the functions entering the exponential representation of f1 and
f 2 are expressed in terms of hyperbolic functions. In section 9 we consider a
model where the functions f1 and f 2 are expressed in terms of exponential and
hypergeometric functions. In Sections 10 and 11, the functions f1 and f 2 are
expressed in terms of confluent functions (Abramowitz and Stegun [1], Lebedev
[33], Whittaker and Watson [53]). Finally in Sections 12 and 13, the functions f1
and f 2 are expressed in terms of the confluent Heun functions (Decarreau et al
[9], Forsyth [17], Heun [24], Maier [38] and [39], Ronveau [46], Slavyanov and
Lay [51], Whittaker and Watson [53]). In the above models, the probability
functions P1 and P2 are expressed in terms of the characteristic functions
through the Gil-Pelaez formula (Gil-Pelaez [20], Gurland [22], Heston [23],
Lukacs [37], Scott [49], Sephard [50]). The improper integrals entering to the Gil-
Pelaez formula can in principle be evaluated using the Adamchik-Marichev
algorithm (Adamchik and Marichev [2], Roach [44] and [45]) using Meijer’s G-
function (Meijer [41], Mathai and Saxena [40], Roach [44] and [45]), implemented
in many Symbolic Languages.
In all the models we consider, the argument of the functions depend on τ ≡ T − t
(T is the maturity time). We thus claim that the time-dependent coefficients we

3
consider, might capture more detailed behavior of the movement of the derivative
price under stochastic volatility.
In a forthcoming paper, we shall consider (to the models we have examined in this
paper) some supplementary issues like jumps (in the sense of Andersen and
Andreasen [3], Das [7], Das and Foresi [8], Duffie, Pan and Singleton [10], Kou
[30], Kou, Petrella and Wang [31], Kou and Wang [32], Lewis [35]). We shall
also consider stochastic interest rates (as in Jiang [26] and Grzelak and Oosterlee
[21]).

2. Stochastic Volatility Models.
The PDE for Pricing Options under Stochastic Volatility.
2.1. The general form of the Models.
We consider continuous stochastic volatility (SV) models which consist of the
following:
A number of stocks S t which follow a geometric Brownian motion
dS t = µ S t dt + σ t S t dWt                                   (2.1)
The drift µ is considered to be constant, while the volatility σ t is a positive
function of an underlying stochastic process Yt ,
σ t = f (Yt )                                                   (2.2)
satisfying the stochastic differential equation
dYt = a ( t , Yt ) dt + b( t , Yt ) dZ t                        (2.3)
while Wt and Z t are two correlated Brownian motions
dWt dZ t = ρ dt                                                 (2.4)
where ρ is the correlation coefficient, considered to be a constant.

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2.2. The PDE for Option Pricing under SV
In order to price SV models, we make use of no-arbitrage considerations. The risk-
free portfolio is constructed as in the Black-Scholes case. However the
construction method is quite different.
In the SV option pricing model, there is only one risk asset S, but two random
sources Wt and Z t . Therefore the market is non-complete and we cannot have a
perfect replication of the underlying stock. However no-arbitrage considerations
are not enough to evaluate the option price. It is known that the market can be
completed by adding an option written on a stock S. The market is complete as
long as we have two traded assets, the underlying stock S and one benchmark
option G. All other options can become replicated from these two traded assets.
A risk-free portfolio Π consists of an option F we want to make pricing, − ∆1
units of the underlying asset S and − ∆ 2 units of the benchmark option G.
Therefore
Π = F − ∆1 ⋅ S − ∆ 2 ⋅ G                                     (2.5)
The portfolio is self-financing, therefore
dΠ = dF − ∆1 ⋅ dS − ∆ 2 ⋅ dG                                 (2.6)
The F and G are functions of the variables t, S t and Yt . By applying the 2-
dimensional Itô formula, we obtain (see any standard textbook, like Elliot and
Kopp [13], Karatzas and Shreve [28], Klebaner [29] and Øksendal [43] )
∂F 1 2        ∂ 2F 1 2 ∂ 2F                ∂2 F 
dΠ =    + f (Y) S2     + b       + ρ f (Y) S b         dt +
 ∂t 2         ∂ S2 2 ∂ Υ 2                ∂ S ∂Y 
                                                 
∂F      ∂F
+      dS +    dY − ∆1 ⋅ dS −
∂S      ∂Y

 ∂ G 1 2
                ∂ 2G 1 2 ∂ 2G                ∂2 G 
− ∆ 2 ⋅     + f (Y) S2     + b       + ρ f (Y) S b         dt +
 ∂ t 2
               ∂ S2 2 ∂ Υ 2                ∂ S ∂Y 


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∂G      ∂G    
+      dS +    dY                                           (2.7)
∂S      ∂Y 

In the previous relation, we rearrange the various terms so as to appear the
coefficients of dt , dS and dY :
∂F 1 2        ∂ 2F 1 2 ∂ 2F                ∂2 F 
dΠ =    + f (Y) S2
 ∂t 2             + b       + ρ f (Y) S b         dt −
              ∂ S2 2 ∂ Υ 2                ∂ S ∂Y 


∂G 1 2       2 ∂ G
2
1 2 ∂ 2G                ∂2 G 
− ∆2 ⋅
   + f (Y) S       + b       + ρ f (Y) S b         dt +
 ∂t 2          ∂ S2 2 ∂ Υ 2                ∂ S ∂Y 

 ∂F        ∂G             ∂F        ∂G   
+    − ∆2 ⋅    − ∆1  dS +     − ∆2 ⋅       dY                   (2.8)
 ∂S        ∂S             ∂Y        ∂Y   
In order to have a risk-free portfolio, we take the coefficients of dS and dY to
be zero:
∂F        ∂G
− ∆2 ⋅    − ∆1 = 0                                          (2.9)
∂S        ∂S

∂F        ∂G
− ∆2 ⋅    =0                                                (2.10)
∂Y        ∂Y
We solve the previous system with respect to ∆1 and            ∆ 2 , and we get the
following expressions
∂F ∂G ∂F  ∂G
∆1 =     −   ⋅                                                (2.11)
∂S  ∂S ∂ Y  ∂ Y
        
∂F     ∂G
∆2 =                                                            (2.12)
∂Y     ∂Y
This portfolio is risk-free as long as we substitute the two previous expressions for
the ∆1 and ∆ 2 into (2.8). On the other side, the portfolio should gain a risk-free
interest, to avoid arbitrage opportunities. Therefore we should have
dΠ = r Π dt
which means that

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∂F 1 2       2 ∂ F
2
1 2 ∂2F                ∂2 F 
dΠ = 
   + f (Y) S       + b      + ρ f (Y) S b         dt −
 ∂t 2          ∂ S2 2 ∂ Υ 2               ∂ S ∂Y 

∂F
∂Y ∂G 1 2
           2 ∂ G 1 2 ∂ G
2        2
∂2 G  dt
−    ⋅   + f (Y) S        + b     + ρ f (Y) S b
∂G  ∂t 2          ∂S 2  2 ∂ Υ2               ∂ S ∂Y 
                                               
∂Y
= r Π dt                                                            (2.13)
where Π is given by (2.5), with ∆1 and ∆ 2 given by (2.11) and (2.12)
respectively:
      ∂G ∂F               ∂F
         ⋅  
 ∂ F ∂S ∂ Y               ∂Y
Π = F−      −         ⋅S −            ⋅G                                (2.14)
 ∂S     ∂G                ∂G
        ∂Y                ∂Y
            
Combining the previous two equations, (2.13) and (2.14), we obtain
∂F 1 2           2 ∂ F
2
1 2 ∂ 2F
 ∂ t + 2 f (Y ) S ∂ S 2 + 2 b ∂ Υ 2 +



∂2 F            ∂F
      ∂F
+ ρ f (Y) S b        − r F+ rS             =
∂ S ∂Y           ∂S
      ∂Y

∂G 1 2        ∂ 2G 1 2 ∂ 2G
=   + f (Y) S2
 ∂t 2             + b       +
              ∂ S2 2 ∂ Υ 2

∂2 G            ∂G 
   ∂G
+ ρ f (Y) S b          − rG + rS                         (2.15)
∂ S ∂Y           ∂S 
   ∂Y

The left hand side of the previous equality is a function of F while the right hand
side is a function of G. In order this equation holds true, the two members should
be equal to a function − k ( t , S, Y) , which is independent of any option.
The equation for F can now be written as
∂F 1 2       ∂2F 1 2 ∂ 2F                ∂2 F
+ f (Y) S2     + b      + ρ f (Y) S b        −rF+
∂t 2         ∂ S2 2 ∂ Υ 2               ∂ S ∂Y

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∂F                     ∂F
+ rS      = − k ( t , S, Y) ⋅                                 (2.16)
∂S                     ∂Y
We write the function k ( t , S, Y) to the form
k ( t , S, Y) = a ( t , Y) − b( t , Y) ⋅ Λ( t , S, Y)         (2.17)
where a ( t , Y) is the drift term of the process Yt while Λ( t , S, Y) is called the
market price of the volatility risk.
The PDE for the option takes the form
∂F 1 2      2 ∂ F
2
1 2 ∂ 2F                ∂2 F
+ f (Y) S       + b       + ρ f (Y) S b        −rF+
∂t 2          ∂ S2 2 ∂ Υ 2                ∂ S ∂Y
∂F                ∂F
+ rS      + (a − b Λ ) ⋅    =0                                (2.18)
∂S                ∂Y
Given the terminal condition for F, the PDE (2.18) is solved under the driving
process Yt . The function Λ( t , S, Y) cannot be determined from the arbitrage
theory. In theory it is determined by the benchmark option G. In practice it is the
market which determines the value of the volatility risk.

3. Solving the PDE for the Option Pricing
Considering F to be the price of a European call option, we have (Heston [23],

F = S ⋅ P1 − K ⋅ e − r (T − t ) P2                            (3.1)
where K is the strike price and T is the maturity.
Using this form of solution, we may substitute the partial derivatives of F into the
equation (2.18). That would result into two partial differential equations for the
functions P1 and P2 . In fact we find, using (3.1), the following expressions for the

partial derivatives
∂F    ∂ P1                                            ∂ P2
=S      − K r e − r (T − t ) P2 − K e − r (T − t )
∂t     ∂t                                              ∂t

8
∂F          ∂ P1                    ∂ P2
= P1 + S      − K e − r (T − t )
∂S          ∂S                       ∂S

∂2 F          ∂ P1        ∂ 2 P1               − r (T − t )   ∂ 2 P2
=2          +S            − Ke
∂S2             ∂S         ∂S 2                               ∂S 2

∂F    ∂ P1                    ∂ P2
=S      − K e − r (T − t )
∂Y    ∂Y                      ∂Y

∂2 F            ∂ 2 P1          − r (T − t )   ∂ 2 P2
=S            − Ke
∂Y2             ∂Y2                            ∂Y2

∂ 2 F ∂ P1    ∂ 2 P1     − r (T − t ) ∂ P2
2
=     +S        −Ke
∂S ∂ Y ∂ Y     ∂S ∂ Y                  ∂S ∂ Y
Substituting the partial derivatives calculated above into equation (2.18), we find
∂ P1                                                       ∂ P2
S          − K r e − r (T − t ) P2 − K e − r (T − t )                  +
∂t                                                          ∂t

                                       
2  ∂ P1    ∂ 2 P1     − r (T − t ) ∂ P 2 
2
1 2
+ f ( Y ) S 2      +S        −Ke                    +
2            ∂S
         ∂S 2                     ∂S 2 


1 2  ∂ P1                       
− r (T − t ) ∂ P2 
2                     2

+ b S      −Ke                   +
2  ∂Y 2                    ∂Y 2 
                            
 ∂ P1    ∂ 2 P1                         
− r (T − t ) ∂ P2 
2

+ ρ f (Y) b S       +S        −Ke                     −
 ∂Y

∂S ∂ Y                  ∂S ∂ Y 


− r {S ⋅ P1 − K ⋅ e − r ( T − t ) P2 } +

       ∂ P1                   ∂ P2 
+ r S P1 + S      − K e− r (T − t )      +
       ∂S                      ∂S 

 ∂ P1                   ∂ P2 
+ (a − b Λ ) S     − K e− r (T − t )      =0
 ∂Y                     ∂Y 

The previous equation can be written as

9
 ∂ P1           ∂ P1 1 2      2 ∂ P1 1 2 ∂ P1
2        2

S      + f (Y) S
2
+ f (Y) S       + b       +
 ∂t

∂S 2            ∂S 2 2   ∂ Y2

∂ P1                    ∂ 2 P1          ∂ P1                   ∂ P1 

+ ρ f (Y) b                 + ρ f (Y) b S             + rS           + (a − b Λ )        −
∂Y                     ∂S ∂ Y              ∂S                  ∂Y 

 ∂ P2 1
2 ∂ P2  1 2 ∂ P2
2        2
− r (T − t ) 
− Ke                   + f (Y) S
2
+ b       +
 ∂t 2
                 ∂S2 2 ∂ Y 2

∂ 2 P2           ∂ P2∂ P2 

+ ρ f (Y) b S        + rS    + (a − b Λ )      =0
∂S ∂ Y      ∂S              ∂Y 

The above equation should be true for any K and any S. Therefore we obtain the
following two equations satisfied by P1 and P2 :

∂ P1             1 2      2 ∂ P1 1 2 ∂ P1
∂ P12        2
+ f (Y) S
2
+ f (Y) S       + b       +
∂t           ∂S 2          ∂S2 2    ∂ Y2

∂ P1                         ∂ 2 P1          ∂ P1                  ∂ P1
+ ρ f (Y ) b             + ρ f (Y) b S               + rS          + (a − b Λ )          =0      (3.2)
∂Y                        ∂S ∂ Y             ∂S                   ∂Y
and

∂ P21 2      2 ∂ P2
2
1 2 ∂ P2
2
+ f (Y) S        + b       +
∂t 2           ∂S 2 2   ∂ Y2

∂ 2 P2              ∂ P2                  ∂ P2
+ ρ f (Y) b S                   + rS          + (a − b Λ )          =0                           (3.3)
∂S ∂ Y              ∂S                    ∂Y
We can further transform the two previous equations. Under the substitution
x = ln S                                                                                (3.4)
we have the following transformation formulas for the partial derivatives
∂P ∂P
S      =
∂S ∂ x

10
∂2 P        ∂2 P         ∂P
S2
=             −
∂S 2
∂x   2       ∂x

∂2 P   ∂2 P
S       =
∂S ∂ Y ∂ x ∂ Y
Equation (3.2) then becomes

∂ P1           1 2     ∂ 2 P ∂ P1  1 2 ∂ 2 P1
∂ P1
+ f (Y)
2
+ f (Y)  2 −        + b         +
∂t         ∂x 2      ∂x     ∂x  2     ∂ Y2
            

∂ P1                       ∂ 2 P1        ∂ P1                  ∂ P1
+ ρ f (Y) b          + ρ f (Y) b                  +r          + (a − b Λ )          =0
∂Y                        ∂x ∂Y           ∂x                    ∂Y
which is equivalent to

∂ P11 2    ∂ 2 P1  1 2         ∂ P1 1 2 ∂ P1
2
+ f (Y)       +  r + f (Y)      + b       +
∂t 2      ∂ x2  2             ∂x 2     ∂ Y2

∂ 2 P1                                       ∂ P1
+ ρ b f (Y)                     + { ρ b f (Y ) + (a − b Λ)}            =0         (3.5)
∂x ∂Y                                        ∂Y
Equation (3.3) transforms into

∂ P21 2     ∂ 2 P2 ∂ P2  1 ∂ 2 P2
+ f (Y)        −      + b2     +
∂t 2       ∂ x2    ∂ x  2 ∂ Y2
             

∂ 2 P2         ∂ P2                  ∂ P2
+ ρ f (Y ) b                   +r          + (a − b Λ )          =0
∂ x ∂Y          ∂x                    ∂Y
which is equivalent to

∂ P2 1 2    ∂ 2 P2  1 2          ∂ P2 1 2 ∂ P2
2
+ f (Y)       +  r − f (Y )      + b       +
∂t  2      ∂ x2        2        ∂ x 2 ∂ Y2

∂ 2 P2                       ∂ P2
+ ρ b f (Y)                     + (a − b Λ )           =0                         (3.6)
∂x ∂Y                        ∂Y

11
The characteristic functions f 1 and                           f 2 satisfy the same partial differential

equations with (3.5) and (3.6). We thus have the following two partial differential
equations for the functions f 1 and f 2 :

∂f 11 2     ∂2 f1  1 2         ∂f 1 1 2 ∂ f 1
2
+ f (Y )      +  r + f (Y)      + b        +
∂t 2       ∂ x2  2            ∂x 2 ∂ Y2

∂2 f1                                               ∂f 1
+ ρ b f (Y )                  + { ρ b f (Y ) + (a − b Λ )}                    =0   (3.7)
∂ x ∂Y                                               ∂Y
and

∂f 21 2    ∂2 f 2  1 2         ∂f 2 1 2 ∂ f 2
2
+ f (Y)       +  r − f (Y)      + b        +
∂t 2      ∂ x2        2       ∂ x 2 ∂ Y2

∂2f 2                         ∂f 2
+ ρ b f (Y)                   + (a − b Λ )               =0                        (3.8)
∂x ∂Y                          ∂Y
The risk-free probabilities P1 and P2 can then be evaluated using the Fourier
inversion formula, the so-called Gil-Pelaez formula (Gil-Pelaez [20], Gurland
[22], Heston [23], Lukacs [37], Scott [49] and Sephard [50])
∞        e −i φ ln K f1 ( x , Y, t; φ) 
1 1
P1 ( x , Y, t , ln K ) =     +         ∫    Re                                 dφ         (3.9)
2 π        0       
              iφ               

∞       e −i φ ln K f 2 ( x , Y, t; φ) 
1 1
P2 ( x , Y, t , ln K ) = +              ∫   Re                                  dφ        (3.10)
2 π             0      
              iφ                

The above improper integrals can in principle be evaluated using the Adamchik-
Marichev algorithm (Adamchik and Marichev [2], Roach [44] and [45]) using
Meijer’s G-function (Meijer [41], Mathai and Saxena [40], Roach [44] and [45]).

