Matematisk statistik
Matematikcentrum
Lunds tekniska högskola
Tentamen: 2007–05–25 kl 800 –1300
FMS170/MAS232—Prissättning av Derivattillgångar, 6 p
Solution.
1. Let Xt = f (t, Wt ) = et/2 cos(Wt ). We now calculate dXt with Itôs formula:
∂f
1 ∂ 2f
∂f
dXt = df (t, Wt ) =
+
dt + 1 dWt
2
∂t
2 ∂x
∂x
1 t/2
1 t/2
e cos(Wt ) − e cos(Wt ) dt − et/2 sin(Wt )dWt
=
2
2
= −et/2 sin(Wt )dWt
This process has no drift term moreover we have that the diffusion part has finite second moment
using the Itô isometry, i.e.
Z t
Z t
Z t
s/2
2
s
2
E[( −e sin(Ws )dWs ) ] =
E[e sin(Ws ) ]ds ≤
E[es ]ds = et − 1 < ∞.
0
0
0
This gives that X (t) is a Martingale.
Alternative solution:
We can directly calculate E[Xt|Fs ] for s < t as
E[Xt|Fs ] = E[et/2 cos(Wt )|Fs ] = E[e(t−s)/2 cos(Wt − Ws + Ws )es/2 |Fs ]
= E[e(t−s)/2 (cos(Wt − Ws ) cos(Ws )es/2 − sin(Wt − Ws ) sin(Ws )es/2 )|Fs ]
2
= es/2 cos(Ws )(e(t−s)/2 e−(t−s) /2 ) + sin(Ws )es/2 0
= es/2 cos(Ws ) = Xs .
Finally we need to establish that E[|Xt |] < ∞ which is easily done since |Xt | < et/2 < ∞.
2. Let Π (t) be the price of the derivative X for t ≤ T , moreover let PK (t) and CK (t) be the price
at time t of a put option and call option respectively both with maturity T and strike price K .
The price Π (t) is given by any of the following equivalent portfolios (there are more possible
equivalent portfolios).
Π (t) = P2K (t) − PK (t)
= K B(t)/B(T ) − CK (t) + C2K (t)
= S(t) + PK (t) − 2CK (t) + C2K (t)
Looking at the payoffs at time T show that the portfolios have the same payoff as the contract
X , i.e.

S(T ) ≤ K
 K
2K − S(T ) K ≤ S(T ) ≤ 2K

0
S(T ) ≥ 2K
1
3. Looking at the pay-offs we see that
ΦC (S(T))
K1−K2
0
ΦC (S(T))
2
0
K2
1
K1
S(T)
the pay-off for the derivative C2 dominate the payoff for C1 for all possible values of S(T ).
Therefore the price of C2 , ΠC2 (t), must for all 0 ≤ t ≤ T be higher than the price of C1 , ΠC1 (t),
to avoid arbitrage. To see this assume that there exist a time t such that ΠC1 (t) ≥ ΠC2 (t). Now
we can form a zero cost portfolio consisting of a long position in C2 a short position in C1 and
finally we put ΠC1 (t) − ΠC2 (t) in the bank-account. At time T our portfolio will be worth
(ΠC1 (t) − ΠC2 (t))B(T )/B(t) + (ΠC2 (T ) − ΠC1 (T )),
and since both these terms are positive this is an arbitrage opportunity. Therefore we must have
that ΠC1 (t) ≤ ΠC2 (t) for 0 ≤ t ≤ T to avoid arbitrage.
4. The price of the ZCB, p(t, T ), is given by
h RT
i
p(t, T ) = E e− t r(s)ds .
The short rate r(s) for s > t is given by
Z s
Z s
Z s
σ2 2
2
−s
−t
dW (u) .
r(s) = r +
Θ(u)du +
σdW (u) = r − c(e − e ) + (s − t ) + σ
2
t
t
t
The integral of the short rate r(s) between t and T is then given by
Z T
Z T
Z s
σ2 2
−s
−t
2
−
r(s)ds = −
r − c(e − e ) + (s − t ) + σ
dW (u) ds
2
t
t
t
Z TZ s
σ 2 3 3 σ2 2
−T
−t
−t
= −r(T −t)−c(e −e )−ce (T −t)− (T −t )+ t (T −t)−σ
dW (u)ds .
