Lecture 7a Deriving Black-Scholes Formula

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					MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang                        1



          Lecture 7a: Deriving Black-Scholes Formula
   The price of a European call satisfies the following Black-
Scholes Equation with a final condition
               
                ∂c 1 2 2 ∂ 2c        ∂c
                    + σ S      2
                                 + rS    − rc = 0,                               (1)
                c(S, T2 = max(S − K, 0).
                 ∂t
                       )
                            ∂S        ∂S

The forward-price transformation

                                  c = c · er(T −t),           S = S · er(T −t)

coverts equation (1) to a new Black-Scholes problem without
interest rate r                       
                                       ∂c    1 2 2 ∂ 2c
                                            + σ S        2 = 0,                  (2)
                                              2
                                       c∂t , T ) = max(S − K, 0).
                                                     ∂S
                                          (S
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang                               2


The price-moneyness transformation
                                    1 σ 2 τ +x
                    S = Ke          2            ,        τ = T − t,   c = Kf (x, τ )

converts equation (2) into a standard heat equation with an
initial condition
                                         
                                          ∂f 1 2 ∂ 2f
                                               − σ        = 0,                          (3)
                                           ∂τ 2 ∂x2 x
                                          f (x, 0) = max(e − 1, 0).

   With the two-step transformation above, the Black-Scholes
option pricing problem has been converted to an initial value
problem of the standard heat equation.
   We also know that the Green’s function of the standard heat
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang                                     3


equation is
                                                                    1              (x−x0 )2
                                                                               −
                                       G(x, τ ; x0) = √                    e         2σ 2 τ
                                                                  2πσ 2τ
   The solution of the standard heat equation with an arbitrary
initial condition f (x, 0) can be written in terms of Green’s func-
tion
                                                      +∞
                                f (x, τ ) =                   f (x0, 0)G(x, τ ; x0)dx0.       (4)
                                                    −∞
To understand the methodology here, we may treat Green’s
function G(x, τ ; x0) as the price at τ , of a δ option that gives
payoff δ(x − x0). We decompose a European call into a portfolio
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang                                          4


of δ options
                                                       +∞
                                  f (x, 0) =                  f (x0, 0)δ(x − x0)dx0.
                                                     −∞

Then we price the European call by summing up the price of
all the δ options, that is equation (4).
   Then the solution of equation (3) is
                                          +∞
                f (x, τ ) =                    max(ex0 − 1, 0)G(x, τ ; x0)dx0
                                      −∞
                                       +∞
                              =                (ex0 − 1)G(x, τ ; x0)dx0
                                      0
                                          +∞                                  +∞
                                                x0
                              =                e G(x, τ ; x0)dx0 −                 G(x, τ ; x0)dx0
                                      0                                   0
                              = B − A,
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang                                                   5


where
                                 +∞                               +∞                        (x−x0 )2
                                                                            1           −
                  A=                  G(x, τ ; x0)dx0 =                √            e         2σ 2 τ   dx0.
                             0                                0            2πσ 2τ
                      x − x0               √
Changing variable y =    √ , dx0 = −σ τ dy, and when x0 =
                       σ τ
                               x
+∞, y = −∞; when x0 = 0, y = √ , then
                             σ τ
                         −∞              2 √
                               1      − y2
                A = −       √        e σ τ dy
                         x
                         √    2πσ 2τ
                        σ τ
                                                    x
                                                    √
                                                  1 − y2
                                                   σ τ
                                         =      √ e 2 dy
                                            −∞    2π
                                                x
                                         = N    √ .
                                               σ τ
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang                                                          6


                                           +∞                                   (x−x0 )2
                                                              1               −
                          B =                   e x0 √                    e       2σ 2 τ    dx0
                                       0                  2πσ 2τ
                                           +∞                             x2 −2xx0 −2σ 2 τ x0 +x2
                                                      1               −                         0
                                =               √                 e               2σ 2τ               dx0
                                       0            2πσ 2τ
                                           +∞                             (x+σ 2 τ −x0 )2 −σ 2 τ (2x+σ 2 τ )
                                                      1               −
                                =               √                 e                    2σ 2 τ                  dx0
                                       0            2πσ 2τ
                                                     +∞                                    (x+σ 2 τ −x0 )2
                                       x+ 1 σ 2 τ                         1            −
                                = e       2                   √                    e           2σ 2 τ        dx0.
                                                     0            2πσ 2τ
                      x + σ 2 τ − x0               1
Changing variable y =      √         , then dy = − √ dx0, dx0 =
                          σ τ                     σ τ
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang                                    7