3.1. Evaluation of the function f1 ( x, Y, t ; φ)
Suppose that the characteristic function f 1 admits a representation of the form

f1 ( x , Y , t ; φ ) = e C ( t , T ; φ ) + D ( t , T ; φ ) Y + i φ x               (3.11)

12
subject to

f1 ( x , Y, T ; φ) = ei φ x                                (3.12)
The previous condition is satisfied as long as
C(T, T ; φ) = 0 and D(T, T ; φ) = 0                        (3.13)
Equation (3.7), because of (3.11), becomes
∂C ∂ D    1                   1            1
−      −    Y − φ 2 f 2 (Y) + i φ  r + f 2 (Y)  + b 2 D 2 +
∂t ∂t     2                      2         2
+ { ( ρ + i φ ρ) b f (Y) + (a − b Λ )} D = 0                         (3.14)
Differentiating the previous equation with respect to Y, we find
∂D 1 2 ∂            1       ∂            1     ∂
−       − φ   (f 2 (Y)) + (i φ)    (f 2 (Y)) + D 2    (b 2 ) +
∂ t 2 ∂Y            2      ∂Y            2    ∂Y
∂
+D        { ( ρ + i φ ρ) b f (Y) + (a − b Λ )} = 0
∂Y
Differentiating again the above equation with respect to Y, we find

1   ∂2              1      ∂2           1     ∂2
− φ2      (f 2 (Y)) + (i φ) 2 (f 2 (Y)) + D 2       (b 2 ) +
2 ∂Y   2            2     ∂Y            2    ∂Y  2

∂2
+D          { ( ρ + i φ ρ) b f (Y) + (a − b Λ)} = 0
∂Y 2
which is equivalent to the equation

1             ∂2              1     ∂2
− (φ 2 − i φ)       (f 2 (Y)) + D 2       (b 2 ) +
2            ∂Y  2            2    ∂Y  2

∂2
+D          { ( ρ + i φ ρ) b f (Y ) + (a − b Λ)} = 0       (3.15)
∂Y 2
The last two terms of the previous equation contain the function D which
depends on the variable T. This variable however does not appear in the first term.
Therefore the only possibility (3.15) to be true is the first term to be zero and the
sum of the other two terms to be zero as well:

13
1 2         ∂2
− (φ − i φ)       (f 2 (Y)) = 0                                         (3.16)
2          ∂Y  2

and

1 2 ∂2               ∂2
D      (b 2 ) + D      { ( ρ + i φ ρ) b f (Y) + (a − b Λ)} = 0        (3.17)
2   ∂Y 2            ∂Y 2
From (3.16) we obtain

∂2
(f 2 (Y)) = 0                                                 (3.18)
∂Y    2

From equation (3.17), dividing through by D, we obtain the equation

1 ∂2           ∂2
D     (b ) +
2
{ ( ρ + i φ ρ) b f (Y) + (a − b Λ)} = 0              (3.19)
2 ∂Y 2        ∂Y 2
From the previous equation, since the variable T is only contained in the first
term, the only possibility to be true is each term to be zero:

1 ∂2
D 2 (b 2 ) = 0                                                         (3.20)
2 ∂Y

and

∂2
{ ( ρ + i φ ρ) b f (Y) + (a − b Λ)} = 0                       (3.21)
∂Y 2
From equation (3.20) we obtain

∂2
(b 2 ) = 0                                        (3.22)
∂Y    2

Instead of using equation (3.21), we prefer to use the next two equations

∂2                          ∂2
(b f (Y)) = 0 and         (a − b Λ ) = 0          (3.23)
∂Y 2                         ∂Y2

From relations (3.18), (3.22) and (3.23) we conclude that f 2 (Y) , b 2 , b f (Y) and
a − b Λ are linear functions with respect to Y.
Therefore we have

14
f 2 ( Y ) = h 1 ( t ) ⋅ Y + g1 ( t )                                                    (3.24)

b 2 = h 2 (t) ⋅ Y + g 2 (t )                                                            (3.25)
b f ( Y) = h ( t ) ⋅ Y + g ( t )                                                        (3.26)
a − b Λ = δ( t ) ⋅ Y + ε ( t )                                                          (3.27)
The time-dependent functions appearing in (3.24)-(3.26) are not independent each
other. Compatibility of these equations ensures that

f 2 ( Y ) b 2 = ( b f ( Y )) 2 ⇔

⇔ ( h1 ( t ) ⋅ Y + g1 ( t )) ( h 2 ( t ) ⋅ Y + g 2 ( t )) = ( h ( t ) ⋅ Y + g ( t )) 2 ⇔

⇔ h1 ( t ) h 2 ( t ) ⋅ Y 2 + ( h1 ( t ) g 2 ( t ) + h 2 ( t ) g1 ( t )) Y + g1 ( t ) g 2 ( t ) =

= h 2 (t ) ⋅ Y 2 + 2 h (t) g(t ) Y + g 2 (t )
Equating the coefficients of similar powers of Y, we obtain the following
compatibility conditions

h1 ( t ) h 2 ( t ) = h 2 ( t )                                                                     (3.28)
h 1 ( t ) g 2 ( t ) + h 2 ( t ) g1 ( t ) = 2 h ( t ) g ( t )                                       (3.29)

g1 ( t ) g 2 ( t ) = g 2 ( t )                                                                     (3.30)
Equation (3.14) gives, because of (3.24)-(3.27)
∂C ∂ D    1                                       1
−     −    Y − (φ 2 − i φ) (h1 ( t ) ⋅ Y + g1 ( t )) + (h 2 ( t ) ⋅ Y + g 2 ( t )) D 2 +
∂t ∂t     2                                       2
+ i r φ + (ρ + i φ ρ) D (h ( t ) ⋅ Y + g ( t )) + (δ( t ) ⋅ Y + ε( t ))D = 0
which is equivalent to
 ∂C 1                                                      1 2                          
 ∂ t + 2 g 2 ( t ) D + [ (ρ + i φ ρ) g ( t ) + ε( t )] D − 2 (φ − i φ) g1 ( t ) + i r φ  +
−                   2

                                                                                        

 ∂D 1                                                      1 2                  
 ∂ t + 2 h 2 ( t ) D + [ (ρ + i φ ρ) h ( t ) + δ( t )] D − 2 (φ − i φ) h1 ( t )  Y = 0
+ −                   2

                                                                                
Since the above equation should be true for any Y, we obtain from the above
relation the two equations

15
∂C 1                                                   1
−     + g 2 ( t ) D 2 + [ (ρ + i φ ρ) g ( t ) + ε( t )] D − (φ 2 − i φ) g 1 ( t ) + i r φ = 0   (3.31)
∂t 2                                                   2
and
∂D 1                                                   1
−     + h 2 ( t ) D 2 + [ (ρ + i φ ρ) h ( t ) + δ( t )] D − (φ 2 − i φ) h1 ( t ) = 0           (3.32)
∂t 2                                                   2
This is a system of two independent first order ordinary differential equations with
unknown functions C( t , T ; φ) and D( t , T ; φ) subject to the conditions
C(T, T ; φ) = 0 and D(T, T ; φ) = 0
Equation (3.32) can be written as
∂D 1                                                   1
= h 2 ( t ) D 2 + [ (ρ + i φ ρ) h ( t ) + δ( t )] D − (φ 2 − i φ) h1 ( t )           (3.33)
∂t 2                                                   2
which is a Riccati equation (see for example Forsyth [16], Ince [25], Kamke [27]
and Murphy [42]).
This equation can be transformed further, considering the substitution
τ=T−t                                                                                   (3.34)
where T is the time of maturity. Since
∂D ∂D ∂t   ∂D
=   ⋅ =−                                                                             (3.35)
∂τ ∂ t ∂τ  ∂t
equation (3.33) takes on the form
∂D    1                                             1
= − h 2 (τ) D 2 − [ (ρ + i φ ρ) h (τ) + δ(τ)] D + (φ 2 − i φ) h1 (τ)                (3.36)
∂τ    2                                             2
The previous equation, under the substitution
2     w ′(τ)
D=           ⋅                                                                 (3.37)
h 2 (τ) w (τ)
takes the form of a second order ordinary differential equation
 h ′ ( τ)                                   
w ′′( τ) −  2
 h ( τ)   − [ (ρ + i ρ φ) h ( τ) + δ( τ)] +  w ′( τ) −

 2                                          
1
− (φ 2 − i φ) h1 (τ) h 2 (τ) w (τ) = 0                                (3.38)
4

16
This equation can be solved by assigning values to the functions h ( τ) , δ(τ) and
h1 (τ) , h 2 (τ) so as equation admits a closed-form solution. In the sequel we shall
consider the choice
δ(τ) = h (τ)                                                        (3.39)
We also have the compatibility condition (3.28)

h 1 ( τ ) h 2 ( τ) = h 2 ( τ)
Therefore equation (3.38) takes on the form
 h ′ (τ)                                      1
w ′′(τ) −  2
 h (τ)   − [1 + ρ (1 + i φ)] ⋅ h (τ)  w ′(τ) − (φ 2 − i φ) h 2 (τ) w (τ) = 0
                                          (3.40)
 2                                            4

The previous equation can be solved by assigning values to the functions h ( τ) and
h 2 (τ) so as equation (3.37) to have a closed-form solution. We set

ζ = 1 + ρ (1 + i φ) , η = φ 2 − i φ and Σ = ζ 2 + η                          (3.41)
After determining the function w (τ) from (3.40), we would be able to determine
the function D(τ ; φ) from (3.37) subject to the condition D(0 ; φ) = 0 .
Once we have found a closed form solution of the function D(τ ; φ) , we can
determine the function C(τ ; φ) which is solution to the differential equation
(3.31). The solution of this equation is given by (after changing variable t to τ )
τ
1
C( τ ; φ) = − ∫ g 2 (s) D 2 (s; φ) ds −
20
τ
− ∫ [ (ρ + i φ ρ) g (s) + ε(s)] D(s; φ) ds +
0

τ
1
+ (φ 2 − i φ) ∫ g1 (s) ds − i r φ τ                                          (3.42)
2            0

The above form of the solution satisfies the condition C(0 ; φ) = 0 .

17
3.2. Evaluation of the function f 2 ( x , Y, t ; φ)
Suppose next that the characteristic function f 2 admits a representation of the

form

f 2 ( x , Y , t ; φ) = e G ( t , T ; φ ) + H ( t , T ; φ ) Y + i φ x   (3.43)
subject to

f 2 ( x , Y, T ; φ) = ei φ x                                           (3.44)
The previous condition is satisfied as long as
G (T, T ; φ) = 0 and H(T, T ; φ) = 0                                   (3.45)
Equation (3.8) then, because of (3.43), becomes
∂G ∂ H     1                   1            1
−       −    Y − φ 2 f 2 (Y) + i φ  r − f 2 (Y)  + b 2 H 2 +
∂t   ∂t    2                   2            2
+ i φ ρ b f ( Y ) H + (a − b Λ ) H = 0                                 (3.46)
Differentiating the previous equation with respect to Y, we find
∂H 1 2 ∂            1     ∂
−       − φ   (f 2 (Y)) + H 2    (b 2 ) −
∂ t 2 ∂Y            2    ∂Y
1       ∂                      ∂                ∂
− (i φ)    (f 2 (Y)) + i φ ρ H    (b f (Y)) + H    (a − b Λ ) = 0
2      ∂Y                     ∂Y               ∂Y
Differentiating again the above equation with respect to Y, we find

1 2 ∂2            1 2 ∂2         1      ∂2
− φ       (f (Y)) + H
2
(b ) − (i φ) 2 (f 2 (Y)) +
2
2 ∂Y   2          2  ∂Y  2       2     ∂Y

∂2                             ∂2
+ i φρ H               (b f (Y)) + H                  (a − b Λ ) = 0
∂Y 2                            ∂Y 2
which is equivalent to the equation

1 2         ∂2            1 2 ∂2
− (φ + i φ)       (f (Y)) + H
2
(b 2 ) +
2          ∂Y  2          2  ∂Y  2

∂2
+H              {(i φ ρ)(b f (Y)) + (a − b Λ )} = 0                    (3.47)
∂Y 2

18
From the previous equation we obtain, as in section 3.1, the relations (3.18), (3.22)
and (3.23) and then (3.24)-(3.27), together with the compatibility conditions
(3.28)-(3.30).
Equation (3.47) gives, because of (3.24)-(3.27)
∂G ∂H    1                                       1
−     −   Y − (φ 2 + i φ) (h1 ( t ) ⋅ Y + g1 ( t )) + (h 2 ( t ) ⋅ Y + g 2 ( t )) H 2 +
∂t ∂t    2                                       2
+ i r φ + i φ ρ H (h ( t ) ⋅ Y + g ( t )) + (δ( t ) ⋅ Y + ε( t )) H = 0
which is equivalent to
 ∂G 1                                                  1 2                          
 ∂ t + 2 g 2 ( t ) H + [ (i φ ρ) g ( t ) + ε( t )] H − 2 (φ + i φ) g1 ( t ) + i r φ  +
−                   2

                                                                                    
 ∂H 1                                                 1 2                  
 ∂ t + 2 h 2 ( t ) H + [(i φ ρ) h ( t ) + δ( t )] H − 2 (φ + i φ) h1 ( t )  Y = 0
+ −                   2

                                                                           
Since the above equation should be true for any Y, we obtain from the above
relation the two equations
∂G 1                                               1
−     + g 2 ( t ) H 2 + [ (i φ ρ) g ( t ) + ε( t )] H − (φ 2 + i φ) g1 ( t ) + i r φ = 0     (3.48)
∂t 2                                               2
and
∂H 1                                              1
−     + h 2 ( t ) H 2 + [(i φ ρ) h ( t ) + δ( t )] H − (φ 2 + i φ) h1 ( t ) = 0              (3.49)
∂t 2                                              2
This is a system of two first order ordinary differential equations with unknown
functions G ( t , T ; φ) and H( t , T ; φ) , subject to the conditions (3.45)
G (T, T ; φ) = 0 and H(T, T ; φ) = 0
Equation (3.49) can be written as
∂H 1                                              1
= h 2 ( t ) H 2 + [(i φ ρ) h ( t ) + δ( t )] H − (φ 2 + i φ) h1 ( t )               (3.50)
∂t 2                                              2
which is a Riccati equation (see for example Forsyth [16], Ince [25], Kamke [27]
and Murphy [42]). This equation can be transformed further, considering the
substitution

19
τ=T−t                                                                                   (3.51)
where T is the time of maturity.
Since
∂D ∂D ∂t   ∂D
=   ⋅ =−                                                                              (3.52)
∂τ ∂ t ∂τ  ∂t
equation (3.50) takes on the form
∂H    1                                        1
= − h 2 (τ) H 2 − [(i φ ρ) h (τ) + δ(τ)] H + (φ 2 + i φ) h1 (τ)                      (3.53)
∂τ    2                                        2
This equation, under the substitution
2     z ′(τ)
H(τ) =           ⋅                                                              (3.54)
h 2 (τ) z(τ)
takes the form
 h ′ ( τ)                                      1 2
 h ( τ) − [ (i ρ φ) h ( τ) + δ( τ)]  z ′( τ) − 4 (φ + i φ) h1 ( τ) h 2 ( τ) z( τ) = 0 (3.55)
z ′′( τ) −  2                                  
 2                                  
This is a second order ordinary differential equation which can be solved by
assigning values to the functions h ( τ) , δ(τ) and h1 (τ) , h 2 (τ) so as equation
admits a closed-form solution. In our case we shall consider the choice
δ(τ) = h (τ)                                                           (3.56)
We also have the compatibility condition (3.28)

h1 ( t ) h 2 ( t ) = h 2 ( t )
Therefore equation (3.55) takes on the form
 h ′ ( τ)                                  1 2
 h ( τ) − ( 1 + i ρ φ) ⋅ h ( τ)  z ′( τ) − 4 (φ + i φ) h ( τ) z ( τ) = 0
z ′′( τ) −  2                              
2
(3.57)
 2                              
The previous equation can be solved by assigning values to the functions h (τ) and
h 2 (τ) so as equation (3.57) has a closed-form solution.
We set

θ = 1 + i ρ φ , ω = φ2 + i φ and ∆ = θ2 + ω                                     (3.58)

20
After determining the function z( τ) from (3.51), we would be able to determine
the function H (τ ; φ) from (3.54) subject to the condition H(0 ; φ) = 0 .
Once we have found a closed form solution for the function H(τ ; φ) , we can
determine the function G (τ ; φ) which is solution to the differential equation
(3.48). The solution of this equation is given by (after changing variable t to τ )
τ
1
G ( τ ; φ) = −     ∫ g 2 (s) H (s; φ) ds −
2
20
τ
− ∫ [ (i φ ρ) g (s) + ε(s)] H (s; φ) ds +
0

τ
1
+ (φ 2 + i φ) ∫ g1 (s) ds − i r φ τ                                      (3.59)
2            0

The above form of solution satisfies the condition G (0 ; φ) = 0 .

Conclusion. Equation (2.18), because of (3.24)-(3.27), takes on the form
∂F 1                             2 ∂ F 1
2
∂ 2F
+ [ h1 ( t ) ⋅ Y + g1 ( t )] S       + [ h 2 ( t ) ⋅ Y + g 2 ( t )]      +
∂t 2                               ∂ S2 2                             ∂ Υ2

∂2 F            ∂F                            ∂F
+ ρ S[ h ( t ) ⋅ Y + g ( t )]        − r F+ rS    + [ δ( t ) ⋅ Y + ε( t )] ⋅    =0   (3.60)
∂ S ∂Y           ∂S                            ∂Y
where the functions h ( τ) , h1 (τ) , h 2 (τ) and g(τ) , g1 (τ) , g 2 (τ) satisfy the
compatibility conditions (3.28)-(3.30). In building models, we suppose that
δ(τ) = h (τ) and we start by assigning values to the functions h ( τ) and h 2 (τ) so as
equations (3.40) and (3.57) admit closed-form solutions. After determining the
functions w (τ) and z(τ) we determine functions D(τ; φ) and H(τ; φ) from (3.37)
and (3.54) respectively. Assigning values to the functions g(τ) , g1 (τ) , g 2 (τ) and
ε(τ) , we determine C( τ; φ) and G (τ; φ) from (3.42) and (3.59) respectively. We
then determine f1 ( x, Y, t ; φ) and f 2 ( x, Y, t ; φ) from (3.11) and (3.43)
respectively.

21
Finally we determine P1 ( x , Y, t , ln K ) and P2 ( x , Y, t , ln K ) using (3.9) and
(3.10) respectively. The value of the option is then determined from (3.1).
The time-dependent coefficients we consider, might capture more detailed
behavior of the movement of the derivative price under stochastic volatility.
Using the above procedure, we are considering some models next.

4. A simple model with constant coefficients
Using the assignment

h ( t ) = ξ and h 2 ( t ) = ξ 2 , ξ > 0                             (4.1)
we obtain from the compatibility condition (3.28) that
h1 ( t ) = 1                                                        (4.2)
Equation (3.33) then takes on the form (taking into account δ( t ) = h ( t ) = ξ )
∂D 1 2 2                           1
= ξ D + [ (ρ + i φ ρ) ξ + ξ ] D − (φ 2 − i φ)                              (4.3)
∂t 2                               2
which is a Riccati equation.
Equation (4.3) under the substitution
2 w ′( t )
D=−        ⋅                                                        (4.4)
ξ2 w(t)
becomes
1
w ′′( t ) − [ 1 + ρ (1 + i φ)] ξ w ′( t ) − (φ 2 − i φ) ξ 2 w ( t ) = 0      (4.5)
4
Using further the definition of parameters introduces in (3.38)
ζ = 1 + ρ (1 + i φ) , η = φ 2 − i φ and Σ = ζ 2 + η                          (4.6)
equation (4.5) can be written as
1
w ′′( t ) − (ζ ξ) w ′( t ) − (η ξ 2 ) w ( t ) = 0                            (4.7)
4
The above equation is a second order ordinary differential equation with general
solution

22
 ( Σ − ζ) ξ             ( Σ + ζ ) ξ 
w ( t ) = C1 exp −          t  + C 2 exp            t            (4.8)
     2                      2       
Since
 ( Σ − ζ) ξ       ( Σ − ζ) ξ 
w ′( t ) = C1 −            exp −          t +
     2                2      
( Σ + ζ ) ξ      ( Σ + ζ ) ξ 
+ C2              exp            t
    2                2       
We obtain from (4.4)
2
D ( t , T ; φ) = −        ×
ξ2

 ( Σ − ζ) ξ       ( Σ − ζ) ξ          ( Σ + ζ ) ξ      ( Σ + ζ ) ξ 
C1 −            exp −           t + C2                exp            t
     2                2                    2               2       
×
 ( Σ − ζ) ξ              ( Σ + ζ ) ξ 
C1 exp −            t  + C 2 exp             t
     2                       2        
 ( Σ − ζ) ξ 
Multiplying numerator and denominator by exp            t  , we obtain from the
     2       
above formula
2
D ( t , T ; φ) = −        ×
ξ2

 ( Σ − ζ) ξ       ( Σ + ζ ) ξ 
C1 −            + C2              exp( Σ ξ t )
     2                2       
×                                                        (4.9)
C1 + C 2 exp( Σ ξ t )
The condition D(T, T ; ϕ) = 0 is satisfied as long as

 ( Σ − ζ) ξ             ( Σ + ζ ) ξ 
C1 −            + C2                    exp( Σ ξ T) = 0
     2                      2       
from which we can determine the ratio of the two constants:
C1   Σ +ζ
=      × exp ( ξ Σ T )                                            (4.10)
C2   Σ −ζ

23
Equation (4.9) can be written as
2
D ( t , T ; φ) = −        ×
ξ2

C1    ( Σ − ζ) ξ   ( Σ + ζ ) ξ 
−           +              exp( Σ ξ t )
C2        2             2      
×
C1
+ exp( Σ ξ t )
C2
Upon substituting the ratio of the constants given by (4.10) into the previous
equation, we obtain the following expression for the function D( t , T ; φ) :
2
D ( t , T ; φ) = −        ×
ξ2

Σ +ζ                  ( Σ − ζ) ξ             ( Σ + ζ ) ξ 
× exp ( ξ Σ T )  −          +                        exp(ξ Σ t )
Σ −ζ                      2                      2       
×
Σ +ζ
× exp ( ξ Σ T ) + exp(ξ Σ t )
Σ −ζ
or
Σ +ζ     exp [ ξ Σ (T − t ) ] − 1
D( t , T ; φ) =              ×
ξ     Σ +ζ
× exp [ ξ Σ (T − t ) ] + 1
Σ −ζ
or
ξ Σ                    ξ Σ            
exp       (T − t )  − exp −      (T − t ) 
Σ +ζ          2                        2           
D( t , T ; φ) =      ×
ξ     Σ +ζ       ξ Σ                    ξ Σ            
× exp       (T − t )  + exp −      (T − t ) 
Σ −ζ        2                       2            
or
ξ Σ               
sinh          (T − t ) 
2 ( Σ + ζ)                           2                
D ( t , T ; φ) =            ×
ξ                      ξ Σ                           ξ Σ          
( m + 1) cosh     (T − t )  + ( m − 1) sinh     (T − t ) 
 2                             2           

