MSM 3G11/4G11 Exercise Sheet 4. Solutions
1. The Black Scholes equation is given by
∂V
1
∂ 2V
∂V
+ σ 2 S 2 2 + rS
− rV = 0.
∂t
2
∂S
∂S
We wish to change from independent variables (S, t) to (x, τ ) with S = Eex , t = T − σ2τ2 ,
2
2
∂x
∂τ
= 0, ∂S
= S1 , ∂τ
= − σ2 and ∂S
= 0.
i.e., x = log(S/E), τ = σ2 (T − t). It follows that ∂x
∂t
∂t
Using the chain rule
∂V ∂x ∂V ∂τ
1 ∂V
∂V
=
+
= − σ2
∂t
∂x ∂t
∂τ ∂t
2 ∂τ
and
∂V ∂x ∂V ∂τ
1 ∂V
∂V
=
+
=
.
∂S
∂x ∂S
∂τ ∂S
S ∂x
Hence,
∂ 2V
1 ∂V
1 ∂V
∂
1 ∂ 2V
=
−
=
+
.
∂S 2
∂S S ∂x
S 2 ∂x
S 2 ∂x2
This gives
1 ∂V
1 ∂V
1 2 2
1 ∂ 2V
1 2 ∂V
+ σ S − 2
+ 2 2 + rS
− rV = 0
− σ
2 ∂τ
2
S ∂x
S ∂x
S ∂x
or
∂V
∂V
∂ 2V
∂V
+
−
−k
+ kV = 0.
2
∂τ
∂x
∂x
∂x
If
(k − 1)x (k + 1)2 τ
V = Eu(x, τ ) exp −
−
= u(x, τ )F (x, τ ),
2
4
(k−1)x
(k+1)2 τ
with F = E exp − 2 − 4
, then
∂V
∂u
(k + 1)2
=
F+ −
uF ,
∂τ
∂τ
4
∂u
(k − 1)
∂V
=
F−
uF
∂x
∂x
2
and
∂ 2u
∂u
(k − 1)2
∂ 2V
=
F
−
(k
−
1)
F
+
uF.
∂x2
∂x2
∂x
4
Now divide through by F to give
∂u
(k + 1)2 (k − 1) (k − 1)2 k(k − 1)
∂u
∂ 2u
+u −
−
−
+
+ k + (1+(k −1)−k)− 2 = 0
∂τ
4
2
4
2
∂x
∂x
or
∂ 2u
∂u
=
.
∂τ
∂x2
1
2. If the pay-off function of an option is V (S, T ) = g(S) we can use the transformation
(k − 1)x (k + 1)2 τ
V (S, t) = Eu(x, τ ) exp −
−
2
4
to calculate V (S, T ) by noting τ = 0 when t = T :
(k − 1)x
.
g(S) = Eu(x, 0) exp −
2
Denoting u0 (x) = u(x, 0) we have
Hence,
g(S = Eex )
exp
u0 (x) =
E
1
(k − 1)x .
2
g (S = Ees )
u0 (s) =
exp
E
1
(k − 1)s .
2
(a) Vanilla call: g(S) = max(S − E, 0) and
u0 (s) = (es − 1, 0) exp
= max exp
1
(k
2
1
(k
2
− 1)s
+ 1)s − exp
1
(k
2
− 1)s , 0 .
(b) Vanilla put: g(S) = max(E − S, 0) and
1
1
(k − 1)s − exp
(k + 1)s , 0 .
u0 (s) = max exp
2
2
(c) Bullish vertical spread:
g(S) = max(S − E1 , 0) − max(S − E2 , 0)
Set E = E1 . Then,
1
1
u0 (s) = max exp
(k + 1)s − exp
(k − 1)s , 0
2
2
E2
1
1
(k + 1)s −
(k − 1)s , 0 .
exp
− max exp
2
E1
2
(d) Cash or nothing call: g(S) = BH(S − E) and
B
u0 (s) = H (Ees − E) exp
E
1
(k − 1)s .
2
(e) Supershare: g(S) = d1 (H(S − E) − H(S − E − d)) and
1
(H (Ees − E) − H (Ees − E − d)) exp
u0 (s) =
Ed
2
1
(k − 1)s .
2
(f) Asset or nothing call: g(S) = SH(S − E) and
s
s
u0 (s) = e H (Ee − E) exp
1
(k − 1)s .
2
3. This is an optional question. The full solution is available on pages 78-79 of the recommended text and involves tedious but elementary integration and algebraic manipulation.
