# MATH 5010CONTINUOUS-TIME FINANCE ANSWER KEYS FOR HOMEWORK ASSIGNMENT 3 by sij18839

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```									                   MATH 5010 CONTINUOUS-TIME FINANCE
ANSWER KEYS FOR HOMEWORK ASSIGNMENT 3

(1) Note that a3 − b3 = (a − b)3 + 3b(a − b)2 + 3b2 (a − b). Substitute a = Wj+1 , b = Wj ,
3
Wj+1 − Wj3 = (Wj+1 − Wj )3 + 3Wj (Wj+1 − Wj )2 + 3Wj2 (Wj+1 − Wj ).

W (T )3 = W 3 (T ) − W 3 (0)
n−1
=              3
(Wj+1 − Wj3 )
j=0
n−1
=            {(Wj+1 − Wj )3 + 3Wj (Wj+1 − Wj )2 + 3Wj2 (Wj+1 − Wj )}.
j=0
The ﬁrst term
n−1                                  n−1
|         {(Wj+1 − Wj )3 }| ≤                  |(Wj+1 − Wj )|(Wj+1 − Wj )2
j=0                                  j=0
n−1
≤         (|Wj+1 − Wj |) · (tj+1 − tj )
j=0
n−1
≤    max |Wj+1 − Wj | ·                (tj+1 − tj )
0≤j≤n1
j=0
=    max |Wj+1 − Wj | · T,
0≤j≤n1
n−1                 2
goes to zero when the partition P → 0. The second term                                j=0 3Wj (Wj+1 −Wj )       converges
T
to   0    3W (t)dt as P → 0 by the deﬁnition. Note that the integral involved only use
n−1   2
the left end points of each interval. The third term                                  j=0 3Wj (Wj+1       − Wj ) converges
T
to   0    3W (t)2 dW (t) by the deﬁnition of Ito integral as P → 0. Therefore we obtain
T                       T
W (t)3 = 3             W (t)dt + 3             W 2 (t)dW (t).
0                       0
1
2                              MATH 5010 CONTINUOUS-TIME FINANACE

(2) (i) Applying the Ito-Doeblin formula for f (x) = ln x with x = S(t) and
1                    1
ft = 0, fx =               ,        fxx = −      ,
x                    x2

1
d ln S(t) = ft dt + fx dx + fxx dx · dx
2
1          1 1
= 0 · dt +      dS(t) −         dS(t) · dS(t)
S(t)         2 S(t)2
σ(t)2
= µ(t)dt + σ(t)dW (t) −         dt
2
1
= (µ(t) − σ 2 (t))dt + σ(t)dW (t).
2
Thus the Ito integral for the process is given by the deﬁnition,
t                         t
1
ln S(t) − ln S(0) =            d ln S(t) =               (µ(s) − σ 2 (s))ds + σ(s)dW (s).
0                         0              2
Therefore
t                                    t
1
S(t) = S(0) exp{                 (µ(s) − σ 2 (s))ds +                 σ(s)dW (s)}.
0              2                     0

(ii) Applying the Ito-Doeblin formula for f (x) = xp with x = S(t) and

ft = 0, fx = pxp−1 ,                   fxx = p(p − 1)xp−2 ,

1
dS(t)p = pS(t)p−1 dS(t) + p(p − 1)S(t)p−2 dS(t) · dS(t)
2
1
= pS(t) (µ(t)dt + σ(t)dW (t)) + p(p − 1)S(t)p σ 2 (t)dt
p
2
p(p − 1) 2
= S(t)p (pµ(t) +         σ (t))dt + S(t)p pσ(t)dW (t).
2
(3) (i) Applying the Ito-Doeblin formula for f (x) = eβt x with x = R(t) and

ft = βeβt x, fx = eβt ,                     fxx = 0,

1
d(eβt R(t)) = ft dt + fx dx + fxx dx · dx
2
βt
= βe R(t)dt + eβt dR(t)