24
or
ξ Σ              
tanh         (T − t ) 
2 ( Σ + ζ)                   2               
D ( t , T ; φ) =            ×                                                        (4.11)
ξ                                 ξ Σ               
( m + 1) + ( m − 1) tanh          (T − t ) 
 2                
where
Σ +ζ
m=                                                                                   (4.12)
Σ −ζ
Equation (4.11) can also be written as
ξ Σ            
tanh       (T − t ) 
2η           2             
D ( t , T ; φ) =    ×                                                                (4.13)
ξ              ξ Σ            
Σ + ζ tanh       (T − t ) 
 2             
Under the choice g 2 ( t ) = 0 , we obtain from (3.30) that g(t ) = 0 . We then have
from (3.29) that g1 ( t ) = 0 . Therefore we get from (3.31) by integration
T
C( t , T ; φ) = − ∫ ε(s) D(s, T; φ) ds − i r φ (T − t )                     (4.14)
t

The previous expression can be further simplified using (4.13) in place of
D( t , T ; φ) . We find that

          ξ Σ                 
     tanh       (T − s )      
T
2           2                 
C( t , T ; φ) = − η   ∫   ε (s )  ×                               ds − i r φ (T − t )   (4.15)
t          ξ             ξ Σ
Σ + ζ tanh 

(T − s ) 
                2               
                             
Since we have for the integral
        ξ Σ               
   tanh      (T − s )     
           2              
∫                                ds =
 Σ + ζ tanh ξ Σ          
     (T − s) 
                            
             2           

25
ζ            2              ξ Σ                   ξ Σ          
=     (T − s) −    × ln  Σ cosh     (T − s)  + ζ sinh     ( T − s)  
η           ηξ      
         2                     2           
the choice
1 2
ε( t ) = −     ξ                                                                 (4.16)
4
converts (4.15) into
ξζ
C ( t , T ; φ) = −      (T − t ) − i r φ (T − t ) +
2
ξ Σ           ζ        ξ Σ          
+ ln cosh     (T − t )  +   sinh     (T − t )                       (4.17)
 2              Σ       2           
We can similarly, using the same procedure, and under the same choice of
functions h ( t ) and h 2 ( t ) , determine the functions H( t , T ; φ) and G ( t , T ; φ) . We
find, using (3.50)
ξ ∆            
tanh       (T − t ) 
2ω           2             
H ( t , T ; φ) =    ×                                                            (4.18)
ξ              ξ ∆            
∆ + θ tanh       (T − t ) 
 2             
where θ , ω and ∆ are defined in (3.58).
Under the choice g 2 ( t ) = 0 , we obtain from (3.30) g( t ) = 0 . We then have from
(3.29) that g1 ( t ) = 0 . Therefore we get from (3.48) that the function G ( t , T ; φ) is
evaluated using the formula
T
G ( t , T ; φ) = − ∫ ε(s) H (s, T; φ) ds − i r φ (T − t )                        (4.19)
t

The previous expression can be further simplified using either (4.18) in place of
H( t , T ; φ) . We find that

26
          ξ ∆                 
     tanh       (T − s )      
T
2           2                 
G ( t , T ; φ) = − ω ∫ ε(s)  ×                               ds − i r φ (T − t )    (4.20)
 ξ            ξ ∆            
t          ∆ + θ tanh       (T − s ) 
                                
               2             
The choice
1
ε( t ) = − ξ 2                                                                  (4.21)
4
converts (4.20) into
ξθ
G ( t , T ; φ) = −      (T − t ) − i r φ (T − t ) +
2
ξ ∆           θ        ξ ∆          
+ ln cosh     (T − t )  +   sinh     (T − t )                      (4.22)
 2              ∆       2           
We are now in a position to determine the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ)
using (3.11) and (3.43) respectively:

f1 ( x , Y , t ; φ ) = e C ( t , T ; φ ) + D ( t , T ; φ ) Y + i φ x

f 2 ( x , Y , t ; φ) = e G ( t , T ; φ ) + H ( t , T ; φ ) Y + i φ x
We have the following expressions

ξ Σ           ζ        ξ Σ          
f1 ( x , Y, t ; φ) = cosh     (T − t )  +   sinh     (T − t )  ×
 2              Σ       2           
                                                     ξ Σ                 
                                                tanh      (T − t )       
 ξζ                                      2η            2                 
× exp  −  ( T − t ) − i r φ (T − t ) + i φ x +    ×                             Y
 2                                        ξ              ξ Σ           
Σ + ζ tanh      (T − t )  
                                                                           
                                                          2             
and
ξ ∆           θ        ξ ∆          
f 2 ( x , Y, t ; φ) = cosh     (T − t )  +   sinh     (T − t )  ×
 2              ∆       2           

27
                                                    ξ ∆                  
                                               tanh       (T − t )       
 ξθ                                     2ω            2                  
× exp  −  (T − t ) − i r φ (T − t ) + i φ x +    ×                              Y
 2                                       ξ              ξ ∆            
∆ + θ tanh       (T − t )  
                                                                           
                                                         2              
The risk-free probabilities P1 and P2 can then be evaluated using the Fourier
inversion formulas (3.9) and (3.10) respectively. Therefore we obtain
∞
1 1            R 1 ( x , Y , t ; φ) 
P1 ( x , Y, t , ln K ) = +     ∫ Re                          dφ
2 π    0              iφ            
∞
1 1             R 2 ( x , Y , t ; φ) 
P2 ( x , Y, t , ln K ) = +      ∫ Re                          dφ
2 π     0              iφ            
where

ξ Σ           ζ        ξ Σ          
R 1 ( x , Y, t ; φ) = cosh     (T − t )  +   sinh     (T − t )  ×
 2              Σ       2           

                                                               ξ Σ                 
                                                          tanh      (T − t )       
 ξζ                                                2η            2                 
× exp  −  (T − t ) − i r φ (T − t ) + i φ ( x − ln K ) +    ×                             Y
 2                                                  ξ              ξ Σ           
Σ + ζ tanh      (T − t )  
                                                                                     
                                                                    2             
and
ξ ∆           θ        ξ ∆          
R 2 ( x , Y, t ; φ) = cosh     (T − t )  +   sinh     (T − t )  ×
 2              ∆       2           
                                                               ξ ∆                  
                                                          tanh       (T − t )       
 ξθ                                                2ω            2                  
× exp  −  (T − t ) − i r φ (T − t ) + i φ ( x − ln K ) +    ×                              Y
 2                                                  ξ              ξ ∆            
∆ + θ tanh       (T − t )  
                                                                                      
                                                                    2              
The price F of a European call option is given by (3.1)

F = S ⋅ P1 − K ⋅ e − r (T − t ) P2
where K is the strike price and T is the maturity.

28
The above model is very similar to the Heston model (Heston [23]) and it is the
simplest stochastic volatility model we have introduced. Once we start considering
time-dependent functions h ( t ) , h 2 ( t ) , etc. models become more complicated,
requiring the use of special functions. We are considering some of these models
next.

5. Model II. The choice
h (τ) = τ and h 2 (τ) = τ                                        (5.1)
converts equation (3.40) into the equation
1                                  1
w ′′(τ) −  − [ 1 + ρ (1 + i φ)] ⋅ τ  w ′(τ) − (φ 2 − i φ) τ 2 w (τ) = 0    (5.2)
τ                                  4
with general solution
      Σ +ζ 2            Σ −ζ 2 
w ( t ) = C1 exp  −        τ  + C 2 exp      τ                           (5.3)
       4                4      
                               
The parameters ζ , η and Σ are defined by

ζ = 1 + ρ (1 + i φ) , η = φ 2 − i φ and Σ = ζ 2 + η              (5.4)
Therefore equation (3.36) has a solution given by (3.37)
2     w ′(τ)
D(τ) =            ⋅        ,   h 2 (τ) = τ                       (5.5)
h 2 (τ) w (τ)
subject to the condition D(0 ; φ) = 0 .
Since
      Σ +ζ                Σ +ζ 2       Σ −ζ        Σ −ζ 2
w ′(τ) = C1  −        τ  exp  −           τ  + C2      τ  exp      τ 
       2                   4           2           4     
                                                         
equation (5.5) can be written as

29
   Σ +ζ           Σ +ζ 2         Σ −ζ         Σ −ζ 2
C1  −     τ  exp  −       τ  + C2        τ  exp      τ 
    2              4             2            4     
D ( τ) = ⋅                                                       
2
τ                       Σ +ζ 2             Σ −ζ 2
C1 exp  −        τ  + C 2 exp        τ 
      4                 4        
                                 
 Σ +ζ 2
or, multiplying numerator and denominator by exp      τ  and simplifying,
 4     
       
we obtain
   Σ +ζ            Σ −ζ        Σ 2
C1  −          + C2          exp    τ 
    2              2           2   
D(τ; ϕ) = 2                                      
(5.6)
 Σ 2
C1 + C 2 exp    τ 
 2     
       
The condition D(0; ϕ) = 0 is satisfied, taking into account (5.6), as long as

   Σ +ζ            Σ −ζ   
C1  −           + C2         = 0
    2              2      
                          
From the previous equality we can determine the ratio of the two constants:

C1   Σ −ζ
=                                                              (5.7)
C2   Σ +ζ
Equation (5.6) can be written as

C1 
−
Σ +ζ     Σ −ζ 
+
 Σ 2
 exp    τ 
C2 
   2       2 
         
 2



D(τ; ϕ) = 2
C1        Σ 2
+ exp    τ 
C2        2    
      
which in turn can also be written, taking into account (5.7), as

Σ −ζ   Σ +ζ   Σ −ζ          Σ 2
−       +         exp    τ 
Σ +ζ   2      2            2   
D(τ; ϕ) = 2                                     
Σ −ζ        Σ 2
+ exp    τ 
Σ +ζ         2     
       

30
or, after the necessary simplifications,
 Σ 2
− ( Σ − ζ ) + ( Σ − ζ ) exp    τ 
 2   
D(τ; ϕ) =                                  
Σ −ζ           Σ 2
+ exp      τ 
Σ +ζ           2      
        
or
 Σ 2
− 1 + exp     τ 
 2    
D(τ; ϕ) = ( Σ − ζ)                    
Σ −ζ          Σ 2
+ exp    τ 
Σ +ζ          2   
     
or
 Σ 2             Σ 2
exp      τ  − exp  −    τ 
 4             4      
D(τ; ϕ) = ( Σ − ζ )                               
Σ −ζ          Σ 2         Σ 2
exp  −    τ  + exp     τ 
Σ +ζ       4              4    
                     
or
 Σ 2
2 ( Σ − ζ) sinh     τ 
 4     
D(τ; ϕ) =                                  
(5.8)
 Σ 2                  Σ 2
(1 + m) cosh    τ  + (1 − m) sinh    τ 
 4                    4   
                          
where
Σ −ζ
m=                                                          (5.9)
Σ +ζ
We thus obtain
 Σ 2
2 ( Σ − ζ) tanh      τ 
 4    
D(τ; ϕ) =                                                 (5.10)
 Σ 2
(1 + m) + (1 − m) tanh    τ 
 4   
     

31
We are now in a position to determine the function C(τ ; φ) given by equation
(3.42) by assigning values to the functions g 1 ( t ) , g 2 ( t ) , g( t ) and ε( t ) , taking into
account the compatibility conditions (3.28)-(3.30).
Let g 2 ( t ) = 0 . We then obtain from (3.30) that g( t ) = 0 . According to the assigned
values to the functions h( t ) and h 2 ( t ) , we get from (3.28) that h1 (τ) = τ . We
then obtain from (3.29) that g1 ( t ) = 0 . Therefore we have from (3.42) the
following expression for the function C(τ ; φ) :
τ
C(τ ; φ) = − ∫ ε(s) D(s; φ) ds − i r φ τ                                 (5.11)
0

At this stage, given the form of the function D( τ; φ) expressed by (5.10), we have
to assign a function to ε( t ) , so that to obtain a closed form solution for the
function C( τ; φ) . Writing (5.10) as

 Σ 2
2 ( Σ − ζ ) τ ⋅ tanh      τ 
 4     
1
D(τ; ϕ) = ⋅                               
(5.12)
τ                            Σ 2
(1 + m) + (1 − m) tanh       τ 
 4     
       
we see that the choice
ε ( τ) = τ                                                                        (5.13)
 Σ 2
2 ( Σ − ζ ) s tanh     s 
τ                          4     
C(τ ; φ) = − ∫                                  ds − i r φ τ
 Σ 2
0
(1 + m) + (1 − m) tanh      s 
 4     
       
where the integral can be evaluated explicitly.
In fact, the above expression can also be written as (using the definition of m)

32
 Σ 2
s ⋅ tanh     s 
τ              4    
C ( τ ; φ) = − η                          ds − i r φ τ
∫                 Σ 2
0
Σ + ζ ⋅ tanh    s 
 4   
     
Calculating the integral,
 Σ 2
s ⋅ tanh     s 
 4    
       ds =
∫                   Σ 2
Σ + ζ ⋅ tanh    s 
 4   
     

ζ 2 2                     Σ 2              Σ 2
=−       ⋅ s + ⋅ ln       Σ cosh
 4 s  + ζ ⋅ sinh 


 4 s 
2η      η                                       
we arrive at

ζ 2             Σ 2 ζ            Σ 2
C( τ ; φ) =     τ − 2 ln cosh                      
 4 τ  + Σ ⋅ sinh  4 τ  − i r φ τ                     (5.14)
2                                    
Using equations (3.57), (3.54) and (3.59), under the same choice of functions h( t ) ,
h 2 ( t ) , g 2 ( t ) , δ( t ) and ε( t ) , we can determine the functions H (τ ; φ) and G (τ ; φ) .
We obtain
 ∆ 2
2 ( ∆ − θ) ⋅ tanh         τ 
 4     
H( τ; ϕ) =                               
(5.15)
 ∆ 2
(1 + n ) + (1 − n ) tanh     τ 
 4     
       
where
∆ −θ
n=                                                                                (5.16)
∆ +θ
We see that the choice
ε(τ) = τ

33
             ∆ 2
2 ( ∆ − θ) s ⋅ tanh                  s 
τ                                        4   
G (τ ; φ) = − ∫                                                      ds − i r φ τ
              ∆ 2
0
(1 + n ) + (1 − n ) tanh                s 
             4   
                 
where the integral can be evaluated explicitly.
In fact, the above expression can also be written as (using the definition of n)
 ∆ 2
s ⋅ tanh     s 
τ                 4    
G ( τ ; φ) = − ω                              ds − i r φ τ
∫                    ∆ 2
0
∆ + θ ⋅ tanh    s 
 4   
     
Calculating the integral,
 ∆ 2
s ⋅ tanh     s 
 4    
       ds =
∫                  ∆ 2
∆ + θ ⋅ tanh    s 
 4   
     

θ 2 2                          ∆ 2              ∆ 2
=−      ⋅ s + ln              ∆ cosh
 4 s  + θ ⋅ sinh 


 4 s 
2ω      ω                                            
we arrive at

θ                    ∆ 2 θ            ∆ 2
G (τ ; φ) = ⋅ τ 2 − 2 ⋅ ln cosh                      
 4 τ  + ∆ ⋅ sinh  4 τ  − i r φ τ                       (5.17)
2                                         
We are now in a position to determine the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ)
using (3.11) and (3.43) respectively:

f1 ( x , Y , t ; φ ) = e C ( t , T ; φ ) + D ( t , T ; φ ) Y + i φ x

f 2 ( x , Y , t ; φ) = e G ( t , T ; φ ) + H ( t , T ; φ ) Y + i φ x
Using the previously determined quantities, we find the following expressions for
the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ) :

34
1
f1 ( x , Y, t; φ) =                                                    2
×
 Σ 2 ζ            Σ 2
cosh                       
 4 τ  + Σ ⋅ sinh  4 τ 
                      

                                            Σ 2     
                                  η ⋅ tanh     τ    
                                            4    
ζ
× exp i φ x − i r φ τ + τ 2 +                          Y

                  2                            Σ 2 
                                 Σ + ζ ⋅ tanh    τ  
 4   
                                                    
1
f 2 ( x , Y , t ; φ) =                                                      2
×
 ∆ 2 θ            ∆ 2
cosh                      
 4 τ  + ∆ ⋅ sinh  4 τ 
                      

                                            ∆ 2     
                                  ω ⋅ tanh     τ    
                                            4    
θ
× exp i φ x − i r φ τ + τ 2 +                          Y

                  2                            ∆ 2 
                                 ∆ + θ ⋅ tanh    τ  
 4   
                                                    
The risk-free probabilities P1 and P2 can then be evaluated using the Fourier
inversion formulas (3.9) and (3.10) respectively. We find, using the previously
determined quantities f1 and f 2
∞
1 1               R 1 ( x , Y, K , t; φ) 
P1 ( x , Y, t , ln K ) = +       ∫ Re                             dφ
2 π      0                  iφ           
∞
1 1            R 2 ( x , Y , K , t ; φ) 
P2 ( x , Y, t , ln K ) =    +    ∫ Re                               dφ
2 π   0                  iφ             
where R 1 ( x , Y, K , t; φ) and R 2 ( x , Y, K, t; φ) are given by
1
R 1 ( x , Y , t ; φ) =                                                  2
×
 Σ 2 ζ           Σ 2
cosh 
 4 τ +
    ⋅ sinh      
 4 τ 
       Σ             

35
                                                       Σ 2     
                                             η ⋅ tanh     τ    
                                                       4    
ζ
× exp i φ ( x − ln K ) − i r φ τ + τ 2 +                          Y

                             2                            Σ 2 
                                            Σ + ζ ⋅ tanh    τ  
 4   
                                                               
1
R 2 ( x , Y, t; φ) =                                            2
×
 ∆ 2 θ           ∆ 2
cosh 
 4 τ +
    ⋅ sinh      
 4 τ 
       ∆             

                                                       ∆ 2     
                                             ω ⋅ tanh     τ    
                                                       4    
θ
× exp i φ ( x − ln K ) − i r φ τ + τ 2 +                          Y

                             2                            ∆ 2 
                                            ∆ + θ ⋅ tanh    τ  
 4   
                                                               
The integration appearing in the Gill-Pelaez inversion formulas for the
determination of P1 and P2 , can be performed in principle using the Adamchik-

Marichev algorithm (Adamchik and Marichev [2], Roach [44] and [45]). This
algorithm uses the known Meijer function (Meijer [41], Mathai and Saxena [40]).
The price F of a European call option is given by (3.1)

F = S ⋅ P1 − K ⋅ e − r (T − t ) P2
where K is the strike price and T is the maturity.