R∞
4. The delta function can be defined by δ(x) = 0 if x 6= 0 and −∞ δ(x)dx = 1. Other
R∞
properties of the delta function are −∞ δ(x)F (x)dx = F (0) and if the solutions of the
equation h(x) = 0 are x = x1 , x2 , . . . , xk and h′ (xi ) 6= 0, then
Z
Here we have g(S) =
1
d
∞
δ(h(x))F (x)dx =
−∞
k
X
i=1
1
|h′ (x
i )|
F (xi ).
(1)
(H(S − E) − H(S − E − d)). Graphically we have
V
1/d
S
E
E+d
R∞
For all d it is clear that −∞ g(S)ds = 1. Also it is intuitive that in the limit as d → 0+ ,
g(S) = 0 except at S = E, where it is infinity. Hence, g(S) = δ(S − E).
5. We calculate the value of the supershare in the limit as d → 0 by writing g(S) = δ(S −E).
This gives
u0 (s) = E1 δ(S − E) exp 12 (k − 1)s
=
1
δ(E
E
(es − 1)) exp
1
(k
2
− 1)s .
Using the solution to the diffusion equation we have
Z ∞
1
1
1
1
s
2
δ(E (e − 1)) exp
(k − 1)s exp − (x − s) ds.
u(x, τ ) = √
2
4τ
2 τ π −∞ E
3
We use (1). The only root of E(es − 1) is s = 0. Also
u(x, τ ) =
d
(E(es
ds
− 1))|s=0 = E. Hence,
2
1
1
− x4τ
√
e
.
E2 2 τ π
6. As derived during lectures the appropriate form of the Black-Scholes equation for a European vanilla call option where the underlying S has a constant dividend yield D0 is
∂C
1 2 2 ∂ 2 C ∂C
σ S
+
− rC + (r − D0 )S
= 0.
2
2
∂S
∂t
∂S
(2)
Our goal is to use the substitution C(S, t) = e−D0 (T −t) C1 (S, t) to transform the above into
an equation which we already know the answer to. Note that at the exercise time t = T
we have
C(S, T ) = C1 (S, T ) = max(S − E, 0)
(i.e C and C1 both have the same pay-off function as a standard European vanilla call).
We now calculate
∂C
∂C1
= e−D0 (T −t)
,
∂S
∂S
2
∂ 2C
−D0 (T −t) ∂ C1
=
e
∂S 2
∂S 2
and
∂C1
∂C
= D0 e−D0 (T −t) C1 + e−D0 (T −t)
.
∂t
∂t
Substituting into equation (2) we obtain
∂C1 1 2 2 −D0 (T −t) ∂ 2 C1
+ σ S e
∂t
2
∂S 2
∂C1
− re−D0 (T −t) C1 = 0
+(r − D0 )Se−D0 (T −t)
∂S
D0 e−D0 (T −t) C1 + e−D0 (T −t)
Dividing through by e−D0 (T −t) and combining like terms gives
∂C1
1 2 2 ∂ 2 C1 ∂C1
σ S
+
− (r − D0 )C1 + (r − D0 )S
= 0.
2
2
∂S
∂t
∂S
or
1 2 2 ∂ 2 C1 ∂C1
∂C1
σ S
− r ∗ C1 + r ∗ S
= 0.
+
2
2
∂S
∂t
∂S
where r∗ = r − D0 . This equation must be solved subject to the pay-off function
C1 (S, T ) = max(S − E, 0). Note that is identically our standard Black Scholes equation for a European vanilla call except that r has been replaced by r∗ = r − D0 . Thus we
can immediately write
C1 = SN (d1 ) − E exp(−r∗ (T − t))N (d2 )
4
where
1
N (q) = √
2π
Z
q
1 2
exp − s ds,
2
−∞
2
ln(S/E) + r∗ + σ2 (T − t)
√
d1 =
σ T −t
and
σ2
∗
ln(S/E) + r − 2 (T − t)
√
d2 =
.
σ T −t
Our final answer is then
C(S, t) = e−D0 (T −t) C1
= e−D0 (T −t) (SN (d1 ) − E exp(−r∗ (T − t))N (d2 ))
7. Substitute numbers directly into the formula from question 6 to give r∗ = 0.1,
d1 =
ln(1.3) + (0.1 + 0.08)(4)
√
= 1.227959...,
0.4 4
d2 = 0.4279553306 and
C(S, t) = exp(−D0 (T − t))(SN (d1 ) − E exp(−r∗ (T − t))N (d2 )) = 0.4766944475.
8. (a) By arbitrage we have
and
−
−
S t+
d = S td − dy S td
(3)
+
−
+
(4)
Cd S t−
d , td = Cd S td , td
+
−
−
as re= Cd S t−
Combining (3) and (4) gives Cd S t−
d − dy S td , td
d , td
quired.
