Financial Modeling I
February 16, 2015 – I part only
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Oral exam choice:
Time to complete: 1.5 hours
t
Z
February 18
February 23
Ws dWs . Compute:
1. Let Wt be a Wiener process and Xt = sin
0
i) d(Xt2 ) ;
Solution: First of all, let Yt =
Rt
0
ii) d(tXt ) .
Ws dWs , i.e. dYt = 0 dt + Wt dWt .
i) Consider function F (t, y) = sin2 (y); then
∂F
=0 ,
∂t
∂2F
= 2[cos2 (y) − sin2 (y)] .
∂y 2
∂F
= 2 sin(y) cos(y) ,
∂y
Then Xt2 = sin2 (Yt ) and by Itô’s Lemma
1
d(Xt2 ) = dF (t, Yt ) = 0 + 2 sin(Yt ) cos(Yt ) × 0 + Wt2 × 2[cos2 (Yt ) − sin2 (Yt )] dt + 2 sin(Yt ) cos(Yt )Wt dWt
2
= [cos2 (Yt ) − sin2 (Yt )]Wt2 dt + 2 sin(Yt ) cos(Yt )Wt dWt .
Remembering that cos2 (Yt ) − sin2 (Yt ) = cos(2Yt ) and 2 sin(Yt ) cos(Yt ) = sin(2Yt ), we can write the stochastic
differential in the form
Z t
Z t
2
2
d(Xt ) = cos 2
Ws dWs Wt dt + sin 2
Ws dWs Wt dWt .
0
0
ii) Consider function G(t, y) = t sin(y); then
∂2F
= −t sin(y) .
∂y 2
∂F
= t cos(y) ,
∂y
∂F
= sin(y) ,
∂t
Then tXt = G(t, Yt ) and by Itô’s Lemma
1 2
d(tXt ) = dG(t, Yt ) = sin(Yt ) + t cos(Yt ) × 0 − t sin(Yt ) × Wt dt + t cos(Yt )Wt dWt
2
Z t
Z t
t 2
= sin
Ws dWs tWt dWt .
Ws dWs
1 − Wt dt + cos
2
0
0
t
Z
2. Let Wt be a Wiener process and Xt =
Ws dWs . Compute:
0
i) d(Wt2 Xt2 ) ;
ii) d (exp(Xt )) .
Solution: First of all, dXt = 0 dt + Wt dWt ; since trivially dWt = 0 dt + 1 dWt , we consider the vector process
Wt
0
1 0
Wt
d
=
dt +
d
Xt
0
0 Wt
Wt
and let
C=
1
0
0
Wt
1
0
0
Wt
>
1
=
0
0
Wt2
i) Consider function F (t, w, x) = w2 x2 ; then
∂
F (t, w, x) = 0 ,
∂t
∂2
F (t, w, x) = 2x2 ,
∂w2
∂
F (t, w, x) = 2wx2 ,
∂w
∂2
F (t, w, x) = 2w2 ,
∂x2
∂
F (t, w, x) = 2w2 x ,
∂x
∂2
F (t, w, x) = 4wx .
∂w∂x
Now Wt2 Xt2 = F (t, Wt , Xt ) and by the multidimensional Itô Lemma
1
1
d(Wt2 Xt2 ) = dF (t, Wt , Xt ) = 0 + 2Wt Xt2 × 0 + 2Wt2 Xt × 0 + × 1 × 2Xt2 + 0 × 4Wt Xt + × Wt2 × 2Wt2 dt
2
2
+ 2Wt Xt2 × 1 dWt + 2Wt2 Xt × Wt dWt
= (Xt2 + Wt4 ) dt + 2Wt Xt (Xt + Wt2 ) dWt .
ii) Consider function G(t, x) = exp(x); then
∂2
G(t, x) = exp(x) .