6. Model III. The choice
h (τ) = τ and h 2 (τ) = τ                                         (6.1)
converts equation (3.40) into the equation
 1                                1
w ′′(τ) −  − [1 + ρ (1 + i φ)] ⋅ τ  w ′(τ) − (φ 2 − i φ) τ w (τ) = 0         (6.2)
 2τ                               4
with general solution
       Σ + ζ 3/ 2         Σ − ζ 3/ 2 
w ( t ) = C1 exp  −          τ  + C 2 exp       τ                          (6.3)
        3                 3          
                                     

36
The parameters ζ , η and Σ are defined by

ζ = 1 + ρ (1 + i φ) , η = φ 2 − i φ and Σ = ζ 2 + η          (6.4)
Therefore equation (3.36) has a solution given by (3.37)
2     w ′(τ)
D(τ) =            ⋅        ,   h 2 (τ) = τ                   (6.5)
h 2 (τ) w (τ)
subject to the condition D(0 ; φ) = 0 .
Since
       Σ +ζ               Σ + ζ 3/ 2 
w ′( τ) = C1  −          τ  exp  −          τ +
        2                  3         
                                     
 Σ −ζ         Σ − ζ 3/ 2 
+ C2       τ  exp       τ 
 2            3          
                         
equation (6.5) can be written as
2
D ( τ) =       ×
τ
   Σ +ζ             Σ + ζ 3/ 2      Σ −ζ           Σ − ζ 3/ 2 
C1  −       τ  exp  −        τ  + C2          τ  exp       τ 
    2                3              2              3          
×                                                                 
     Σ + ζ 3/ 2          Σ − ζ 3/ 2 
C1 exp  −         τ  + C 2 exp        τ 
      3                  3          
                                    
 Σ + ζ 3/ 2 
or, multiplying numerator and denominator by exp       τ  and
 3          
            
simplifying, we obtain
   Σ +ζ                     Σ −ζ        2 Σ 3/ 2 
C1  −                   + C2          exp     τ 
    2                       2           3        
D(τ) = 2 ×                                                    
(6.6)
 2 Σ 3/ 2 
C1 + C 2 exp     τ 
 3        
          
The condition D(0; ϕ) = 0 is satisfied, taking into account (6.6), as long as

37
   Σ +ζ            Σ −ζ   
C1  −           + C2         = 0
    2              2      
                          
From the previous equality we can determine the ratio of the two constants:

C1   Σ −ζ
=                                                           (6.7)
C2   Σ +ζ
Equation (6.6) can be written as

C1 
−
Σ +ζ      Σ −ζ 
+
 2 Σ 3/ 2 
 exp     τ 
C2 
   2        2 
         
 3



D(τ; ϕ) = 2
C1        2 Σ 3/ 2 
+ exp     τ 
C2        3        
          
which in turn can also be written, taking into account (6.7), as

Σ −ζ   Σ +ζ   Σ −ζ         2 Σ 3/ 2 
−      +         exp     τ 
Σ +ζ   2     2 
         
 3



D(τ; ϕ) = 2
Σ −ζ        2 Σ 3/ 2 
+ exp     τ 
Σ +ζ        3        
          
or, after the necessary simplifications,
 2 Σ 3/ 2 
− ( Σ − ζ) + ( Σ − ζ) exp       τ 
 3        
D(τ; ϕ) =                                     
Σ −ζ          2 Σ 3/ 2 
+ exp       τ 
Σ +ζ          3        
          
or
 2 Σ 3/ 2 
− 1 + exp       τ 
 3        
D(τ; ϕ) = ( Σ − ζ)                        
Σ −ζ          2 Σ 3/ 2 
+ exp     τ 
Σ +ζ          3        
          
or

38
 Σ 3/ 2          Σ 3/ 2 
exp      τ  − exp  −   τ 
 3              3       
D(τ; ϕ) = ( Σ − ζ )                                     
Σ −ζ         Σ 3/ 2      Σ 3/ 2 
exp  −   τ  + exp     τ 
Σ +ζ       3             3      
                      
or
 Σ 3/ 2 
2 ( Σ − ζ) sinh    τ 
 3      
D(τ; ϕ) =                                    
(6.8)
 Σ 3/ 2                Σ 3/ 2 
(1 + m) cosh    τ  + (1 − m) sinh     τ 
 3                     3      
                              
where
Σ −ζ
m=                                                                                (6.9)
Σ +ζ
We thus obtain
 Σ 3/ 2 
2 ( Σ − ζ ) tanh     τ 
 3      
D(τ; ϕ) =                             
(6.10)
 Σ 3/ 2 
(1 + m) + (1 − m) tanh    τ 
 3      
        
We are now in a position to determine the function C(τ ; φ) given by equation
(3.39) by assigning values to the functions g 1 ( t ) , g 2 ( t ) , g( t ) and ε( t ) , taking into
account the compatibility conditions (3.28)-(3.30).
Let g 2 ( t ) = 0 . We then obtain from (3.30) that g( t ) = 0 . According to the assigned

values to the functions h ( t ) and h 2 ( t ) , we get from (3.28) that h1 (τ) = τ . We

then obtain from (3.29) that g1 ( t ) = 0 . Therefore we have from (3.42) the
following expression for the function C(τ ; φ) :
τ
C(τ ; φ) = − ∫ ε(s) D(s; φ) ds − i r φ τ                                  (6.11)
0

39
At this stage, given the form of the function D( τ; φ) expressed by (6.10), we have
to assign a function to ε( t ) , so that to obtain a closed form solution for the
function C( τ; φ) . Writing (6.10) as

       Σ 3/ 2 
2 ( Σ − ζ ) τ ⋅ tanh         τ 
      3       
D(τ; ϕ) =
1
⋅                                     
(6.12)
τ                                Σ 3/ 2 
(1 + m) + (1 − m) tanh        τ 
     3       
             
we see that the choice

ε(τ) = τ                                                                     (6.13)
 Σ 3/ 2 
τ
2 ( Σ − ζ ) s tanh  3 s 

C ( τ ; φ) = − ∫                                 ds − i r φ τ
 Σ 3/ 2 
0    (1 + m) + (1 − m) tanh 
 3 s 

        
where the integral can be evaluated explicitly.
In fact, the above expression can also be written as (using the definition of m)
 Σ 3/ 2 
s ⋅ tanh     s 
τ               3      
C ( τ ; φ) = − η                           ds − i r φ τ
∫                 Σ 3/ 2 
0
Σ + ζ ⋅ tanh    s 
 3      
        
Calculating the integral,
 Σ 3/ 2 
s ⋅ tanh     s 
 3      
         ds =
∫                    Σ 3/ 2 
Σ + ζ ⋅ tanh    s 
 3      
        

2ζ 3/ 2 2                  Σ 3/ 2           Σ 3/ 2 
=−        ⋅s  + ln         Σ cosh                 
 3 s  + ζ ⋅ sinh  3 s 

3η      η                                          

40
we arrive at

2ζ 3/ 2             Σ 3/ 2  ζ          Σ 3/ 2 
C( τ ; φ) =      ⋅τ   − 2 ln cosh
 3 τ +     ⋅ sinh         
 3 τ  − i r φ τ (6.14)
3                           Σ                
Using equations (3.57), (3.54) and (3.59), under the same choice of functions h( t ) ,
h 2 ( t ) , g 2 ( t ) , δ( t ) and ε( t ) , we can determine the functions H (τ ; φ) and G (τ ; φ) .
We obtain
 ∆ 3/ 2 
2 ( ∆ − θ) ⋅ tanh         τ 
 3      
H( τ; ϕ) =                                
(6.15)
 ∆ 3/ 2 
(1 + n ) + (1 − n ) tanh     τ 
 3      
        
where
∆ −θ
n=                                                                                (6.16)
∆ +θ
We see that the choice

ε(τ) = τ                                                                           (6.17)
    ∆ 3/ 2 
2 ( ∆ − θ) s ⋅ tanh           s 
τ                               3       
G (τ ; φ) = − ∫                                           ds − i r φ τ
    ∆ 3/ 2 
0
(1 + n ) + (1 − n ) tanh      s 
   3       
           
where the integral can be evaluated explicitly.
In fact, the above expression can also be written as (using the definition of n)
 ∆ 3/ 2 
s ⋅ tanh     s 
τ                3      
G ( τ ; φ) = − ω                            ds − i r φ τ
∫                  ∆ 3/ 2 
0
∆ + θ ⋅ tanh    s 
 3      
        
Calculating the integral,

41
 ∆ 3/ 2 
s ⋅ tanh     s 
 3      
         ds =
∫                  ∆ 3/ 2 
∆ + θ ⋅ tanh    s 
 3      
        

2θ 3/ 2 2                              ∆ 3/ 2           ∆ 3/ 2 
=−       ⋅s  + ⋅ ln                   ∆ cosh
 3 s  + θ ⋅ sinh 
          3 s 

3ω      ω                                                      
we arrive at

2θ 3/ 2               ∆ 3/ 2  θ        ∆ 3/ 2 
G ( τ ; φ) =     ⋅τ    − 2 ⋅ ln cosh                         
 3 τ  + ∆ ⋅ sinh  3 τ  − i r φ τ (6.18)
3                                             
We are now in a position to determine the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ)
using (3.11) and (3.43) respectively:

f1 ( x , Y , t ; φ ) = e C ( t , T ; φ ) + D ( t , T ; φ ) Y + i φ x

f 2 ( x , Y , t ; φ) = e G ( t , T ; φ ) + H ( t , T ; φ ) Y + i φ x
Using the previously determined quantities, we find the following expressions for
the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ) :
1
f1 ( x , Y , t ; φ ) =                                                                   2
×
 Σ 3/ 2  ζ          Σ 3/ 2 
cosh 
 3 τ +     ⋅ sinh 
 3 τ 

          Σ                

                                                               Σ 3/ 2     
                                                     η ⋅ tanh     τ       
                                                               3      
× exp i φ x − i r φ τ +
2 ζ 3/ 2
⋅τ    +                                          Y

                    3                                             Σ 3/ 2  
                                                    Σ + ζ ⋅ tanh    τ  
 3      
                                                                          
1
f 2 ( x , Y, t; φ) =                                                                     2
×
 ∆ 3/ 2  θ         ∆ 3/ 2 
cosh 
 3 τ +    ⋅ sinh 
 3 τ 

         ∆                

42
                                                   ∆ 3/ 2     
                                         ω ⋅ tanh     τ       
                                                   3      
× exp i φ x − i r φ τ +
2θ 3/ 2
⋅τ   +                               Y

                    3                                 ∆ 3/ 2  
                                        ∆ + θ ⋅ tanh    τ  
 3      
                                                              
The risk-free probabilities P1 and P2 can then be evaluated using the Fourier
inversion formulas (3.9) and (3.10) respectively.
We find, using the previously determined quantities f1 and f 2
∞
1 1            R 1 ( x , Y, K , t; φ) 
P1 ( x , Y, t , ln K ) = +       ∫ Re                          dφ
2 π      0               iφ           
∞
1 1         R 2 ( x , Y , K , t ; φ) 
P2 ( x , Y, t , ln K ) =    +    ∫ Re                            dφ
2 π   0               iφ             
where R 1 ( x , Y, K , t; φ) and R 2 ( x , Y, K, t; φ) are given by
1
R 1 ( x , Y , t ; φ) =                                                      2
×
 Σ 3/ 2  ζ          Σ 3/ 2 
cosh 
 3 τ +     ⋅ sinh 
 3 τ 

          Σ                

                                                            Σ 3/ 2     
                                                  η ⋅ tanh     τ       
                                                            3      
× exp i φ ( x − ln K ) − i r φ τ +
2ζ 3/ 2
⋅τ   +                             Y

                               3                               Σ 3/ 2  
                                                 Σ + ζ ⋅ tanh    τ  
 3      
                                                                       
1
R 2 ( x , Y, t; φ) =                                                           2
×
 ∆ 3/ 2  θ         ∆ 3/ 2 
cosh 
 3 τ +    ⋅ sinh 
 3 τ 

         ∆                

                                                            ∆ 3/ 2     
                                                  ω ⋅ tanh     τ       
                                                            3      
× exp i φ ( x − ln K ) − i r φ τ +
2θ 3/ 2
⋅τ   +                             Y

                               3                               ∆ 3/ 2  
                                                 ∆ + θ ⋅ tanh    τ  
 3      
                                                                       

43
The integration appearing in the Gill-Pelaez inversion formulas for the
determination of P1 and P2 , can be performed in principle using the Adamchik-

Marichev algorithm (Adamchik and Marichev [2], Roach [44] and [45]) This
algorithm uses the known Meijer function (Meijer [41], Mathai and Saxena [40]).
The price F of a European call option is given by (3.1)

F = S ⋅ P1 − K ⋅ e − r (T − t ) P2
where K is the strike price and T is the maturity.

7. Model IV. The choice
h 2 (τ) = τ 2 and h ( τ) = τ2                                      (7.1)
converts equation (3.40) into the equation
2                                    1
w ′′(τ) −  − [ 1 + ρ (1 + i φ)] ⋅ τ 2  w ′(τ) − (φ 2 − i φ) τ 4 w (τ) = 0   (7.2)
τ                                    4
with general solution
        Σ +ζ 3            Σ −ζ 3
w (τ) = C1 exp  −          τ  + C 2 exp      τ                         (7.3)
         6                6     
                                
where

ζ = 1 + ρ (1 + i φ) , η = φ 2 − i φ and Σ = ζ 2 + η
Therefore equation (3.36) has a solution given by (3.37)
2     w ′(τ)
D(τ) =          ⋅        ,           h 2 ( τ) = τ 2                (7.4)
h 2 (τ) w (τ)
subject to the condition D(0 ; φ) = 0 .
Since
     Σ +ζ 2                  Σ +ζ 3       Σ −ζ 2      Σ −ζ 3
w ′( τ) = C1  −       τ  exp  −               τ  + C2      τ  exp      τ 
      2                       6           2           6     
                                                            
equation (7.4) can be written as

44
   Σ +ζ 2           Σ +ζ 3         Σ −ζ 2       Σ −ζ 3
C1  −       τ  exp  −       τ  + C2        τ  exp        τ 
    2                6             2            6        
D ( τ) = 2 ⋅                                                            
2
τ                           Σ +ζ 3             Σ −ζ 3
C1 exp  −       τ  + C 2 exp       τ 
     6                 6      
                              
 Σ +ζ 3
or, multiplying numerator and denominator by exp         τ  and simplifying,
 6        
          
we obtain
   Σ +ζ            Σ −ζ        Σ 3
C1  −          + C2          exp    τ 
    2              2           3   
D(τ; ϕ) = 2                                      
(7.5)
 Σ 3
C1 + C 2 exp    τ 
 3    
      
The condition D(0; ϕ) = 0 is satisfied, taking into account (7.5), as long as

   Σ +ζ            Σ −ζ   
C1  −           + C2         = 0
    2              2      
                          
From the previous equality we can determine the ratio of the two constants:

C1   Σ −ζ
=                                                             (7.6)
C2   Σ +ζ
Equation (7.5) can be written as

C1 
−
Σ +ζ      Σ −ζ 
+
 Σ 3
 exp    τ 
C2 
   2        2 
         
 3



D(τ; ϕ) = 2
C1        Σ 3
+ exp    τ 
C2        3    
      
which in turn can also be written, taking into account (7.6), as

Σ −ζ   Σ +ζ   Σ −ζ          Σ 3
−       +         exp    τ 
Σ +ζ   2      2 
         
 3



D(τ; ϕ) = 2
Σ −ζ        Σ 3
+ exp    τ 
Σ +ζ         3    
      
or, after the necessary simplifications,

45
 Σ 3
− ( Σ − ζ ) + ( Σ − ζ ) exp    τ 
 3   
D(τ; ϕ) =                                  
Σ −ζ           Σ 3
+ exp      τ 
Σ +ζ           3      
        
or
 Σ 3
− 1 + exp     τ 
 3    
D(τ; ϕ) = ( Σ − ζ )                   
Σ −ζ          Σ 3
+ exp    τ 
Σ +ζ          3   
     
or
 Σ 3             Σ 3
exp     τ  − exp  −     τ 
 6            6       
D(τ; ϕ) = ( Σ − ζ )                               
Σ −ζ         Σ 3          Σ 3
exp  −     τ  + exp      τ 
Σ +ζ      6               6     
                       
or
 Σ 3
2 ( Σ − ζ) sinh     τ 
 6     
D(τ; ϕ) =                                  
(7.7)
 Σ 3                  Σ 3
(1 + m) cosh    τ  + (1 − m) sinh    τ 
 6                    6   
                          
where
Σ −ζ
m=                                                           (7.8)
Σ +ζ
We thus obtain
 Σ 3
2 ( Σ − ζ) tanh      τ 
 6    
D(τ; ϕ) =                                                  (7.9)
 Σ 3
(1 + m) + (1 − m) tanh    τ 
 6   
     

46
We are now in a position to determine the function C(τ ; φ) given by equation
(3.42) by assigning values to the functions g 1 ( t ) , g 2 ( t ) , g( t ) and ε( t ) , taking into
account the compatibility conditions (3.28)-(3.30).
Let g 2 ( t ) = 0 . We then obtain from (3.30) that g( t ) = 0 . According to the assigned

values to the functions h ( t ) and h 2 ( t ) , we get from (3.28) that h1 (τ) = τ 2 . We
then obtain from (3.29) that g1 ( t ) = 0 . Therefore we have from (3.42) the
following expression for the function C(τ ; φ) :
τ
C( τ ; φ) = − ∫ ε(s) D(s; φ) ds − i r φ τ                                (7.10)
0

At this stage, given the form of the function D(τ; φ) expressed by (7.9), we have to
assign a function to ε( t ) , so that to obtain a closed form solution for the function
C(τ; φ) . Writing (7.9) as

      Σ 3
2 ( Σ − ζ ) τ 2 ⋅ tanh        τ 
     6   
1
D(τ; ϕ) = 2 ⋅                                  
(7.11)
τ                                  Σ 3
(1 + m) + (1 − m) tanh         τ 
    6   
        
we see that the choice

ε ( τ) = τ 2                                                                      (7.12)
 Σ 3
2 ( Σ − ζ ) s 2 tanh      s 
τ                           6     
C(τ ; φ) = − ∫                                    ds − i r φ τ
 Σ 3
0
(1 + m) + (1 − m) tanh       s 
 6     
       
where the integral can be evaluated explicitly.
In fact, the above expression can also be written as (using the definition of m)

47
 Σ 3
s 2 ⋅ tanh     s 
τ               6    
C ( τ ; φ) = − η                         ds − i r φ τ
∫                 Σ 3
0
Σ + ζ ⋅ tanh     s 
 6    
      
Calculating the integral,
 Σ 3
s 2 ⋅ tanh     s 
 6    
       ds =
∫                   Σ 3
Σ + ζ ⋅ tanh     s 
 6    
      

ζ 3 2                   Σ 3              Σ 3
=−       s + ⋅ ln       Σ cosh
 6 s  + ζ ⋅ sinh 


 6 s 
3η    η                                       
we arrive at

ζ 3             Σ 3 ζ            Σ 3
C( τ ; φ) =     τ − 2 ln cosh                      
 6 τ  + Σ ⋅ sinh  6 τ  − i r φ τ                       (7.13)
3                                    
Using equations (3.57), (3.54) and (3.59), under the same choice of functions h( t ) ,
h 2 ( t ) , g 2 ( t ) , δ( t ) and ε( t ) , we can determine the functions H (τ ; φ) and G (τ ; φ) .
We obtain
 ∆ 3
2 ( ∆ − θ) ⋅ tanh         τ 
 6     
H( τ; ϕ) =                               
(7.14)
 ∆ 3
(1 + n ) + (1 − n ) tanh     τ 
 6     
       
where
∆ −θ
n=                                                                                (7.15)
∆ +θ
We see that the choice

ε(τ) = τ 2

48
              ∆ 3
2 ( ∆ − θ) s 2 ⋅ tanh                  s 
τ                                        6   
G (τ ; φ) = − ∫                                                       ds − i r φ τ
              ∆ 3
0
(1 + n ) + (1 − n ) tanh                s 
             6   
                 
where the integral can be evaluated explicitly.
In fact, the above expression can also be written as (using the definition of n)
 ∆ 3
s 2 ⋅ tanh     s 
τ                  6    
G ( τ ; φ) = − ω                              ds − i r φ τ
∫                    ∆ 3
0
∆ + θ ⋅ tanh     s 
 6    
      
Calculating the integral,
 ∆ 3
s 2 ⋅ tanh     s 
 6    
       ds =
∫                  ∆ 3
∆ + θ ⋅ tanh     s 
 6    
      

θ 3 2                           ∆ 3              ∆ 3
=−      s + ⋅ ln               ∆ cosh
 6 s  + θ ⋅ sinh 


 6 s 
3ω    ω                                               
we arrive at

θ 3               ∆ 3 θ           ∆ 3
G ( τ ; φ) =     τ − 2 ⋅ ln cosh
 6 τ +
    ⋅ sinh      
 6 τ  −irφτ                               (7.16)
3                       ∆             
We are now in a position to determine the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ)
using (3.11) and (3.43) respectively:

f1 ( x , Y , t ; φ ) = e C ( t , T ; φ ) + D ( t , T ; φ ) Y + i φ x

f 2 ( x , Y , t ; φ) = e G ( t , T ; φ ) + H ( t , T ; φ ) Y + i φ x
Using the previously determined quantities, we find the following expressions for
the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ) :

49
1
f1 ( x , Y , t ; φ ) =                                                 2
×
 Σ 3 ζ            Σ 3
cosh                       
 6 τ  + Σ ⋅ sinh  6 τ 
                      

                                           Σ 3     
                                 η ⋅ tanh     τ    
                                           6    
ζ
× exp i φ x − i r φ τ + τ3 +                          Y

                  3                           Σ 3 
                                Σ + ζ ⋅ tanh    τ  
 6   
                                                   
1
f 2 ( x , Y, t; φ) =                                                       2
×
 ∆ 3 θ            ∆ 3
cosh                       
 6 τ  + ∆ ⋅ sinh  6 τ 
                      

                                            ∆ 3     
                                  ω ⋅ tanh     τ    
                                            6    
θ
× exp i φ x − i r φ τ + τ3 +                           Y

                  3                            ∆ 3 
                                 ∆ + θ ⋅ tanh    τ  
 6   
                                                    
The risk-free probabilities P1 and P2 can then be evaluated using the Fourier
inversion formulas (3.9) and (3.10) respectively.
We find, using the previously determined quantities f1 and f 2
∞
1 1               R 1 ( x , Y, K , t; φ) 
P1 ( x , Y, t , ln K ) = +       ∫ Re                             dφ
2 π      0                  iφ           
∞
1 1            R 2 ( x , Y , K , t ; φ) 
P2 ( x , Y, t , ln K ) =    +    ∫ Re                               dφ
2 π   0                  iφ             
where R 1 ( x , Y, K , t; φ) and R 2 ( x , Y, K, t; φ) are given by
1
R 1 ( x , Y , t ; φ) =                                                   2
×
 Σ 3 ζ           Σ 3
cosh
 6 τ +
    ⋅ sinh      
 6 τ 
       Σ             

50
                                                        Σ 3     
                                              η ⋅ tanh     τ    
                                                        6    
ζ
× exp i φ ( x − ln K ) − i r φ τ + τ3 +                            Y

                             3                             Σ 3 
                                             Σ + ζ ⋅ tanh    τ  
 6   
                                                                
1
R 2 ( x , Y , t ; φ) =                                           2
×
 ∆ 3 θ           ∆ 3
cosh
 6 τ +
    ⋅ sinh      
 6 τ 
       ∆             

                                                        ∆ 3     
                                              ω ⋅ tanh     τ    
                                                        6    
θ
× exp i φ ( x − ln K ) − i r φ τ + τ 3 +                           Y

                             3                             ∆ 3 
                                             ∆ + θ ⋅ tanh    τ  
 6   
                                                                
The integration appearing in the Gill-Pelaez inversion formulas for the
determination of P1 and P2 , can be performed in principle using the Adamchik-

Marichev algorithm (Adamchik and Marichev [2], Roach [44] and [45]) This
algorithm uses the known Meijer function (Meijer [41], Mathai and Saxena [40]).
The price F of a European call option is given by (3.1)

F = S ⋅ P1 − K ⋅ e − r (T − t ) P2
where K is the strike price and T is the maturity.