(b) Since there is no dividend payment between t+
d ≤ t ≤ T then there will be no
dividend jump condition and our lognormal random variable behaves in the usual
way. It follows that Cd (S, t) = C(S, t; E) for all t+
d ≤ t ≤ T (i.e., Cd (S, t) and
C(S, t; E) satisfies the same PDE and boundary conditions for t+
d ≤ t ≤ T ).
(c) This is just the jump condition from (a).
(d) The pay-off function for C(S − dy S, t; E) is simply
E
g(S) = max((S − dy S − E), 0) = (1 − dy ) max S −
,0
1 − dy
by factoring out the term 1 − dy .
5
(e) We now combine the above results to give
−
−
Cd S td , td = max S −
E
, 0 (1 − dy ).
1 − dy
We treat this as a new pay-off function for Cd evaluated at time t = t−
d , so for
−
0 ≤ t ≤ td we value Cd by determining the standard value of a call option with
E
strike price 1−d
and expiry date t−
d and then multiply by a factor 1 − dy .
y
9. If F = S(t) exp(r(T − t)) and V (S, t′ ) = S(t′ ) − F exp(−r(T − t′ )), then
V (S, t′ ) = S(t′ ) − S(t) exp(r(T − t)) exp(−r(T − t′ )) = S(t′ ) − S(t) exp(r(t′ − t)).
Thus
(i) V (S, t) = 0, so it costs nothing to enter a forward contract.
(ii) V (S, T ) = S(T ) − S(t) exp(r(T − t)) = S(T ) − F . The holder of the forward contact
is obliged to purchase S at price F . If S(T ) > F the holder makes a profit. If
S(T ) < F the holder makes a loss.
10. We begin by noting that the value of an option V (S, t) which only depends on the value
of the underlying stock S and time t satisfies the Black Scholes equation
∂V
∂V
1 2 2 ∂ 2V
σ S
+
− rV + rS
= 0.
2
2
∂S
∂t
∂S
Our goal is to transform the Black-Scholes equation for the value of an option V (S, t)
into a suitable form to determine the value of a call option C(F, t) which is a function of
the forward price F and time t. To answer this question we must use the chain rule for
a function of two variables. Suppose V (S, t) = C(F (S, t), τ (S, t)). We then use the chain
rule to give
∂V
∂C ∂F
∂C ∂τ
=
+
∂S
∂F ∂S
∂τ ∂S
and
∂C ∂F
∂C ∂τ
∂V
=
+
.
∂t
∂F ∂t
∂τ ∂t
In our case we have
F (S, t) = Ser(T −t)
and
τ (S, t) = t
which gives
∂F
= er(T −t)
∂S
∂F
= −rSer(T −t)
∂t
∂τ
=0
∂S
6
∂τ
= 1.
∂t
This gives
∂V
∂C
= er(T −t)
∂S
∂F
and
We now calculate
∂C ∂C
∂V
= −rSer(T −t)
+
.
∂t
∂F
∂τ
∂ ∂V
∂
∂ 2V
r(T −t) ∂C
=
=
e
∂S 2
∂S ∂S
∂S
∂F
2
2
∂ C ∂F
∂ 2 C ∂τ
2r(T −t) ∂ C
= er(T −t)
=
e
+
∂F 2 ∂S
∂F ∂τ ∂S
∂F 2
Substituting for all of the derivatives calculated using the chain rule, and using S =
F e−r(T −t) the Black-Scholes equation becomes
2
∂ 2C
∂C
1 2
σ F e−r(T −t) e2r(T −t) 2 + −rF e−r(T −t) er(T −t)
2
∂F
∂F
∂C
∂C
− rC + rF e−e(T −t) er(T −t)
= 0.
+
∂τ
∂F
Combining all like terms and noting that τ = t we arrive at the desired result
1 2 2 ∂ 2 C ∂C
σ F
+
− rC = 0.
2
∂F 2
∂t
Recall that the Black-Scholes equation with continuous dividends is
∂V
∂V
1 2 2 ∂ 2V
σ S
+
− rV + (r − D0 )S
= 0.
2
2
∂S
∂t
∂S
Thus our partial differential equation for a Euorpean Call option on a future with a
forward price F is identical to the Black-Scholes equations for an option on a stock S
which pays continuous dividends for the case when r = D0 (i.e. in the above replace V
with C, F with S and let r = D0 ). From question 6 we know how to determine the value
of a European Call option for a stock which pays continuous dividends. We then use this
result to determine the value of our call option on a futures contract to give
C(F, t) = e−r(T −t) (F N (d1 ) − EN (d2 ))
where
1
N (q) = √
2π
Z
q
1 2
exp − s ds,
2
−∞
2
ln(F/E) + σ2 (T − t)
√
d1 =
σ T −t
7
and
2
ln(F/E) − σ2 (T − t)
√
d2 =
.
σ T −t
8