= eβt (βR(t)dt + (α − βR(t))dt + σdW (t))

= eβt (αdt + σdW (t).
MATH 5010 CONTINUOUS-TIME FINANACE                                                   3

(ii) Based on (i), we have
t                       t                              t
eβt R(t) − R(0) =             d(eβt R(t)) =           eβs αds +                      eβs σdW (s).
0                       0                              0
t
α
R(t) = e−βt R(0) + e−βt eβs |t + σe−βt
0                                         eβs dW (s)
β                                       0
t
α
= e−βt R(0) + (1 − e−βt ) + σe−βt                                     eβs dW (s).
β                                               0
t βs
Note that   eβs   is a non-random function and            0 e dW (s)             is a normal distribution.
(4) (i) Note that
1   2
N (y) = √ e−y /2 .
2π
1   2
N (d− ) =          √ e−d− /2
2π
1 −(d+ −σ√τ )2 /2
=   √ e
2π
1 −d2 /2+d+ σ√τ −σ2 τ /2
=   √ e +
2π
√
τ −σ 2 τ /2
= N (d+ )ed+ σ
x
= N (d+ )eln K +rτ .
√
The ﬁrst equality follows from the basic fact, the second from d− = d+ − σ τ , the third
from algebra of (a − b)2 = a− 2ab + b2 and the fourth from the basic fact and the last equality
follows from the deﬁnition
x
d+ στ = ln        + rτ + σ 2 τ /2.
K
Now we have
x
Ke−rτ N (d− ) = Ke−rτ N (d+ )eln K +rτ = xN (d+ ).

(ii)
∂c
cx =
∂x
∂d+                 ∂d−
= N (d+ ) + xN (d+ )     − Ke−rτ N (d− )
∂x                  ∂x
∂d+            ∂d−
= N (d+ ) + xN (d+ )     − xN (d+ )
∂x             ∂x
= N (d+ ).
4                            MATH 5010 CONTINUOUS-TIME FINANACE

The ﬁrst identity follows from the product rule and chain rule for the partial derivative with
∂d−       ∂d+
respect to x, the second from (i) and the last from the identity    ∂x   =    ∂x   by the deﬁnition
of d± .
(iii)
∂d+                                    ∂d−
ct = xN (d+ )         − Ke−rτ rN (d− ) − Ke−rτ N (d− )
∂t                                     ∂t
∂d−               σ                                 ∂d−
= xN (d+ )       − xN (d+ ) √ − Ke−rτ rN (d− ) − Ke−rτ N (d− )
∂t             2 τ                                  ∂t
σ       −rτ
= −xN (d+ ) √ − Ke rN (d− )
2 τ
σx
= −rKe−rτ N (d− ) − √ N (d+ ).
2 τ
The ﬁrst equality follows from chain rule and product rule, the second from
∂d−   ∂d+   σ
=     + √ ,
∂t    ∂t  2 τ
the third and the fourth from (i).
(iv) Using (ii), we have
∂cx           ∂d+     1
cxx =       = N (d+ )     =   √ N (d+ ).
∂x             ∂x   σx τ
By applying results from (ii) and (iii), the left hand side of the Black-Scholes equation
ct + rxcx + 1 σ 2 x2 cxx equals to
2
σx                      1        1
= {−rKe−rτ N (d− ) − √ N (d+ )} + rxN (d+ ) + σ 2 x2 √ N (d+ )
2 τ                      2      σx τ
−rτ
= −rKe N (d− ) + rxN (d+ )

= r(xN (d+ ) − Ke−rτ N (d− ))