8. Model V. The choice
h 2 (τ) = τ a and h (τ) = τ a , a + 1 ≠ 0                         (8.1)
converts equation (3.40) into the equation
a                                   1
w ′′(τ) −  − [1 + ρ (1 + i φ)] ⋅ τ a  w ′(τ) − (φ 2 − i φ) τ 2a w (τ) = 0    (8.2)
τ                                   4
with general solution
    Σ + ζ a +1           Σ − ζ a +1 
w (τ) = C1 exp  −         τ  + C 2 exp            τ                     (8.3)
 2 (a + 1)               2 (a + 1)   
                                     

51
where

ζ = 1 + ρ (1 + i φ) , η = φ 2 − i φ and Σ = ζ 2 + η                        (8.4)
Therefore equation (3.33) has a solution given by (3.34)
2     w ′(τ)
D(τ) =           ⋅        ,   h 2 ( τ) = τ a                               (8.5)
h 2 (τ) w (τ)
subject to the condition D(0 ; φ) = 0 .
Since
       Σ +ζ a         Σ + ζ a +1 
w ′( τ) = C1  −         τ  exp  −         τ +
        2          2 (a + 1)     
                                 
 Σ −ζ a      Σ − ζ a +1 
+ C2      τ  exp            τ 
 2           2 (a + 1)   
                         
equation (8.5) can be written as
2
D(τ) =        ×
τa
  Σ +ζ a                   Σ + ζ a +1                 Σ −ζ            Σ −ζ             
C1 −     τ            exp  −         τ   +C                   τa    exp            τ a +1 
   2                    2 (a + 1)        2            2               2 (a + 1)        
                                                                                       
×
    Σ + ζ a +1           Σ −ζ             
C 1 exp  −         τ    + C exp            τ a +1 
 2 (a + 1)         2     2 (a + 1)        
                                          
 Σ + ζ a +1 
or, multiplying numerator and denominator by exp                 
 2(a + 1) τ  and
                
simplifying, we obtain
   Σ +ζ                Σ −ζ          Σ a +1 
C1  −              + C2            exp       τ 
    2                  2             a +1   
D(τ; ϕ) = 2                                               
(8.6)
 Σ a +1 
C1 + C 2 exp       τ 
 a +1      
           
The condition D(0; ϕ) = 0 is satisfied, taking into account (8.6), as long as

52
   Σ +ζ            Σ −ζ   
C1  −           + C2         = 0
    2              2      
                          
From the previous equality we can determine the ratio of the two constants:

C1   Σ −ζ
=                                                           (8.7)
C2   Σ +ζ
Equation (8.6) can be written as

C1 
−
Σ +ζ      Σ −ζ 
+
 Σ a +1 
 exp       τ 
C2 
   2        2 
           
 a +1



D(τ; ϕ) = 2
C1        Σ a +1 
+ exp       τ 
C2        a +1      
           
which in turn can also be written, taking into account (8.7), as

Σ −ζ   Σ +ζ   Σ −ζ           Σ a +1 
−      +           exp       τ 
Σ +ζ   2     2 
           
 a +1



D(τ; ϕ) = 2
Σ −ζ        Σ a +1 
+ exp       τ 
Σ +ζ        a +1      
           
or, after the necessary simplifications,
 Σ a +1 
− ( Σ − ζ ) + ( Σ − ζ) exp        τ 
 a +1   
D(τ; ϕ) =                                     
Σ −ζ           Σ a +1 
+ exp       τ 
Σ +ζ           a +1      
           
or
 Σ a +1 
− 1 + exp       τ 
 a +1     
D(τ; ϕ) = ( Σ − ζ)                        
Σ −ζ          Σ a +1 
+ exp       τ 
Σ +ζ          a +1    
         
or

53
      Σ                            Σ           
exp            τ a +1  − exp  −             τ a +1 
 2 (a + 1)                   2 (a + 1)         
D(τ; ϕ) = ( Σ − ζ )                                                            
Σ −ζ               Σ                          Σ          
exp  −              τ a +1  + exp            τa +1 
Σ +ζ        2 (a + 1)                    2 (a + 1)       
                                              
or
      Σ           
2 ( Σ − ζ) sinh                τ a +1 
 2 (a + 1)        
D(τ; ϕ) =                                                     
(8.8)
      Σ                                   Σ           
(1 + m) cosh            τ a +1  + (1 − m) sinh              τ a +1 
 2 (a + 1)                           2 (a + 1)        
                                                      
where
Σ −ζ
m=                                                                                  (8.9)
Σ +ζ
We thus obtain
      Σ           
2 ( Σ − ζ ) tanh            τ a +1 
 2 (a + 1)        
D(τ; ϕ) =                                       
(8.10)
      Σ            
(1 + m) + (1 − m) tanh             τ a +1 
 2 (a + 1)         
                   
We are now in a position to determine the function C(τ ; φ) given by equation
(3.42) by assigning values to the functions g 1 ( t ) , g 2 ( t ) , g( t ) and ε( t ) , taking into
account the compatibility conditions (3.28)-(3.30).
Let g 2 ( t ) = 0 . We then obtain from (3.30) that g( t ) = 0 . According to the assigned

values to the functions h ( t ) and h 2 ( t ) , we get from (3.28) that h1 (τ) = τ a . We
then obtain from (3.29) that g1 ( t ) = 0 . Therefore we have from (3.42) the
following expression for the function C(τ ; φ) :
τ
C(τ ; φ) = − ∫ ε(s) D(s; φ) ds − i r φ τ                                    (8.11)
0

54
At this stage, given the form of the function D( τ; φ) expressed by (8.10), we have
to assign a function to ε( t ) , so that to obtain a closed form solution for the
function C( τ; φ) . Writing (8.10) as

      Σ           
2 ( Σ − ζ) τ a ⋅ tanh            τ a +1 
 2 (a + 1)        
1
D(τ; ϕ) = a ⋅                                          
(8.12)
τ                                  Σ           
(1 + m) + (1 − m) tanh             τ a +1 
 2 (a + 1)        
                  
we see that the choice

ε ( τ) = τ a                                                                            (8.13)
      Σ           
2 ( Σ − ζ) s a tanh            s a +1 
τ                          2 (a + 1)        
C(τ ; φ) = − ∫                                               ds − i r φ τ
      Σ           
0
(1 + m) + (1 − m) tanh             s a +1 
 2 (a + 1)        
                  
where the integral can be evaluated explicitly.
In fact, the above expression can also be written as (using the definition of m)
      Σ            
s a ⋅ tanh            s a +1 
τ               2 (a + 1)         
C ( τ ; φ) = − η                                       ds − i r φ τ
∫                       Σ            
0
Σ + ζ ⋅ tanh              s a +1 
 2 (a + 1)         
                   
Calculating the integral,
      Σ            
s a ⋅ tanh            s a +1 
 2 (a + 1)         
                    ds =
∫                          Σ            
Σ + ζ ⋅ tanh              s a +1 
 2 (a + 1)         
                   

ζ              2                     Σ                             Σ           
=−               s a +1 + ⋅ ln     Σ cosh           s a +1  + ζ ⋅ sinh            s a +1 
(a + 1) η         η                2 (a + 1)                     2 (a + 1)        
                                                

55
we arrive at

      Σ            ζ                Σ           
C(τ ; φ) = − 2 ⋅ ln cosh           τ a +1  +   ⋅ sinh            τ a +1 
 2 (a + 1)           Σ         2 (a + 1)        
                                                
ζ a +1
+        τ −irφτ                                                (8.14)
a +1
Using equations (3.57), (3.54) and (3.59), under the same choice of functions h ( t ) ,
h 2 ( t ) , g 2 ( t ) , δ( t ) and ε( t ) , we can determine the functions H (τ ; φ) and G (τ ; φ) .
We obtain
      ∆           
2 ( ∆ − θ) ⋅ tanh               τ a +1 
 2 (a + 1)        
H( τ; ϕ) =                                          
(8.15)
      ∆           
(1 + n ) + (1 − n ) tanh            τ a +1 
 2 (a + 1)        
                  
where
∆ −θ
n=                                                                                (8.16)
∆ +θ
We see that the choice

ε(τ) = τa
      ∆            
2 ( ∆ − θ) s a ⋅ tanh               s a +1 
τ                            2 (a + 1)         
G (τ ; φ) = − ∫                                               ds − i r φ τ
      ∆           
0
(1 + n ) + (1 − n ) tanh            s a +1 
 2 (a + 1)        
                  
where the integral can be evaluated explicitly.
In fact, the above expression can also be written as (using the definition of n)

56
      ∆            
s a ⋅ tanh            s a +1 
τ                  2 (a + 1)         
G ( τ ; φ) = − ω                                           ds − i r φ τ
∫                          ∆            
0
∆ + θ ⋅ tanh              s a +1 
 2 (a + 1)         
                   
Calculating the integral,
      ∆            
s a ⋅ tanh            s a +1 
 2 (a + 1)         
                    ds =
∫                        ∆            
∆ + θ ⋅ tanh              s a +1 
 2 (a + 1)         
                   

θ             2                               ∆                             ∆           
=−             s a +1 + ⋅ ln               ∆ cosh           s a +1  + θ ⋅ sinh            s a +1 
(a + 1) ω         ω                          2 (a + 1)                     2 (a + 1)        
                                                
we arrive at

      ∆            θ                ∆           
G (τ ; φ) = − 2 ⋅ ln cosh           τ a +1  +   ⋅ sinh            τ a +1 
 2 (a + 1)           ∆         2 (a + 1)        
                                                
θ a +1
+        τ −irφτ                                                        (8.17)
a +1
We are now in a position to determine the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ)
using (3.11) and (3.43) respectively:

f1 ( x , Y , t ; φ ) = e C ( t , T ; φ ) + D ( t , T ; φ ) Y + i φ x

f 2 ( x , Y , t ; φ) = e G ( t , T ; φ ) + H ( t , T ; φ ) Y + i φ x
Using the previously determined quantities, we find the following expressions for
the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ) :
1
f1 ( x , Y, t; φ) =                                                                      2
×
      Σ            ζ                Σ           
cosh            τ a +1  +   ⋅ sinh            τ a +1 
 2 (a + 1)           Σ         2 (a + 1)        
                                                

57
                                                           Σ                
                                            η ⋅ tanh            τ a +1      
                                                      2 (a + 1)        
× exp i φ x − i r φ τ +
ζ a +1
τ +                                             Y

                   a +1                                       Σ       a +1  
                                           Σ + ζ ⋅ tanh             τ  
 2 (a + 1)         
                                                                            
1
f 2 ( x , Y , t ; φ) =                                                                   2
×
      ∆            θ                ∆           
cosh            τ a +1  +   ⋅ sinh            τ a +1 
 2 (a + 1)           ∆         2 (a + 1)        
                                                

                                                           ∆                
                                            ω ⋅ tanh            τ a +1      
                                                      2 (a + 1)        
× exp i φ x − i r φ τ +
θ a +1
τ +                                             Y

                   a +1                                       ∆             
                                           ∆ + θ ⋅ tanh             τ a +1  
 2 (a + 1)         
                                                                            
The risk-free probabilities P1 and P2 can then be evaluated using the Fourier
inversion formulas (3.9) and (3.10) respectively.
We find, using the previously determined quantities f1 and f 2
∞
1 1            R 1 ( x , Y, K , t; φ) 
P1 ( x , Y, t , ln K ) =      +      ∫ Re                           dφ
2 π     0                iφ           
∞
1 1                 R 2 ( x , Y , K , t ; φ) 
P2 ( x , Y, t , ln K ) = +           ∫ Re                             dφ
2 π          0                iφ             
where R 1 ( x , Y, K , t; φ) and R 2 ( x , Y, K, t; φ) are given by
1
R 1 ( x , Y, t; φ) =                                                                    2
×
      Σ            ζ                Σ           
cosh            τ a +1  +   ⋅ sinh            τ a +1 
 2 (a + 1)           Σ         2 (a + 1)        
                                                

                                                                      Σ                
                                                       η ⋅ tanh            τ a +1      
                                                                 2 (a + 1)        
× exp i φ ( x − ln K ) − i r φ τ +
ζ a +1
τ +                                             Y

                              a +1                                       Σ             
                                                      Σ + ζ ⋅ tanh             τ a +1  
 2 (a + 1)         
                                                                                       

58
1
R 2 ( x , Y , t ; φ) =                                                               2
×
      ∆            θ                ∆           
cosh            τ a +1  +   ⋅ sinh            τ a +1 
 2 (a + 1)           ∆         2 (a + 1)        
                                                

                                                                  ∆                
                                                   ω ⋅ tanh            τ a +1      
                                                             2 (a + 1)        
× exp i φ ( x − ln K ) − i r φ τ +
θ a +1
τ +                                         Y

                              a +1                                   ∆             
                                                  ∆ + θ ⋅ tanh             τ a +1  
 2 (a + 1)         
                                                                                   
The integration appearing in the Gill-Pelaez inversion formulas for the
determination of P1 and P2 , can be performed in principle using the Adamchik-

Marichev algorithm (Adamchik and Marichev [2], Roach [44] and [45]) This
algorithm uses the known Meijer function (Meijer [41], Mathai and Saxena [40]).
The price F of a European call option is given by (3.1)

F = S ⋅ P1 − K ⋅ e − r (T − t ) P2
where K is the strike price and T is the maturity.

9. Model VI. The choice
1              1
h 2 (τ) =     and h ( τ) =                                                      (9.1)
τ              τ
converts equation (3.37) into the equation
1                     1          1            1
w ′′(τ) +  + [1 + ρ (1 + i φ)] ⋅  w ′( τ) − (φ 2 − i φ) 2 w (τ) = 0                     (9.2)
τ                     τ          4           τ
with general solution
1                    1
− ( Σ +ζ)               ( Σ −ζ )
w (τ) = C1     τ 2            + C2   τ2                                                   (9.3)
where

ζ = 1 + ρ (1 + i φ) , η = φ 2 − i φ and Σ = ζ 2 + η
Therefore equation (3.36) has a solution given by (3.37)

59
2     w ′(τ)                              1
D(τ) =          ⋅        ,          h 2 ( τ) =                                                    (9.4)
h 2 (τ) w (τ)                                τ
subject to the condition D(0 ; φ) = 0 . Since
1                                               1
1              − (               Σ + ζ ) −1    1                 (               Σ − ζ ) −1
w ′( τ) = − ( Σ + ζ ) C1 τ 2                             + ( Σ − ζ) C 2 τ 2
2                                              2
equation (9.4) can be written as
1                                               1
1              − (                 Σ + ζ ) −1     1                 (               Σ − ζ ) −1
− ( Σ + ζ ) C1 τ 2                                + ( Σ − ζ) C 2 τ 2
D(τ) = 2 τ × 2                1
2
1
− ( Σ +ζ )                        ( Σ −ζ )
C1   τ 2                     + C2    τ2
or,
1                                           1
− ( Σ +ζ)                                      ( Σ −ζ )
− ( Σ + ζ) C1     τ 2                + ( Σ − ζ) C 2 τ           2
D(τ) =                         1                          1
− ( Σ +ζ )                    ( Σ −ζ )
C1   τ 2              + C2       τ2
1
( Σ +ζ)
Multiplying numerator and denominator by τ                          2           and simplifying, we obtain
Σ
− ( Σ + ζ) C1 + ( Σ − ζ) C 2 τ
D ( τ) =                                                                                                     (9.5)
Σ
C1 + C 2 τ
The condition D(0; ϕ) = 0 is satisfied, taking into account (9.5), as long as

− ( Σ + ζ ) C1 + ( Σ − ζ ) C 2 = 0

From the previous equality we can determine the ratio of the two constants:

C1   Σ −ζ
=                                                                                              (9.6)
C2   Σ +ζ
Equation (9.5) can be written as
C1                                        Σ
−      ( Σ + ζ) + ( Σ − ζ) τ
C2
D ( τ) =
C1       Σ
+τ
C2

60
which in turn can also be written, taking into account (9.6), as

Σ −ζ                                  Σ
−           ( Σ + ζ) + ( Σ − ζ) τ
Σ +ζ
D ( τ) =
Σ −ζ              Σ
+τ
Σ +ζ
or, after the necessary simplifications,
Σ
− ( Σ − ζ) + ( Σ − ζ ) τ
D(τ; ϕ) =
Σ −ζ              Σ
+τ
Σ +ζ
or
Σ
− 1+ τ
D(τ; ϕ) = ( Σ − ζ )                                                               (9.7)
Σ
m+τ
where
Σ −ζ
m=                                                                                (9.8)
Σ +ζ
We are now in a position to determine the function C(τ ; φ) given by equation
(3.42) by assigning values to the functions g 1 ( t ) , g 2 ( t ) , g( t ) and ε( t ) , taking into
account the compatibility conditions (3.28)-(3.30).
Let g 2 ( t ) = 0 . We then obtain from (3.30) that g( t ) = 0 . According to the assigned
values to the functions h ( t ) and h 2 ( t ) , we get from (3.28) that h1 ( τ) = 1 / τ . We
then obtain from (3.29) that g1 ( t ) = 0 . Therefore we have from (3.39) the
following expression for the function C(τ ; φ) :
τ
C(τ ; φ) = − ∫ ε(s) D(s; φ) ds − i r φ τ                               (9.9)
0

61
At this stage, given the form of the function D( τ; φ) expressed by (9.7), we have to
assign a function to ε( t ) , so that to obtain a closed form solution for the function
C(τ; φ) . We see that the choice
ε ( τ) = 1                                                                         (9.10)
τ                     Σ
− 1+ s
C(τ ; φ) = − ∫ ( Σ − ζ )                   ds − i r φ τ
Σ
0             m+s
where the integral can be evaluated explicitly.
Calculating the integral,
Σ                                                           
−1+ s                   m +1              1       1    1          Σ
∫                ds = s −        s ⋅ 2 F1 1,    ; 1+    ;− s             
Σ               m                Σ       Σ   m              
m+s                                                                  
we arrive at
      m +1           1       1    1              Σ   
C(τ ; φ) = ( Σ − ζ ) τ − 1 +      ⋅ 2 F1 1,    ; 1+    ;− τ                  − i r φ τ (9.11)
m              Σ       Σ   m                  

                                                     
Using equations (3.57), (3.54) and (3.59), under the same choice of functions h ( t ) ,
h 2 ( t ) , g 2 ( t ) , δ( t ) and ε( t ) , we can determine the functions H (τ ; φ) and G (τ ; φ) .
We obtain
∆
− 1+ τ
H( τ; ϕ) = ( ∆ − θ)                                                                 (9.12)
∆
n+τ
where
∆ −θ
n=                                                                                  (9.13)
∆ +θ
We see that the choice
ε(τ) = 1

62
τ                               ∆
− 1+ s
G (τ ; φ) = − ∫ ( ∆ − θ)                              ds − i r φ τ
∆
0                   n +s
where the integral can be evaluated explicitly.
Calculating the integral,
∆                                                                             
−1+ s                      n +1              1       1    1                         ∆
∫                ds = s −           s ⋅ 2 F1 1,    ; 1+    ;− s                            
∆                    n               ∆       ∆   n                             
n +s                                                                                   
we arrive at
       n +1           1       1    1                               ∆   
G ( τ ; φ) = ( ∆ − θ) τ  − 1 +      ⋅ 2 F1 1,    ; 1+    ;− τ                                   − i r φ τ       (9.14)
n             ∆       ∆   n                                   

                                                                       
We are now in a position to determine the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ)
using (3.11) and (3.43) respectively:

f1 ( x , Y , t ; φ ) = e C ( t , T ; φ ) + D ( t , T ; φ ) Y + i φ x

f 2 ( x , Y , t ; φ) = e G ( t , T ; φ ) + H ( t , T ; φ ) Y + i φ x
Using the previously determined quantities, we find the following expressions for
the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ) :

                            m +1           1       1    1                         Σ       
f1 ( x , Y, t; φ) = exp  ( Σ − ζ ) τ          − 1 +      ⋅ 2 F1 1,    ; 1+    ;− τ                                ×
                              m              Σ       Σ   m                                  
                      
                                                                   
                             − 1+ τ
Σ      
× exp  i φ x − i r φ τ + ( Σ − ζ )                                    Y
                                                         Σ       
                             m+τ                                 
                           n +1           1       1    1                         ∆       
f 2 ( x , Y, t; φ) = exp  ( ∆ − θ) τ           −1 +      ⋅ 2 F1 1,    ; 1+    ;− τ                                ×
                              n             ∆       ∆   n                                  
                      
                                                                  
                            − 1+ τ
∆      
× exp i φ x − i r φ τ + ( ∆ − θ)                                      Y
                                                          ∆       
                            n+τ                                   
The risk-free probabilities P1 and P2 can then be evaluated using the Fourier
inversion formulas (3.9) and (3.10) respectively.