∂x2
∂
G(t, x) = exp(x) ,
∂x
∂
G(t, x) = 0 ,
∂t
Now exp(Xt ) = G(t, Xt ) and by the unidimensional Itô Lemma
1
d (exp(Xt )) = dG(t, Xt ) = 0 + 0 × exp(Xt ) + Wt2 exp(Xt ) dt + Wt exp(Xt ) dWt
2
1 2
= Wt exp(Xt ) dt + Wt exp(Xt ) dWt .
2
3. Consider a non-dividend-paying stock evolving according to a 2-period binomial model, with u = 1.1 and d = 0.6.
Q
Assume no arbitrage, the initial price of the stock to be S(0) = 100 and
E [S(2) | S(0)] = 100. Consider also the
simple contingent claim with maturity in T = 2 and payoff X = Φ S(2) = max[S(2) − 80 , 0].
i) Compute the risk-free rate R and verify the computed R to be consistent with the no arbitrage assumption.
ii) Compute EQ [S(2) | S(1)]. [Hint: be careful, S(1) is a random variable and therefore . . . .]
iii) Compute the price at time zero for contingent claim X.
iv) Assume you sold contigent claim X to a costumer. Now you need to hedge it. Do you have to start your hedging
portfolio strategy by buying or by selling short the stock? (Justify your answer!).
Solution:
i) Under the no arbitrage assumption, discounted prices are martingales, hence 100 = S(0) = EQ [S(2) | S(0)](1 +
R)−2 = 100(1 + R)−2 and therefore R = 0. Since d = 0.6 < 1 + 0 < 1.1 = u the necessary and sufficient condition
for no arbitrage is satisfied.
ii) By martingality of discounted prices we know that EQ [S(2) | S(1)] = S(1)(1 + R) = S(1). Since S(1) is a random
variable, with the two possible outcomes Su (1) = S(0)u = 110 (if the first step is an “up”) and Sd (1) = S(0)d = 60
(if the first step is a “down”), also EQ [S(2) | S(1)] is a random variable, with the same two possible outcomes.
iii) Contingent claim X is a call option, with strike 80. Since the three possible terminal values for the stock price are
Suu (2) = S(0)u2 = 121, Sud (2) = Sdu (2) = S(0)ud = 66 and Sdd (2) = S(0)d2 = 36, the three possible outcomes
for the call option terminal payoff are Xuu = max(121 − 80 , 0) = 41, Xud = Xdu = max(66 − 80 , 0) = 0 and
Xdd = max(36 − 80 , 0) = 0. By martingality of discounted prices we have
Π(0, X) = EQ (X)(1 + R)−2 = EQ (X) = q 2 Xuu + 2q(1 − q)Xud + (1 − q)2 Xdd = 41q 2 .
Since
q=
1+R−d
1 − 0.6
4
=
= = 0.8 ,
u−d
1.1 − 0.6
5
we conclude that Π(0, X) = 41 × 0.8 = 26.24.
iv) We know that the binomial market is complete and hence every contingent claim, including X, can be hedged by
a self-financing portfolio h. We also know that the starting asset allocation h1 = (x1 , y1 ) is given by
x1 =
uVdh (1) − dVuh (1)
,
(1 + R)(u − d)
y1 =
Vuh (1) − Vdh (1)
.
S(0)(u − d)
We also know that the value of the hedge is equal to the price of our contingent claim at every node of the tree.
Therefore, by martingality of discounted prices,

ud
Vuh (1) = qXuu +(1−q)X
Q
= 0.8×41+0.2×0
= 32.8 if the first step was an “up”,
1+R
1+0
E
[X|S(1)]
h
=
V (1) =
 h
1+R
dd
= 0.8×0+0.2×0
=0
if the first step was a “down”.
Vd (1) = qXdu +(1−q)X
1+R
1+0
Therefore
1.1 × 0 − 0.6 × 32.8
= −39.36 ,
0.5
32.8 − 0
y1 =
= 0.656 .
100 × 0.5
x1 =
This means that the hedge stats by buying 0.656 stocks and by selling short 39.36 units of the bond.