  √                                                      x + σ 2τ
−σ τ dy. When x0 = +∞, y = −∞; when x0 = 0, y =             √ , then
                                                          σ τ
                             −∞                   √
                 x+ 1 σ 2 τ           1         2
                                             − y2
           B = e 2              2τ
                                   √        e (−σ τ )dx0
                            x+σ
                              √      2πσ 2τ
                             σ τ
                                                         x+σ 2 τ
                                                           √
                                       x+ 1 σ 2 τ     1 − y2
                                                          σ τ
                                = e       2         √ e 2 dy
                                               −∞     2π
                                    x+ 1 σ 2 τ   x + σ 2τ
                                = e 2 N             √     .
                                                  σ τ
   We conclude that the solution to problem (3) is
                                            x+ 1 σ 2 τ             x + σ 2τ        x
                         f (x, τ ) = e         2         N            √       −N   √     .   (5)
                                                                    σ τ            σ τ
   Given the solution of (3), we now recover the solution of (2)
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang                           8


by transforming the variables backward
                                                                     S  1 2
                                c = Kf (x, τ ),               x = ln   − σ τ,
                                                                     K 2
then
                                           x + σ 2τ
                                               x+ 1 σ 2 τ                       x
                   c (S , t) = Ke       N     √   2                  − KN       √
                                            σ τ                             σ τ
                             = S N (d1) − KN (d2),                                  (6)

where
                                                          S    1
                                           x + σ 2τ          + σ 2τ
                                                              ln
                                      d1 =    √ = K √2              ,
                                            σ τ             σ τ
                                                       S     1 2
                                             x      ln    − σ τ
                                      d2 = √ =         K √2      .
                                           σ τ           σ τ
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   Given the solution of (2), we may recover the solution of (1)

                   c = c e−r(T −t) = S e−r(T −t)N (d1) − Ke−r(T −t)N (d2),

We then have the Black-Scholes formula to price an European
call

                     c(S, t) = SN (d1) − Ke−rτ N (d2),     τ = T − t,        (7)
                                  S    1 2       S         1
                               ln    + σ τ    ln + r + σ 2 τ
                                  K √2          K          2
                         d1 =              =            √        ,
                                    σ τ                σ τ
                                  S    1 2       S          1 2
                               ln    − σ τ    ln + r − σ τ
                                  K √2          K           2
                         d2 =              =            √         .
                                    σ τ                σ τ
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang             10


Exercise: Solve the Black-Scholes equation for the price of a
digital call option
                                    ∂V 1 2 2 ∂ 2V         ∂V
                                        + σ S      2
                                                     + rS    − rV = 0
                                    ∂t 2        ∂S        ∂S
                                                1 S≥K
                                    V (S, T ) =            ≡ 1S≥K
                                                0 S<K
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang             11


Exercise: Solve the Black-Scholes equation for the price of a
digital share call option
                                    ∂V 1 2 2 ∂ 2V         ∂V
                                        + σ S      2
                                                     + rS    − rV = 0
                                    ∂t 2        ∂S        ∂S
                                                S S≥K
                                    V (S, T ) =            ≡ S · 1S≥K
                                                0 S<K
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang            12


Exercise: Solve the Black-Scholes equation for the price of a
digital spread
                                   ∂V 1 2 2 ∂ 2V         ∂V
                                       + σ S      2
                                                    + rS    − rV = 0
                                   ∂t 2 ∂S              ∂S
                                              0     0<S<K−
                                                1
                                   V (S, T ) = 2 K − < S < K +
                                                0 K + < S < +∞
                                              
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang               13


Exercise: Find the Green’s function of the Black-Scholes equa-
tion
                                    ∂V 1 2 2 ∂ 2V           ∂V
                                        + σ S        2
                                                       + rS    − rV = 0
                                    ∂t 2          ∂S        ∂S
                                    V (S, T ) = δ(S − K)
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang           14



        Lecture 7b: Understanding Green’s Function
   Given a normal random number xt,

                                                       dxt = σdBt,

where σ is a constant.
   Suppose at time t, we know xt, and we are interested in the
distribution of xT . The conditional probability density function
is denoted as p(xT , T |xt, t). From probability theory, the transi-
tion probability function p(xT , T |xt, t) satisfies the Fokker Planck
equation, which is a backward heat equation,
                   ∂p 1 2 ∂ 2p
                  
                        + σ           = 0,
                     ∂t 2 ∂x2       t
                                                                     (8)
                     p(xT , T |xt, t) = δ(xT − xt).
                  
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang                                   15



   The solution of equation (8) is
                                                                                (xT − xt)2
                                                                              −
                                                                  1               2
                          p(xT , T |xt, t) =                                 e 2σ (T − t) ,
                                                              2πσ 2(T − t)
which is Green’s function of the standard heat equation. It says
that the future uncertain value of xT is normally distributed
with mean xt and variance σ 2(T − t), then

                                                     E(xT |xt) = xt.


   New Concept: xt is a martingale if E(xT |xt) = xt.
MFIN 7003 Mathematical Techniques of Finance I: Jin E Zhang                                        16


   The general solution of a backward heat equation
                       
                        ∂f 1 2 ∂ 2f
                             + σ      2
                                        =0                                                        (9)
                        f (x, T2 = ∂x(x)
                          ∂t
                                ) f0
describes the price of an option written on xt. The solution can
be written as the expectation of future payoff
                                                              ∞
                    f (x, t) = E(f0(xT )) =                        f0(xT )p(xT , T |xt, t)dxT .
                                                              −∞