63
We find, using the previously determined quantities f1 and f 2
∞
1 1          R 1 ( x , Y, K , t; φ) 
P1 ( x , Y, t , ln K ) =    +    ∫ Re                           dφ
2 π   0                iφ           
∞
1 1           R 2 ( x , Y , K , t ; φ) 
P2 ( x , Y, t , ln K ) =    +    ∫ Re                              dφ
2 π   0                 iφ             
where R 1 ( x , Y, K , t; φ) and R 2 ( x , Y, K, t; φ) are given by

                     m +1           1       1    1          Σ       
R 1 ( x , Y, t; φ) = exp  ( Σ − ζ ) τ    −1 +      ⋅ 2 F1 1,    ; 1+    ;− τ                 ×
                       m              Σ       Σ   m                   
                
                                                   
                                        − 1+ τ
Σ    
× exp  i φ ( x − ln K ) − i r φ τ + ( Σ − ζ )                       Y
                                                          Σ    
                                        m+τ                    
                     n +1           1       1    1          ∆       
R 2 ( x , Y, t; φ) = exp  ( ∆ − θ) τ     −1 +      ⋅ 2 F1 1,    ; 1+    ;− τ                 ×
                        n             ∆       ∆   n                   
                
                                                   
                                       − 1+ τ
∆    
× exp i φ ( x − ln K ) − i r φ τ + ( ∆ − θ)                         Y
                                                        ∆       
                                       n+τ                      
The integration appearing in the Gill-Pelaez inversion formulas for the
determination of P1 and P2 , can be performed in principle using the Adamchik-

Marichev algorithm (Adamchik and Marichev [2], Roach [44] and [45]) This
algorithm uses the known Meijer function (Meijer [41], Mathai and Saxena [40]).
The price F of a European call option is given by (3.1)

F = S ⋅ P1 − K ⋅ e − r (T − t ) P2
where K is the strike price and T is the maturity.

10. Model VII. The choice
h 2 (τ) = τ 2 and h (τ) = τ                                               (10.1)
converts equation (3.40) into the equation

64
2                                 1
w ′′(τ) −  − [1 + ρ (1 + i φ)] ⋅ τ  w ′(τ) − (φ 2 − i φ) τ 2 w (τ) = 0     (10.2)
τ                                 4
with general solution

     ζ Σ −Σ 1     Σ 2        ζ Σ −Σ 1     Σ 2 

w (τ) = C1 U       ,− ,−   τ  + C2 M       ,− ,−   τ  ×
      4Σ     2   2            4Σ     2   2    
                                              
 Σ −ζ 2
× exp      τ                                                  (10.3)
 4     
       
where U(a , b, x ) and M(a , b, x ) are Kummer’s hypergeometric functions, two
linearly independent solutions of the confluent hypergeometric second order
differential equation (see for example Abramowitz and Stegun [1], Lebedev [33],
Whittaker and Watson [53])
x y′′ + (b − x ) y′ − a y = 0
The parameters ζ , η and Σ are defined by

ζ = 1 + ρ (1 + i φ) , η = φ 2 − i φ and Σ = ζ 2 + η             (10.4)
Therefore equation (3.36) has a solution given by (3.37)
2     w ′(τ)
D(τ; φ) =          ⋅        , h 2 (τ) = τ 2                     (10.5)
h 2 (τ) w (τ)
subject to the condition D(0 ; φ) = 0 .
Since
d
U (a , b, t ) = − a U(a + 1, b + 1, t )
dt
and
d                 a
M (a , b, t ) = M (a + 1, b + 1, t )
dt                b
we find
 Σ −ζ 
w ′( τ) =      τ  { C1 X (ζ , Σ ; τ 2 ) + 2 C 2 Y (ζ , Σ ; τ 2 )} ×
 4     
       

65
 Σ −ζ 2
× exp      τ                                            (10.6)
 4     
       
where
ζ Σ −Σ 1     Σ 2     ζ Σ + 3Σ 1    Σ 2
X (ζ , Σ ; τ 2 ) = 2 U       ,− ,−   τ  − U         , ,−   τ  (10.7)
 4Σ     2   2          4Σ     2   2   
                                      
and
ζ Σ−Σ 1     Σ 2     ζ Σ + 3Σ 1    Σ 2
Y (ζ , Σ ; τ 2 ) = M      ,− ,−   τ  − M         , ,−   τ  (10.8)
 4Σ    2   2          4Σ     2   2   
                                     
Therefore
 Σ −ζ 1
D(τ; φ) =      ⋅ ×
 4    τ
       

C1 X (ζ, Σ ; τ 2 ) + 2 C 2 Y (ζ, Σ ; τ 2 )
×                                                                                (10.9)
ζ Σ −Σ 1        Σ 2               ζ Σ −Σ 1             Σ 2
C1 U       ,− ,−        τ  + C2 M                    ,− ,−   τ 
 4Σ     2      2                   4Σ             2   2   
                                                          
It is known that (see for example Abramowitz and Stegun [1], Lebedev [33],
Whittaker and Watson [53])
Γ(1 − b)                   Γ(b − 1) 1− b
U(a , b, x ) =                M(a , b, x ) +         x M(a − b + 1, 2 − b, x )
Γ(a − b + 1)                 Γ (a )

Using this relation we can express X (ζ , Σ ; τ 2 ) in terms of the function M as
follows:

π        ζ Σ −Σ 1     Σ 2
X (ζ , Σ ; τ 2 ) =                 M      ,− ,−   τ +
 ζ Σ + 5Σ   4 Σ     2   2   
Γ                            
   4Σ     
          
3/ 2
8 π      Σ                      ζ Σ + 5Σ 5    Σ 2
+              −                  τ3 M          , ,−   τ −
 ζ Σ −Σ   2                         4Σ     2   2   
3Γ                                                   
 4Σ 
        

66
π       ζ Σ + 3Σ 1    Σ 2
−                 M         , ,−   τ +
 ζ Σ + 5Σ      4Σ     2   2   
Γ                              
   4Σ     
          
1/ 2
2 π       Σ                       ζ Σ + 5Σ 3    Σ 2
+                −                     τ M         , ,−   τ    (10.10)
 ζ Σ + 3Σ   2                          4Σ     2   2   
3Γ                                                      
   4Σ     
          
We thus find that

C1 X (ζ , Σ ; τ 2 ) + 2 C 2 Y (ζ, Σ ; τ 2 ) =

                       
                       
        π              
= C1              + 2C 2  × M1 (ζ, Σ, τ 2 ) +
 Γ  ζ Σ + 5Σ 
                  
      4Σ             
                     
1/ 2
   Σ
+ C1 ⋅ 2 π  −
 2 
              τ × M 2 (ζ, Σ, τ 2 )        (10.11)
    
where
ζ Σ −Σ 1     Σ 2
M1 ( ζ , Σ, τ 2 ) = M       ,− ,−   τ −
 4Σ     2   2   
                
 ζ Σ + 3Σ 1    Σ 2 
− M         , ,−   τ                            (10.12)
   4Σ     2   2    
                   
and


      4        Σ  2  ζ Σ + 5Σ 5    Σ 2
M 2 (ζ, Σ, τ ) = 
2                  −   τ M         , ,−   τ +
ζ Σ −Σ  2        4Σ     2   2   
3 Γ                                 
  4Σ 
          

67


1           ζ Σ + 5Σ 3    Σ 2 
+               M          , ,−   τ                        (10.13)
 ζ Σ + 3Σ       4Σ     2   2    

Γ           
   4Σ                              
                                   
The difference
ζ Σ −Σ 1     Σ 2     ζ Σ + 3Σ 1    Σ 2
M      ,− ,−   τ  − M         , ,−   τ 
 4Σ     2   2          4Σ     2   2   
                                      

appearing in (10.12), is proportional to τ 2 , due to the relation
M (a , b, x ) − M (a + 1, b + 1, x ) =

 a a +1    1  a (a + 1) (a + 1)(a + 2)  2
= −      x +  b ( b + 1) − ( b + 1)(b + 2)  x + O( x )

3
 b b + 1   2                               
Therefore, the numerator in the fraction appearing in (10.9), after canceling with
τ , contains a term proportional to τ , which is also true for the term
1/ 2
   Σ
C1 ⋅ 2 π  −
 2 
              × M 2 (ζ, Σ, τ 2 )
    
The condition then D(0; φ) = 0 is satisfied if and only if C1 = 0 . We thus have that
the function D(τ; φ) is given by (as follows from (10.9) for C1 = 0 )

 Σ −ζ 1       2 Y (ζ, Σ ; τ 2 )
D(τ; φ) =      ⋅ ×
 4    τ
        M  ζ Σ − Σ , − 1 , − Σ τ2 
                        
 4Σ          2       2  
                        
or
Σ −ζ
D( τ; φ) =         ×
2τ

ζ Σ −Σ 1                    Σ 2     ζ           Σ + 3Σ 1     Σ 2
M      ,− ,−                  τ  − M                    , ,−   τ 
 4Σ     2                  2                     4Σ     2   2   
×                                                                    (10.14)
ζ                Σ−Σ 1                 Σ 2
M                     ,− ,−                τ 
                4Σ      2             2     
                                            

68
We are now in a position to determine the function C( t , T ; φ) given by equation
(3.42) by assigning values to the functions g 1 ( t ) , g 2 ( t ) , g( t ) and ε( t ) , taking into
account the compatibility conditions (3.28)-(3.30).
Let g 2 ( t ) = 0 . We then obtain from (3.30) that g( t ) = 0 . According to the assigned
values to the functions h( t ) and h 2 ( t ) , we get from (3.28) that h1 ( t ) = 1 . We then
obtain from (3.29) that g 1 ( t ) = 0 . Therefore we have from (3.42) the following
expression for the function C( t , T ; φ) :
τ
C(τ ; φ) = − ∫ ε(s) D(s; φ) ds − i r φ τ                               (10.15)
0

At this stage, given the form of the function D( τ; φ) expressed by (10.14), we have
to assign a function to ε(τ) , so as to obtain a closed form solution for the function
C(τ; φ) . Writing (10.14) as

       d ζ Σ −Σ 1         Σ 2 
         M       ,− ,−      τ 
1  Σ −ζ    dτ  4 Σ
        2    2     

D(τ; φ) = 2 ×     τ+                            
τ  2           ζ Σ −Σ 1      Σ 2 
        M       ,− ,−      τ  
 4Σ     2    2     
                              
we see that the choice

ε(τ) = τ 2
         d ζ Σ −Σ 1         Σ 2 
           M       ,− ,−      s 
τ
 Σ −ζ    ds  4 Σ
        2    2     
  ds − i r φ τ
C( τ ; φ) = − ∫       s+                            
0 
2         ζ Σ −Σ 1      Σ 2  
          M       ,− ,−      s  
 4Σ     2    2     
                                
where the integral can be evaluated explicitly. We thus find

Σ −ζ 2       ζ Σ −Σ 1     Σ 2
C ( τ ; φ) = −        τ − ln M       ,− ,−   τ  −irφτ                                 (10.16)
4            4Σ     2   2   
                

69
Using equations (3.57), (3.54) and (3.59), under the same choice of functions h ( t ) ,
h 2 ( t ) , g 2 ( t ) , δ( t ) and ε( t ) , we can determine the functions H (τ ; φ) and G (τ ; φ) .
We obtain
∆ −θ
H (τ; φ) =            ×
2τ

θ ∆ −∆ 1                              ∆ 2    θ                  ∆ + 3∆ 1     ∆ 2
M      ,− ,−                            τ − M                           , ,−   τ 
 4∆     2                            2                          4∆      2   2   
×                                                                                    (10.17)
θ                         ∆−∆ 1                        ∆ 2
M                               ,− ,−                      τ 
                          4∆     2                    2     
                                                            
Writing (10.17) as
       d θ ∆−∆ 1          ∆ 2 
         M       ,− ,−      τ 
1  ∆ −θ    dτ  4 ∆
        2    2     

H(τ; φ) = 2 ×     τ+                            
τ  2           θ ∆ −∆ 1      ∆ 2  
        M       ,− ,−      τ  
 4∆     2    2     
                              
We see that the choice

ε(τ) = τ2
         d ζ Σ−Σ 1          Σ 2 
           M       ,− ,−      s 
τ
 Σ −ζ    ds  4 Σ
        2    2     
  ds − i r φ τ
G ( τ ; φ) = − ∫       s+                            
0 
2         ζ Σ −Σ 1      Σ 2  
          M       ,− ,−      s  
 4Σ     2    2     
                                
where the integral can be evaluated explicitly. We thus find

∆ −θ 2       θ ∆ −∆ 1     ∆ 2
G (τ ; φ) = −         τ − ln M       ,− ,−   τ  −irφτ                                         (10.18)
4            4∆     2   2   
                
We are now in a position to determine the functions f1 ( x, Y, t; φ) and f 2 ( x, Y, t; φ)
using (3.11) and (3.43) respectively:

f1 ( x , Y , t ; φ ) = e C ( t , T ; φ ) + D ( t , T ; φ ) Y + i φ x

70
f 2 ( x , Y , t ; φ) = e G ( t , T ; φ ) + H ( t , T ; φ ) Y + i φ x
Using the previously determined quantities, we find the following expressions for
the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ) :

                  Σ −ζ 2
exp i φ x − i r φ τ −
                       τ 

                   2      ×
f1 ( x , Y , t ; φ ) =
ζ Σ −Σ 1               Σ 2 
M               ,− ,−     τ 
 4Σ               2   2    
                           

            d ζ Σ −Σ 1         Σ 2  
              M       ,− ,−      τ  
 1  Σ −ζ    dτ   4Σ     2    2      
× exp 2 ×     τ+                           Y 

τ  2           ζ Σ −Σ 1      Σ 2  
             M       ,− ,−      τ   
                 4Σ                
                       2    2         
                  ∆ −θ 2
exp i φ x − i r φ τ −
                       τ 

                   2      ×
f 2 ( x , Y , t ; φ) =
θ ∆−∆ 1                ∆ 2 
M               ,− ,−     τ 
 4∆               2   2    
                           

            d θ ∆−∆ 1          ∆ 2  
              M       ,− ,−      τ  
 1  ∆ −ζ    dτ   4∆     2    2      
× exp 2 ×     τ+                           Y 

τ  2           θ ∆ −∆ 1      ∆ 2  
             M       ,− ,−      τ   
                 4∆                
                       2    2         
The risk-free probabilities P1 and P2 can then be evaluated using the Fourier
inversion formulas (3.9) and (3.10) respectively.
We find, using the previously determined quantities f1 and f 2
∞
1 1                  R 1 ( x , Y, K , t; φ) 
P1 ( x , Y, t , ln K ) =     +         ∫ Re                              dφ
2 π        0                   iφ           
∞
1 1                  R 2 ( x , Y , K , t ; φ) 
P2 ( x , Y, t , ln K ) =      +         ∫ Re                                dφ
2 π        0                   iφ             
where R 1 ( x, Y, K , t; φ) and R 2 ( x, Y, K, t; φ) are given by

71
                                Σ −ζ 2
exp i φ ( x − ln K ) − i r φ τ −
                                     τ 

                                 2     ×
R 1 ( x , Y , t ; φ) =
ζ Σ −Σ 1                  Σ 2 
M                 ,− ,−          τ 
 4Σ              2       2    
                              

            d ζ Σ −Σ 1         Σ 2  
              M       ,− ,−      τ  
 1  Σ −ζ    dτ  4 Σ
        2    2      
 Y 
× exp 2 ×     τ+                            
τ  2           ζ Σ −Σ 1      Σ 2  
             M       ,− ,−      τ   
                 4Σ                
                       2    2         
                                ∆ −θ 2
exp i φ ( x − ln K ) − i r φ τ −
                                     τ 

                                 2     ×
R 2 ( x , Y , t ; φ) =
θ ∆−∆ 1                   ∆ 2 
M                 ,− ,−          τ 
 4∆              2       2    
                              

            d θ ∆−∆ 1          ∆ 2  
              M       ,− ,−      τ  
 1  ∆ −ζ    dτ  4 ∆
        2    2      
 Y 
× exp 2 ×     τ+                            
τ  2           θ ∆ −∆ 1      ∆ 2  
             M       ,− ,−      τ   
                 4∆                
                       2    2         
The integration appearing in the Gill-Pelaez inversion formulas for the
determination of P1 and P2 , can be performed in principle using the Adamchik-

Marichev algorithm (Adamchik and Marichev [2], Roach [44] and [45]) This
algorithm uses the known Meijer function (Meijer [41], Mathai and Saxena [40]).
The price F of a European call option is given by (3.1)

F = S ⋅ P1 − K ⋅ e − r (T − t ) P2
where K is the strike price and T is the maturity.

11. Model VIII. The choice
h 2 (τ) = τ b and h ( t ) = k τ a , a ≠ b , a ≠ −1, b ≠ −1        (11.1)
converts equation (3.40) into the equation

72
b                                     1
w ′′(τ) −  − [1 + ρ (1 + i φ)] ⋅ k τ a  w ′(τ) − (φ 2 − i φ) (k τ a ) 2 w (τ) = 0   (11.2)
τ                                     4
with general solution given by

      (a − b) ( Σ + ζ ) a − b k Σ a +1             
w (τ) = C1 M                   ,     ,     τ               +
      2 (a + 1) Σ        a +1 a +1                 
                                                   

 (a + b + 2) Σ + (a − b) ζ a + b + 2 k Σ a +1           
+ C 2 t b +1 M                                            τ            ×

,         ,
        2 (a + 1) Σ          a +1     a +1
                                                        

 k ( Σ + ζ ) a +1 
× exp  −           τ                                                           (11.3)
   2 (a + 1)      
                  
where M(a, b, x) are Kummer’s hypergeometric functions, two linearly
independent solutions of the confluent hypergeometric second order differential
equation (see for example Abramowitz and Stegun [1], Lebedev [33], Whittaker
and Watson [53])
x y′′ + (b − c) y′ − a y = 0
The parameters ζ , η and Σ are defined by

ζ = 1 + ρ (1 + i φ) , η = φ 2 − i φ and Σ = ζ 2 + η                       (11.4)
Therefore equation (3.36) has a solution given by (3.37)
2     w ′(τ)
D(τ; φ) =          ⋅        , h 2 (τ) = τ b                               (11.5)
h 2 (τ) w (τ)
subject to the condition D(0 ; φ) = 0 .
Since
d                 a
M (a , b, t ) = M (a + 1, b + 1, t )
dt                b
we find
 k ( Σ + ζ ) a +1 
w ′(τ) = C1 τ a X (ζ , Σ ; τ a +1 ) × exp  −           τ +
   2 (a + 1)      
                  

73
 k ( Σ + ζ ) a +1 
+ C 2 k τ a + b +1 Y (ζ, Σ ; τ a +1 ) × exp  −           τ +
   2 (a + 1)      
                  
 k ( Σ + ζ ) a +1 
+ C 2 (b + 1) τ b × M 3 (ζ , Σ ; τ a +1 ) × exp  −           τ                  (11.6)
   2 (a + 1)      
                  
where

k ( Σ + ζ )   (3 a − b + 2) Σ + (a − b) ζ 2 a − b + 1 k Σ a +1
                                                                       
X (ζ, Σ ; τ a +1 ) =                M                            ,           ,      τ                      −
2                 2 (a + 1) Σ            a +1      a +1                        
                                                                        
 (a − b) ( Σ + ζ ) a − b k Σ a +1               
− M                              τ                   (11.7)

,     ,
 2 (a + 1) Σ        a +1 a +1
                                                
( a + b + 2) Σ + ( a − b ) ζ
Y (ζ , Σ ; τ a +1 ) =                                M1 (ζ, Σ ; τ a +1 ) −
2 (a + b + 2)

Σ+ζ
−        M 2 (ζ, Σ ; τ a +1 )                                          (11.8)
2
 (3 a + b + 4) Σ + (a − b) ζ 2 a + b + 3 k Σ a +1                    
M1 (ζ, Σ ; τ a +1 ) = M                             ,           ,      τ                           (11.9)
         2 (a + 1) Σ            a +1      a +1                       
                                                                     