= rc(t, x)

where the ﬁrst equality follows from plugging the results previously, the second from basic
algebra and the last from the deﬁnition of c(t, x).
x
(v) Since σ > 0, ln K > 0 if x > K. One has
ln x     r + σ 2 /2 √               ln x
lim d+ = lim ( √ K +               · T − t) = lim √ K = +∞.
t→T −    t→T − σ T − t       σ                t→T − σ T − t
√
lim d− = lim d+ − lim σ T − t = +∞.
t→T −         t→T −   t→T −
MATH 5010 CONTINUOUS-TIME FINANACE                          5

lim c(t, x) =     lim xN (d+ ) − Ke−r(T −t) N (d− )
t→T −             t→T −
= xN (+∞) − KN (+∞)

= x − K,

where the ﬁrst is from deﬁnition, the second from the above discussion and the last from
N (+∞) = 1.
Similarly, if x < K, then

ln x   r + σ 2 /2 √               ln x
lim d+ = lim ( √ K +            · T − t) = lim √ K = −∞,
t→T −    t→T − σ T − t    σ                t→T − σ T − t

x
due to ln K < 0. limt→T − d− = −∞. Note that N (−∞) = 0. So limt→T − c(t, x) =
xN (−∞) − KN (−∞) = 0. Therefore we get limt→T − c(t, x) = (x − K)+ by the above two
cases.
(vi) For 0 ≤ t < T , limx→0+ d+ = −∞, limx→0+ d− = −∞ since limx→0+ ln x = −∞.

lim c(t, x) = lim xN (d+ ) − Ke−rτ N (d− ) = 0 − Ke−rτ N (−∞) = 0.
x→0+          x→0+

(vii) Since limx→∞ ln x = +∞, limx→∞ d+ = +∞, limx→∞ d− = +∞.

lim [c(t, x) − x + Ke−rτ ] = lim [x(N (d+ ) − 1) + Ke−rτ (1 − N (d− ))].
x→∞                         x→∞

Note that limx→∞ [Ke−rτ (1−N (d− ))] = Ke−rτ (1−N (+∞)) = 0. The limit limx→∞ [x(N (d+ )−
1)] is ∞ · 0 type. We have to use the L’Hospital rule.

N (d+ ) − 1
lim [x(N (d+ ) − 1)] =      lim
x→∞                         x→∞    1/x
x
N (d+ ) σ√τ
=     lim
x→∞     −1/x2
1          x
= − √       lim d2 /2
σ 2πτ x→∞ e +
2
1        eστ d+ −τ (r+σ /2)
= − √       lim         2
σ 2πτ x→∞        ed+ /2
e−rτ                  2
= − √       lim e−(d+ −στ ) /2
σ 2πτ x→∞
= 0.
6                                MATH 5010 CONTINUOUS-TIME FINANACE

(5) (i) Let Y (t) = −θW (t) − (r + θ2 /2)t. Then dY (t) = −(r + θ2 /2)dt − θdW (t) and ζ(t) = eY (t) .
Then
1
dζ(t) = eY (t) dY (t) + eY (t) dY (t) · dY (t)
2
1
= ζ(t)(−(r + θ2 /2)dt − θdW (t) + θ2 dt)
2
= ζ(t)(−rdt − θdW (t)).

(ii)
dX(t) = rX(t)dt + ∆(t)(α − r)S(t)dt + ∆(t)σS(t)dW (t).

d(ζ(t)X(t)) = dζ(t) · X(t) + ζ(t) · dX(t) + dζ(t) · dX(t)

= ζ(t)(−rdt − θdW (t))X(t) + ζ(t) · (rX(t)dt + ∆(t)(α − r)S(t)dt + ∆(t)σS(t)dW (t))

−θζ(t)∆(t)σS(t)dt

= −θζ(t)X(t)dW (t) + σζ(t)S(t)∆(t)dW (t).

Hence ζ(t)X(t) is a martingale by the Ito process property.
(iii) Now we have the strategy X(t) satisﬁes X(T ) = V (T ) and the martingale property
gives
ζ(0)X(0) = E[ζ(T )X(T )] = E[ζ(T )V (T )|F0 ]

Hence the result follows from ζ(0) = 1.

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