 (a − b) ( Σ + ζ ) a − b k Σ a +1                 
M 2 (ζ, Σ ; τ a +1 ) = M                   ,     ,     τ                                          (11.10)
 2 (a + 1) Σ        a +1 a +1                     
                                                  
 (a + b + 2) Σ + (a − b) ζ a + b + 2 k Σ a +1                   
M 3 (ζ, Σ ; τ a +1 ) = M                           ,         ,      τ                             (11.11)
        2 (a + 1) Σ          a +1     a +1                      
                                                                
We thus arrive at the following expression for the function D(τ; φ)
D(τ; φ) = 2 ×

C1 τ a − b X (ζ, Σ ; τ a +1 ) + C 2 k τ a +1 Y(ζ, Σ ; τ a +1 ) + C 2 (b + 1) M 3 (ζ, Σ ; τ a +1 )
×
C1 M 2 (ζ, Σ; τ a +1 ) + C 2 t b +1 M 3 (ζ, Σ; τ a +1 )
(11.12)

The term X (ζ, Σ ; τ a +1 ) contains a difference of Kummer’s functions, which,
due to the relation

74
M (a , b, x ) − M (a + 1, b + 1, x ) =

 a a +1    1  a (a + 1) (a + 1)(a + 2)  2
= −      x +  b ( b + 1) − ( b + 1)(b + 2)  x + O( x )

3
 b b + 1   2                               

is proportional to τ a+1 . Therefore the term proportional to the constant C1 has the

form t 2a − b +1 .
The condition then D(0; φ) = 0 is satisfied if and only if C 2 = 0 with
2a − b +1 > 0
We thus have that the function D( τ; φ) is given by (as follows from (11.12) for
C2 = 0 )

τ a − b { M1 (ζ, Σ ; τ a +1 ) − M 2 (ζ, Σ ; τ a +1 )}
D(τ; φ) = k ( Σ + ζ ) ×                                                               (11.13)
M 2 (ζ, Σ; τ a +1 )
We are now in a position to determine the function C( t , T ; φ) given by equation
(3.42) by assigning values to the functions g 1 ( t ) , g 2 ( t ) , g( t ) and ε( t ) , taking into
account the compatibility conditions (3.28)-(3.30).
Let g 2 ( t ) = 0 . We then obtain from (3.30) that g( t ) = 0 . According to the assigned

values to the functions h ( t ) and h 2 ( t ) , we get from (3.28) that h1 ( t ) = k 2 τ 2a − b .
We then obtain from (3.29) that g 1 ( t ) = 0 . Therefore we have from (3.42) the
following expression for the function C( t , T ; φ) :
τ
C( τ ; φ) = − ∫ ε(s) D(s; φ) ds − i r φ τ                                 (11.14)
0

At this stage, given the form of the function D( τ; φ) expressed by (11.13), we have
to assign a function to ε(τ) , so as to obtain a closed form solution for the function
C(τ; φ) . Writing (11.13) as

75
                       d                       
                          M 2 (ζ , Σ; τ a +1 ) 
D(τ; φ) = τ − b  − k ( Σ + ζ ) τ a + 2 dt                      

                        M 2 (ζ, Σ; τ a +1 )   
                                               
we see that the choice

ε(τ) = τb
                      d  (a − b) ( Σ + ζ ) a − b k Σ a +1  
                        M                    ,      ,    s   
τ
                      ds  2 (a + 1) Σ
                    a +1 a +1     
  ds
C( τ ; φ) = − ∫ − k ( Σ + ζ ) s a + 2                                           
0 
 (a − b) ( Σ + ζ ) a − b k Σ a +1  
                       M                   ,      ,    s   
 2 (a + 1) Σ        a +1 a +1      
                                                            
−irφτ
where the integral can be evaluated explicitly. We thus find
k ( Σ + ζ ) a +1
C ( τ ; φ) =              τ −irφτ −
a +1
2
 (a − b) ( Σ + ζ ) a − b k Σ a +1              
− ln M                   ,     ,     τ                         (11.16)
 2 (a + 1) Σ        a +1 a +1                  
                                               
Using equations (3.57), (3.54) and (3.59), under the same choice of functions h ( t ) ,
h 2 ( t ) , g 2 ( t ) , δ( t ) and ε( t ) , we can determine the functions H (τ ; φ) and G (τ ; φ) .
We obtain

τ a − b { M1 (θ, ∆ ; τ a +1 ) − M 2 (θ, ∆ ; τ a +1 )}
H(τ; φ) = k ( ∆ + θ ) ×                                                            (11.17)
M 2 (θ, ∆; τ a +1 )
Writing (11.17) as
                       d                      
                          M 2 (θ, ∆; τ a +1 ) 
H (τ; φ) = τ − b  − k ( ∆ + θ ) τ a + 2 dt           a +1 


                        M 2 (θ, ∆; τ ) 
                                              
we see that the choice

76
ε(τ) = τb
                   d  (a − b) ( ∆ + θ) a − b k ∆ a +1  
                     M                   ,      ,    s   
τ
                   ds  2 (a + 1) ∆
                   a +1 a +1     
  ds
G ( τ ; φ) = − ∫ − k ( ∆ + θ) s + 2
a

0 
 (a − b) ( ∆ + θ) a − b k ∆ a +1  
                    M                  ,      ,    s   
 2 (a + 1) ∆       a +1 a +1      
                                                        
−irφτ
where the integral can be evaluated explicitly. We thus find
k ( ∆ + θ) a +1
G (τ ; φ) =               τ −irφτ−
a +1
2
 (a − b) ( ∆ + θ) a − b k ∆ a +1                               
− ln M                  ,     ,     τ                                        (11.18)
 2 (a + 1) ∆       a +1 a +1                                   
                                                               
We are now in a position to determine the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ)
using (3.11) and (3.43) respectively:

f1 ( x , Y , t ; φ ) = e C ( t , T ; φ ) + D ( t , T ; φ ) Y + i φ x

f 2 ( x , Y , t ; φ) = e G ( t , T ; φ ) + H ( t , T ; φ ) Y + i φ x
Using the previously determined quantities, we find the following expressions for
the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ) :

                   k ( Σ + ζ ) a +1 
exp i φ x − i r φ τ +
                              τ    
                      a +1          
f1 ( x , Y, t; φ) =                                                                       2
×
 (a − b) ( Σ + ζ ) a − b k Σ a +1                             
M                  ,     ,     τ                               
 2 (a + 1) Σ        a +1 a +1                                 
                                                              

                              d                       
                                 M 2 (ζ, Σ; τ a +1 )  
× exp  τ − b  − k ( Σ + ζ ) τ a + 2 dt           a +1  
 Y
                               M 2 (ζ, Σ; τ ) 
                                                        
                                                      

77
                   k ( ∆ + θ) a +1 
exp i φ x − i r φ τ +
                             τ    
                      a +1         
f 2 ( x , Y , t ; φ) =                                                          2
×
 (a − b) ( ∆ + θ) a − b k ∆ a +1                 
M                 ,     ,     τ                   
 2 (a + 1) ∆       a +1 a +1                     
                                                 

                              d                       
                                 M 2 (θ, ∆; τ a +1 )  
× exp  τ − b  − k ( ∆ + θ ) τ a + 2 dt                      Y
       
                        M 2 (θ, ∆; τ a +1 )  
 

                                                      
The risk-free probabilities P1 and P2 can then be evaluated using the Fourier
inversion formulas (3.9) and (3.10) respectively.
We find, using the previously determined quantities f1 and f 2
∞
1 1            R 1 ( x , Y, K , t; φ) 
P1 ( x , Y, t , ln K ) = +      ∫ Re                           dφ
2 π     0                iφ           
∞
1 1            R 2 ( x , Y , K , t ; φ) 
P2 ( x , Y, t , ln K ) = +      ∫ Re                             dφ
2 π     0                iφ             
where R 1 ( x , Y, K , t; φ) and R 2 ( x , Y, K, t; φ) are given by

                              k ( Σ + ζ ) a +1 
exp i φ ( x − ln K ) − i r φ τ +
                                         τ    
                                 a +1          ×
R 1 ( x , Y, t; φ) =                                                   2
 (a − b) ( Σ + ζ ) a − b k Σ a +1 
M                        ,       ,      τ    
 2 (a + 1) Σ             a +1 a +1         
                                           

                              d                       
                                 M 2 (ζ, Σ; τ a +1 )  
× exp  τ − b  − k ( Σ + ζ ) τ a + 2 dt           a +1  
 Y
                               M 2 (ζ, Σ; τ ) 
                                                        
                                                      

                              k ( ∆ + θ) a +1 
exp i φ ( x − ln K ) − i r φ τ +
                                        τ    
                                 a +1         ×
R 2 ( x , Y, t; φ) =                                                  2
 (a − b) ( ∆ + θ) a − b k ∆ a +1 
M                        ,       ,      τ   
 2 (a + 1) ∆             a +1 a +1        
                                          

78
                              d                       
                                 M 2 (θ, ∆; τ a +1 )  
× exp  τ − b  − k ( ∆ + θ ) τ a + 2 dt                      Y
       
                        M 2 (θ, ∆; τ a +1 )  
 

                                                      
The integration appearing in the Gill-Pelaez inversion formulas for the
determination of P1 and P2 , can be performed in principle using the Adamchik-

Marichev algorithm (Adamchik and Marichev [2], Roach [44] and [45]) This
algorithm uses the known Meijer function (Meijer [41], Mathai and Saxena [40]).
The price F of a European call option is given by (3.1)

F = S ⋅ P1 − K ⋅ e − r (T − t ) P2
where K is the strike price and T is the maturity.

12. Model IX. The choice

h (τ) = k τ and h 2 (τ) = 1 + τ 2                                        (12.1)
converts equation (3.40) into the equation
 2τ                                       1
w ′′(τ) −        − [1 + ρ (1 + i φ)] ⋅ k τ  w ′(τ) − (φ 2 − i φ) (k τ) 2 w (τ) = 0   (12.2)
1+ τ 2
         4
with general solution expressed in terms of the Heun confluent function (see for
example Decarreau et al. [9], Forsyth [17], Heun [24], Maier [38] and [39],
Ronveaux [46], Slavyanov and Lay [51], Whittaker and Watson [53])

         k Σ    1         kζ 6−kζ     2
w (τ) = C1 HeunC                               
 2 , − 2 , − 2, − 8 , 8 , − τ  +

                                       

k Σ 1         kζ 6−kζ          

+ C 2 τ ⋅ HeunC    , , − 2, −
 2 2             ,    , − τ 2  ×

               8   8          

 k (ζ + Σ ) 2 
× exp  −
           τ                                                  (12.3)
      4       
where
ζ = 1 + ρ (1 + i φ) , η = φ 2 − i φ and Σ = ζ 2 + η                      (12.4)

79
The derivative of the function w (τ) is given by

 k ( Σ + ζ)           k Σ    1         kζ 6−kζ     2
w ′(τ) =  −
           τ  C1 HeunC 


 2 , − 2 , − 2, − 8 , 8 , − τ  +
     2                                               

k Σ 1         kζ 6−kζ          
       k (ζ + Σ ) 2 
+ C 2 τ ⋅ HeunC    , , − 2, −
 2 2             ,    , − τ 2  × exp −
                 τ +

               8   8                      4       


                   k Σ    1         kζ 6−kζ        
+ C1 ( − 2 τ) HeunC′ 
 2  , − , − 2, −   ,     , − τ2  +


                          2         8    8         
k Σ 1         kζ 6−kζ        
+ C 2 HeunC    , , − 2, −
 2 2             ,    , − τ2  +

               8   8         

k Σ 1         kζ 6−kζ         

+ C 2 ( − 2τ 2 ) HeunC′    , , − 2, −
 2 2             ,    , − τ 2  ×

               8   8          

 k (ζ + Σ ) 2 
× exp −
           τ 
      4       
The above expression can be written as
 k (ζ + Σ ) 2 
w ′( τ) = { C1 X (ζ , Σ ; τ) + C 2 Z(ζ , Σ ; τ)} × exp −
           τ    (12.5)
      4       
Where
 k ( Σ + ζ)         k Σ    1         kζ 6−kζ        
X (ζ , Σ ; τ) =  −
           τ  HeunC 
        2  , − , − 2, −   ,     , − τ2  +

     2                     2         8    8         
k Σ    1         kζ 6−kζ        
+ (− 2 τ) HeunC′ 
 2  , − , − 2, −   ,     , − τ2 
               (12.6)
       2         8    8         
and
 k ( Σ + ζ)    2         k Σ 1         kζ 6−kζ        
Z ( ζ , Σ ; τ) =  −

 τ ⋅ HeunC 
               , , − 2, −
 2 2             ,    , − τ2  +

     2                                  8   8         
k Σ 1         kζ 6−kζ        
+ C 2 HeunC    , , − 2, −
 2 2            ,     , − τ2  +

              8   8          

80
k Σ 1          kζ 6−kζ     2
+ (− 2 τ 2 ) HeunC′                             
 2 , 2 , − 2, − 8 , 8 , − τ  (12.7)
                            
Equation (3.36) has a solution given by (3.37)
2     w ′(τ)
D(τ; φ) =               ⋅        , h 2 (τ) = 1 + τ 2                      (12.8)
h 2 (τ) w (τ)
subject to the condition D(0 ; φ) = 0 .
We obtain
C1 X (ζ , Σ ;0) + C 2 Z(ζ, Σ ;0)
D(0; φ) = 2 ⋅
C1 + C 2 ⋅ 0
where
X(ζ, Σ ;0) = 0 and Z(ζ, Σ ;0) = 1
Therefore D(0 ; φ) = 0 if and only if C 2 = 0 . We thus obtain
2                               X(ζ, Σ ; τ)
D(τ; φ) =            ⋅
1 + τ2             k Σ    1         kζ 6−kζ     2
HeunC                               
 2 , − 2 , − 2, − 8 , 8 , − τ 
                              
or

2     k ( Σ + ζ) 
D(τ; φ) =         ⋅−          τ +
1 + τ2 
     2      

k Σ    1         kζ 6−kζ     2
(− 2 t ) HeunC′                               
 2 , − 2 , − 2, − 8 , 8 , − t 
+
2
⋅                                                (12.9)
1+ τ 2               k Σ     1         kζ 6−kζ    2
HeunC   2 , − 2 , − 2, − 8 , 8 , − t 

                               
We are now in a position to determine the function C(τ ; φ) given by equation
(3.42) by assigning values to the functions g 1 ( t ) , g 2 ( t ) , g( t ) and ε( t ) , taking into
account the compatibility conditions (3.28)-(3.30).
Let g 2 ( t ) = 0 . We then obtain from (3.30) that g( t ) = 0 . According to the assigned

k 2τ2
values to the functions h ( t ) and h 2 ( t ) , we get from (3.28) that h1 (τ) =              . We
1 + τ2

81
then obtain from (3.29) that g1 ( t ) = 0 . Therefore we have from (3.42) the
following expression for the function C(τ ; φ) :
τ
C(τ ; φ) = − ∫ ε(s) D(s; φ) ds − i r φ τ                              (12.10)
0

At this stage, given the form of the function D( τ; φ) expressed by (12.9), we have
to assign a function to ε( t ) , so that to obtain a closed form solution for the
function C( τ; φ) . We see that the choice

1 + τ2
ε(τ) =                                                                            (12.11)
2
τ
 k ( Σ + ζ) 
C ( τ ; φ) = − ∫  −
           s  ds − i r φ τ −

0      2       

k Σ    1         kζ 6−kζ        2
τ
(− 2 s) HeunC′ 
 2 , − 2 , − 2, − 8 , 8 , − s 

−∫                                                       ds
k Σ     1         kζ 6−kζ         
0         HeunC  2   , − , − 2, −    ,     , − s2 

        2          8     8        
where the integrals can be evaluated explicitly.
We thus arrive at
k ( Σ + ζ) 2
C( τ ; φ) =              τ −irφτ −
4
k Σ    1         kζ 6−kζ     2
− ln HeunC                               
 2 , − 2 , − 2, − 8 , 8 , − τ                          (12.12)
                              
Using equations (3.57), (3.54) and (3.59), under the same choice of functions h ( t ) ,
h 2 ( t ) , g 2 ( t ) , δ( t ) and ε( t ) , we can determine the functions H (τ ; φ) and G (τ ; φ) .
We obtain

2  k ( ∆ + θ) 
H(τ; φ) =        ⋅−
2 
τ +

1+ τ       2      

82
k ∆    1         kθ 6−kθ        2
(− 2 τ) HeunC′ 
 2 , − 2 , − 2, − 8 , 8 , − τ 

+
2
⋅                                                                       (12.13)
1+ τ 2              k ∆     1         kθ 6−kθ         
HeunC  2   , − , − 2, −    ,     , − τ2 

        2          8     8        
We see that the choice

1 + τ2
ε(τ) =
2
τ
 k ( ∆ + θ) 
G (τ ; φ) = − ∫  −
           s  ds − i r φ τ −

0 
2       

k ∆    1         kθ 6−kθ        
τ
(− 2 s) HeunC′ 
 2  , − , − 2, −   ,     , − s2 

       2         8   8           ds
−∫
k ∆     1        kθ 6−kθ      2
0        HeunC  2 , − 2 , − 2, − 8 , 8 , − s 

                                
where the integrals can be evaluated explicitly. We thus arrive at
k ( ∆ + θ) 2
G ( τ ; φ) =             τ −irφτ −
4
k ∆    1         kθ 6−kθ        
− ln HeunC 
 2  , − , − 2, −   ,     , − τ2 
                          (12.14)
       2         8   8          
We are now in a position to determine the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ)
using (3.11) and (3.43) respectively:

f1 ( x , Y , t ; φ ) = e C ( t , T ; φ ) + D ( t , T ; φ ) Y + i φ x

f 2 ( x , Y , t ; φ) = e G ( t , T ; φ ) + H ( t , T ; φ ) Y + i φ x
Using the previously determined quantities, we find the following expressions for
the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ) :

83
                   k ( Σ + ζ) 2 
exp  i φ x − i r φ τ +
                             τ 
                       4        
f 1 ( x , Y , t ; φ) =                                                   ×
k Σ         1           kζ 6−kζ     2
HeunC 
 2 , − 2 , − 2, − 8 , 8 , − τ 

                                     

                             d            k Σ       1        kζ 6−kζ         
                                HeunC           , − , − 2, −    ,    , − τ2   
 2 Y  k ( Σ + ζ)            dτ             2        2        8    8          
× exp                 −           τ+                                                   
2                                                                      
1+ τ            2                       k Σ       1        kζ 6−kζ      2  
HeunC                                       
 2 , − 2 , − 2, − 8 , 8 , − τ   
          
                                                                           
                    k ( ∆ + θ) 2 
exp  i φ x − i r φ τ +
                                τ 

                        4         
f 2 ( x , Y , t ; φ) =                                                      ×
k ∆          1           kθ 6−kθ         2
HeunC 
 2 , − 2 , − 2, − 8 , 8 , − τ 

                                          

                         d        k ∆       1        kθ 6−kθ           
                            HeunC       , − , − 2, −   ,       , − τ2   
 2Y  k ( ∆ + θ)          dτ        2        2        8     8           
× exp        ⋅−            τ+                                               

1+ τ 
2         2                 k ∆       1        kθ 6−kθ        2  
HeunC 
 2 , − 2 , − 2, − 8 , 8 , − τ   

        
                                                                     
The risk-free probabilities P1 and P2 can then be evaluated using the Fourier
inversion formulas (3.9) and (3.10) respectively.
We find, using the previously determined quantities f1 and f 2
∞
1 1           R 1 ( x , Y, K , t; φ) 
P1 ( x , Y, t , ln K ) = +      ∫ Re                          dφ
2 π     0               iφ           
∞
1 1            R 2 ( x , Y , K , t ; φ) 
P2 ( x , Y, t , ln K ) = +      ∫ Re                             dφ
2 π     0                iφ             
where R 1 ( x , Y, K , t; φ) and R 2 ( x , Y, K, t; φ) are given by

                              k ( Σ + ζ) 2 
exp  i φ ( x − ln K ) − i r φ τ +
                                        τ 
                                  4         ×
R 1 ( x , Y , t ; φ) =
k Σ        1            kζ 6−kζ         2
HeunC    2 , − 2 , − 2, − 8 , 8 , − τ 

                                         

84
                                 d            k Σ         1         kζ 6−kζ         
                                   HeunC              , − , − 2, −     ,    , − τ2   
 2 Y  k ( Σ + ζ)                dτ             2          2          8   8          
× exp                −             τ+                                                      
2                                                                           
1+ τ              2                         k Σ         1         kζ 6−kζ      2  
HeunC     2 , − 2 , − 2, − 8 , 8 , − τ   

            
                                                                                  
                               k ( ∆ + θ) 2 
exp  i φ ( x − ln K ) − i r φ τ +
                                            τ 
                                     4          ×
R 2 ( x , Y , t ; φ) =
k ∆         1           kθ 6−kθ              
HeunC    2      , − , − 2, −         ,        , − τ2 

            2             8       8          

                         d         k ∆       1        kθ 6−kθ          
                           HeunC         , − , − 2, −   ,      , − τ2   
 2 Y  k ( ∆ + θ)        dτ          2        2        8     8          
× exp       ⋅−            τ+                                                

1+ τ 
2         2                   k ∆       1        kθ 6−kθ       2  
HeunC  2 , − 2 , − 2, − 8 , 8 , − τ   

       
                                                                     
The integration appearing in the Gill-Pelaez inversion formulas for the
determination of P1 and P2 , can be performed in principle using the Adamchik-

Marichev algorithm (Adamchik and Marichev [2], Roach [44] and [45]). This
algorithm uses the known Meijer function (Meijer [41], Mathai and Saxena [40]).
The price F of a European call option is given by (3.1)

F = S ⋅ P1 − K ⋅ e − r (T − t ) P2
where K is the strike price and T is the maturity.

13. Model X. The choice
h (τ) = k τ and h 2 (τ) = a + b τ 2 , a b ≠ 0                                 (13.1)
converts equation (3.37) into the equation
 2b τ                                          1
w ′′( τ) −           − [ 1 + ρ (1 + i φ)] ⋅ k τ  w ′( τ) − (φ 2 − i φ) ( k τ) 2 w ( τ) = 0   (13.2)
 a + b τ2                                      4
                                     
with general solution expressed in terms of the Heun confluent function (see for
example Decarreau et al. [9], Forsyth [17], Heun [24], Maier [38] and [39],
Ronveaux [46], Slavyanov and Lay [51], Whittaker and Watson [53])

85

         ak Σ    1         a kζ 6b −a kζ b 2 
w ( τ) = C1 HeunC 
 2b , − 2 , − 2, − 8 b ,       ,− τ +

                                   8b     a 

ak Σ 1         a k ζ 6 b − a k ζ b 2 
+ C 2 τ ⋅ HeunC     , , − 2, −
 2b 2               ,           , − τ  ×
                8b       8b        a 


 k ( Σ + ζ) 2 
× exp −
           τ                                             (13.3)
     4        
where
ζ = 1 + ρ (1 + i φ) , η = φ 2 − i φ and Σ = ζ 2 + η              (13.4)
The derivative of the function w (τ) is given by

 k ( Σ + ζ)            ak Σ       1        a k ζ 6b − a k ζ b 2 
w ′(τ) =  −
           τ  C1 HeunC 
            2b    , − , − 2, −      ,          ,− τ +
      2                            2         8b       8b      a 
ak Σ 1           a k ζ 6b − a k ζ b 2          k (ζ + Σ ) 2 
+ C 2 τ ⋅ HeunC 
 2b 2 , , − 2, −      ,           , − τ  × exp −
                τ +

                  8b      8b         a                4      
  2b 
                  ak Σ    1         a k ζ 6b − a k ζ b 2 
+ C1  − τ  HeunC′ 
 2b  , − , − 2, −      ,          ,− τ +
  a 
                          2          8b      8b       a 
ak Σ 1         a k ζ 6b − a k ζ    b 
+ C 2 HeunC    , , − 2, −
 2b 2               ,           , − τ2  +
                8b      8b         a  

 2b 2        ak Σ 1         a k ζ 6b − a k ζ    b   
+ C2  − τ  HeunC′     , , − 2, −
 2b 2               ,           , − τ 2  ×
 a                           8b      8b         a  

 k ( Σ + ζ) 2 
× exp −
           τ 
     4        
The above expression can be written as
 k ( Σ + ζ) 2 
w ′( t ) = { C1 X (ζ, Σ ; τ) + C 2 Z(ζ, Σ ; τ)} × exp −
           τ    (13.5)
     4        
where

86
 k ( Σ + ζ)         ak Σ    1         a k ζ 6b − a k ζ    b 
X (ζ , Σ ; τ) =  −
           τ  HeunC 
        2b  , − , − 2, −      ,           , − τ2  +
     2                      2          8b      8b         a  

 2b          ak Σ     1         a k ζ 6b − a k ζ    b 
+−  τ  HeunC′ 
 2 b , − 2 , − 2, − 8 b ,           , − τ2             (13.6)
 a                                       8b         a  
and
 k ( Σ + ζ) 2           ak Σ 1            a k ζ 6b − a k ζ    b 
Z ( ζ , Σ ; τ) =  −
           τ  ⋅ HeunC 
          2b 2  , , − 2, −      ,           , − τ2  +
      2                                    8b      8b         a  
ak Σ 1           a k ζ 6b − a k ζ    b 
+ HeunC   2 b , 2 , − 2, − 8 b ,           , − τ2  +
                          8b         a   

 2b 2        ak Σ 1         a k ζ 6b − a k ζ b 2 
+−  τ  HeunC′     , , − 2, −
 2b 2               ,          ,− τ                   (13.7)
 a                           8b      8b       a 
Equation (3.36) has a solution given by (3.37)
2     w ′(τ)
D(τ; φ) =            ⋅        , h 2 (τ) = a + b τ 2           (13.8)
h 2 (τ) w (τ)
subject to the condition D(0 ; φ) = 0 . We obtain

C1 X (ζ , Σ ;0) + C 2 Z(ζ, Σ ;0)
D(0; φ) = 2 ⋅
C1 + C 2 ⋅ 0
where
X(ζ, Σ ;0) = 0 and Z(ζ, Σ ;0) = 1
Therefore D(0 ; φ) = 0 if and only if C 2 = 0 . We thus obtain
2                                  X(ζ, Σ ; τ)
D(τ; φ) =              ⋅
a + b τ2             ak Σ    1         a kζ 6b −a kζ b 2 
HeunC 
 2b  , − , − 2, −     ,        ,− τ 
        2          8b     8b     a 
or

2   k ( Σ + ζ) 
D(τ; φ) =          ⋅−
2 
τ +

a + bτ       2      

87
 2b          ak Σ     1         a k ζ 6b − a k ζ b 2 
 − τ  HeunC′ 
 2 b , − 2 , − 2, − 8 b ,          ,− τ 
2        a                                        8b      a 
+          ⋅                                                                              (13.9)
a + bτ 2             ak Σ     1         a k ζ 6b − a k ζ b 2 
HeunC 
 2b   , − , − 2, −       ,         ,− τ 
         2          8b       8b        a 
We are now in a position to determine the function C(τ ; φ) given by equation
(3.42) by assigning values to the functions g 1 ( t ) , g 2 ( t ) , g( t ) and ε( t ) , taking into
account the compatibility conditions (3.28)-(3.30).
Let g 2 ( t ) = 0 . We then obtain from (3.30) that g( t ) = 0 . According to the assigned

k 2τ2
values to the functions h ( t ) and h 2 ( t ) , we get from (3.28) that h1 ( τ) =                .
a + b τ2
We then obtain from (3.29) that g1 ( t ) = 0 . Therefore we have from (3.42) the
following expression for the function C(τ ; φ) :
τ
C( τ ; φ) = − ∫ ε(s) D(s; φ) ds − i r φ τ                                (13.10)
0

At this stage, given the form of the function D( τ; φ) expressed by (13.9), we have
to assign a function to ε( t ) , so that to obtain a closed form solution for the
function C( τ; φ) . We see that the choice

a + b τ2
ε(τ) =                                                                            (13.11)
2
τ
 k ( Σ + ζ) 
C ( τ ; φ) = − ∫  −
           s  ds − i r φ τ

0 
2       

 2b          ak Σ     1         a kζ 6b −a kζ b 2 
 − s  HeunC′ 
 2b   , − , − 2, −     ,        ,− s 
τ
 a                    2          8b     8b     a  ds
−∫
ak Σ     1         a kζ 6b−a kζ b 2 
0        HeunC 
 2b   , − , − 2, −      ,        ,− s 
         2          8b      8b      a 
where the integrals can be evaluated explicitly.

88
We thus arrive at
k ( Σ + ζ) 2
C( τ ; φ) =               τ −irφτ −
4
ak Σ    1         a kζ 6b −a kζ b 2 
− ln HeunC 
 2b  , − , − 2, −     ,        ,− τ                                  (13.12)
        2          8b     8b     a 
Using equations (3.57), (3.54) and (3.59), under the same choice of functions h ( t ) ,
h 2 ( t ) , g 2 ( t ) , δ( t ) and ε( t ) , we can determine the functions H (τ ; φ) and G (τ ; φ) .
We obtain

2  k ( ∆ + θ) 
H(τ; φ) =           ⋅−          τ +
a + b τ2 
     2      


 2b          ak ∆      1        a kθ 6b −a kθ b 2 
 − τ  HeunC′ 
 2b   , − , − 2, −     ,        ,− τ 
2        a                     2         8b     8b     a  (13.13)
+          ⋅
a + bτ 2             ak ∆     1         a kθ 6b −a k θ b 2 
HeunC 
 2 b , − 2 , − 2, − 8 b ,        ,− τ 
                            8b      a 
We see that the choice

a + bτ 2
ε(τ) =
2
τ
 k ( ∆ + θ) 
G ( τ ; φ) = − ∫  −
           s  ds − i r φ τ −

0 
2       

 2b          ak ∆     1         a k θ 6b − a k θ b 2 
 − τ  HeunC′ 
 2b   , − , − 2, −      ,          ,− s 
τ
  a                   2          8b      8b       a  ds
−∫
ak ∆     1         a k θ 6b − a k θ b 2 
0          HeunC 
 2b   , − , − 2, −      ,          ,− s 
         2          8b       8b        a 
where the integrals can be evaluated explicitly. We thus arrive at
k ( ∆ + θ) 2
G ( τ ; φ) =             τ −irφτ −
4

89
ak ∆     1         a kθ 6b −a kθ b 2 
− ln HeunC 
 2 b , − 2 , − 2, − 8 b ,       ,− τ                      (13.14)
                           8b     a 
We are now in a position to determine the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ)
using (3.11) and (3.43) respectively:

f1 ( x , Y , t ; φ ) = e C ( t , T ; φ ) + D ( t , T ; φ ) Y + i φ x

f 2 ( x , Y , t ; φ) = e G ( t , T ; φ ) + H ( t , T ; φ ) Y + i φ x
Using the previously determined quantities, we find the following expressions for
the functions f1 ( x , Y, t; φ) and f 2 ( x , Y, t; φ) :

                   k ( Σ + ζ) 2 
exp  i φ x − i r φ τ +
                             τ 
                       4        
f1 ( x , Y , t ; φ ) =                                                             ×
ak Σ         1           a kζ 6b −a kζ    b 2
HeunC 
 2 b , − 2 , − 2, − 8 b ,               ,− τ 
                                  8b      a 

                              d            ak Σ        1        a kζ 6b −a kζ b 2  
                                 HeunC             , − , − 2, −     ,        , − τ  
 2Y  k ( Σ + ζ )              dτ            2b         2         8b     8b      a  
× exp                   −          τ+                                                         
2                                                                           
 a + bτ           2                      ak Σ        1        a kζ 6b −a kζ b 2   
HeunC     2 b , − 2 , − 2, − 8 b ,          ,− τ 



                                                           8b      a  
 
                    k ( ∆ + θ) 2 
exp  i φ x − i r φ τ +
                               τ 

                        4        
f 2 ( x , Y , t ; φ) =                                                             ×
ak ∆        1            a kθ 6b −a kθ b 2       
HeunC 
 2 b , − 2 , − 2, − 8 b ,                ,− τ 
                                   8b        a  

                          d         ak ∆       1        a kθ 6b −a kθ b 2  
                             HeunC         , − , − 2, −     ,         , − τ  
 2Y  k ( ∆ + θ)           dτ         2b        2         8b     8b       a  
× exp         ⋅ −             τ+                                                   
2                                                                       
 a + bτ          2                ak ∆       1        a kθ 6b − a kθ b 2   
HeunC  2 b , − 2 , − 2, − 8 b ,          ,− τ 



                                                      8b       a  
 
The risk-free probabilities P1 and P2 can then be evaluated using the Fourier
inversion formulas (3.9) and (3.10) respectively.
We find, using the previously determined quantities f1 and f 2

90
∞
1 1              R 1 ( x , Y, K , t; φ) 
P1 ( x , Y, t , ln K ) = +       ∫ Re                            dφ
2 π      0                 iφ           
∞
1 1            R 2 ( x , Y , K , t ; φ) 
P2 ( x , Y, t , ln K ) =    +     ∫ Re                              dφ
2 π    0                 iφ             
where R 1 ( x , Y, K , t; φ) and R 2 ( x , Y, K, t; φ) are given by

                              k ( Σ + ζ) 2 
exp  i φ ( x − ln K ) − i r φ τ +
                                        τ 
                                  4        
R 1 ( x , Y , t ; φ) =                                                              ×
ak Σ           1           a kζ 6b −a kζ       b 2
HeunC 
 2 b , − 2 , − 2, − 8 b ,                  ,− τ 
                                     8b        a 

                                  d            ak Σ         1        a kζ 6b −a kζ b 2  
                                     HeunC              , − , − 2, −      ,        , − τ  
 2Y  k ( Σ + ζ )                  dτ            2b          2          8b      8b     a  
× exp                  −            τ+                                                           
2                                                                               
 a + bτ             2                       ak Σ          1        a kζ 6b −a kζ b 2   
HeunC    2 b , − 2 , − 2, − 8 b ,             ,− τ 



                                                               8b      a  
 
                              k ( ∆ + θ) 2 
exp  i φ ( x − ln K ) − i r φ τ +
                                         τ  
                                    4        
R 2 ( x , Y , t ; φ) =                                                                 ×
ak ∆          1           a kθ 6b − a kθ b 2 
HeunC 
 2 b , − 2 , − 2, − 8 b ,                   ,− τ 
                                      8b         a  

                          d          ak ∆       1        a kθ 6b −a kθ b 2  
                             HeunC          , − , − 2, −     ,        , − τ  
 2Y  k ( ∆ + θ)           dτ          2b        2         8b     8b      a  
× exp         ⋅ −            τ+                                                    
2                                                                       
 a + bτ        2                   ak ∆       1        a kθ 6b − a kθ b 2   
HeunC   2 b , − 2 , − 2, − 8 b ,         ,− τ 



                                                      8b       a  
 
The integration appearing in the Gill-Pelaez inversion formulas for the
determination of P1 and P2 , can be performed in principle using the Adamchik-

Marichev algorithm (Adamchik and Marichev [2], Roach [44] and [45]). This
algorithm uses the known Meijer function (Meijer [41], Mathai and Saxena [40]).
The price F of a European call option is given by (3.1)

F = S ⋅ P1 − K ⋅ e − r (T − t ) P2
where K is the strike price and T is the maturity.

91
14. Summary of the proposed method.
We have introduced a methodology of building stochastic volatility models with
closed-form solutions. The valuation of options is being conducted according to
the following steps:
Step 1. We derive the option pricing PDE (2.18),

∂F 1 2       ∂ 2F 1 2 ∂ 2F                ∂2 F
+ f (Y) S2     + b       + ρ f (Y) S b        −rF+
∂t 2         ∂ S2 2 ∂ Υ 2                ∂ S ∂Y

∂F                ∂F
+ rS       + (a − b Λ ) ⋅    =0
∂S                ∂Y
under SV, i.e. volatility σ t is a function of an underlying stochastic process Yt ,
σ t = f (Yt ) where f (⋅) is a smooth function.
Step 2. We suppose that the pricing function F of a European call option is
expressed through the Heston-Bakshi-Madan (HBM) formula

F = S ⋅ P1 − K ⋅ e − r (T − t ) P2
where the probability functions P1 and P2 are expressed by the Gil-Pelaez

formula (3.9) and (3.10)
∞        e −i φ ln K f1 ( x , Y, t; φ) 
1 1
P1 ( x , Y, t , ln K ) = +                 ∫   Re                                 dφ
2 π                0      
              iφ               

∞       e −i φ ln K f 2 ( x , Y, t; φ) 
1 1
P2 ( x , Y, t , ln K ) =         +         ∫   Re                                  dφ
2 π        0      
              iφ                

Step 3. We prove that the characteristic functions f 1 and f 2 satisfy the partial

differential equations (3.7) and (3.8) respectively. Under the assumption that they
admit the exponential representations (3.11) and (3.43) respectively, i.e.

f1 ( x , Y , t ; φ ) = e C ( t , T ; φ ) + D ( t , T ; φ ) Y + i φ x

f 2 ( x , Y , t ; φ) = e G ( t , T ; φ ) + H ( t , T ; φ ) Y + i φ x

92
we proved that the various functions entering to that representations, satisfy the
equations (3.37), (3.40) and (3.42) for the D( t , T; φ) and C( t , T; φ) ,
2     w ′(τ)
D=          ⋅
h 2 (τ) w (τ)

 h ′ ( τ)                                       1
w ′′(τ) −  2
 h ( τ)   − [ 1 + ρ (1 + i φ)] ⋅ h (τ)  w ′(τ) − (φ 2 − i φ) h 2 (τ) w (τ) = 0

 2                                              4
τ
1
C( τ ; φ) = − ∫ g 2 (s) D 2 (s; φ) ds −
20
τ
− ∫ [ (ρ + i φ ρ) g (s) + ε(s)] D(s; φ) ds +
0

τ
1
+ (φ 2 − i φ) ∫ g1 (s) ds − i r φ τ
2            0

and equations (3.54), (3.57) and (3.59) for the H( t , T; φ) and G ( t , T; φ)
respectively.
2     z ′(τ)
H(τ) =          ⋅
h 2 (τ) z (τ)

 h ′ ( τ)                                  1 2
 h ( τ) − ( 1 + i ρ φ) ⋅ h ( τ)  z ′( τ) − 4 (φ + i φ) h ( τ) z( τ) = 0
z ′′( τ) −  2                              
2
 2                              
τ
1
G (τ ; φ) = − ∫ g 2 (s) H 2 (s; φ) ds −
20
τ
− ∫ [ (i φ ρ) g (s) + ε(s)] H (s; φ) ds +
0

τ
1
+ (φ 2 + i φ) ∫ g1 (s) ds − i r φ τ
2            0

Step 4. In equations (3.40) and (3.57), enter the functions h 2 (τ) and h (τ) . By
assigning appropriate values to those functions we can solve explicitly these
equations. In this way we have introduced a number of models admitting closed-

93
form solutions. In equations (3.42) and (3.59) enter the functions g(τ) , g1 (τ) ,
g 2 (τ) and ε(τ) . These functions have been introduced in (3.24)-(3.27) and obey
the compatibility conditions (3.28)-(3.30). In the first Model, introduced in Section
4, we have assigned constant values to the functions h 2 (τ) and h (τ) , and we have
derived a model similar to Heston’s Model (Heston [23]). In the rest models
(Model II-ModelX), we have considered time dependent functions in place of
h 2 (τ) and h (τ) (see Table 1), and we have derived fairly complicated models
which, we hope, will be able to capture more detailed behavior of the movement
of the derivative price under stochastic volatility.
Model I         h 2 (τ) = ξ 2         h (τ) = ξ       ε(τ) = − ξ 2 / 4   h1 (τ) = 1

Model II        h 2 ( τ) = τ          h (τ) = τ       ε(τ) = τ           h1 (τ) = τ

Model III       h 2 (τ) = τ           h (τ) = τ       ε(τ) = τ           h1 ( τ) = τ

Model IV        h 2 ( τ) = τ 2        h (τ) = τ2      ε(τ) = τ 2         h1 (τ) = τ2

Model V         h 2 ( τ) = τ a        h (τ) = τ a     ε(τ) = τ a         h1 (τ) = τa

Model VI        h 2 (τ) = 1 / τ       h (τ) = 1 / τ   ε(τ) = 1           h1 (τ) = 1 / τ

Model VII       h 2 ( τ) = τ 2        h (τ) = τ       ε(τ) = τ 2         h1 (τ) = 1

Model VIII      h 2 ( τ) = τ b        h (τ) = k τ a   ε(τ) = τb          h1 (τ) = k 2 τ2a − b

Model IX        h 2 (τ) = 1 + τ 2     h (τ) = k τ              1 + τ2                 k 2τ2
ε(τ) =             h 1 ( τ) =
2                   1 + τ2
Model X         h 2 (τ) = a + b τ 2   h (τ) = k τ            a + bτ 2                  k 2 τ2
ε(τ) =             h1 (τ) =
2                     a + bτ 2

Table 1.
Values of the functions h 2 (τ) , h (τ) and ε(τ) of the various Models.
In all the above Models we have put g1 (τ) = g 2 (τ) = g (τ) = 0 and δ(τ) = h (τ) .

94
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