6
2
t
t
The expectation of the stochastic term is
Z T Z s
E
dW (u)ds = 0 ,
t
t
and the variance is
" Z
Z TZ s
V −
dW (u)ds = E −
t
t
T
Z
dW (u)ds
t
"Z
T
t
T
Z
= E
t
Z
=
t
T
2 #
s
" Z
2 #
dsdW (u)
=E
u
2 #
T
(T − u)dW (u)
t
(T − u)3
(T − u) du = −
3
2
2
T
=
t
(T − t)3
.
3
We see that the stochastic term has a normal distribution, and therefore
Z TZ s
(T − t)3
−σ
dW (u)ds ∼ N 0, σ
.
3
t
t
Moreover we have that
h
E e
−σ
RT Rs
t
t
dW (u)ds
i
=e
σ2
(T −t)3
6
.
This gives that p(t, T ) is given by
p(t, T ) = exp −r(T − t) − c(e−T − e−t ) − ce−t (T − t)
σ2 t 2
σ2
σ2 3
3
3
(T − t) + (T − t) .
− (T − t ) +
6
2
6
5. We want to price the derivative X with maturity T and pay-off:
Φ(S1 (T ), S2 (T )) = max(S2 (T ) − S1 (T ), 0).
According to the risk-neutral valuation formula the price at time t, ΠX (t), is given by
ΠX (t) = e−r(T −t) EQ [max(S2 (T ) − S1 (T ), 0)|Ft ].
There are now some different approaches to calculate this expectation using change of numeraire
techniques. The perhaps easiest approach is to use S1 as a numeraire this leads to
1
S2 (T )
QS1
QS1
ΠX (t) = S1 (t)E
max(S2 (T ) − S1 (T ), 0)|Ft = S1 (t)E
− 1, 0)|Ft ,
max(
S1 (T )
S1 (T )
where QS1 is the numeraire measure for S1 . This can now be seen as a European call option
on the ratio S2 /S1 with strike 1. Moreover since S2 /S1 is a ratio between a traded asset and a
numeraire it is automatically a martingale under QS1 . Using that the volatilities does not change
when we change measure we can by calculate the volatilities under Q as
∂ S2
∂ S2
S1 (t)(σ11 dW1 (t) + σ12 dW2 (t)) +
S2 (t)(σ21 dW1 (t) + σ22 dW2 (t))
∂S1 S1
∂S2 S1
S2 (t)
=
((σ21 − σ11 )dW1 (t) + (σ22 − σ12 )dW2 (t))
S1 (t)
This gives that the dynamics for S2 /S1 under QS1 for s ≥ t is given by
S1
S1
S2 (s)
S2 (s) d
=
(σ21 − σ11 )dW1Q (s) + (σ22 − σ12 )dW2Q (s) ,
S1 (s)
S1 (s)
S2 (t)
s2 (t)
=
S1 (t)
s1 (t)
S1
S1
where W1Q (s) and W2Q (s) are independent standard QS1 Brownian motions. We can now
solve this SDE to obtain
S1
S1
s2 (t)
1
S2 (T )
=
exp − (σ21 − σ11 )2 + (σ22 − σ12 )2 (T − t) + (σ21 − σ11 )(W1Q (T ) − W1Q (t))
S1 (T )
s1 (t)
2
S1
S1
+ (σ22 − σ12 )(W2Q (T ) − W2Q (t))
3
This has the same distribution as
√
s2 (t)
1 2
exp − σ (T − t) + σ T − tG ,
s1 (t)
2
where G is a standard Gaussian random variable and where
p
σ = (σ21 − σ11 )2 + (σ22 − σ12 )2 .
We can therefore calculate the price ΠX (t) as
Z ∞
√
s2 (t)
1 2
exp(−g 2 /2)
√
ΠX (t) = s1 (t)
max
exp − σ (T − t) + σ T − tg − 1, 0
dg
s1 (t)
2
2π
−∞
Z ∞
√
s2 (t)
1 2
exp(−g 2 /2)
√
exp − σ (T − t) + σ T − tg − 1
dg
= s1 (t)
s1 (t)
2
2π
−d
Z ∞
√
s2 (t) 1
1 2
exp(−g 2 /2)
2
√ exp − σ (T − t) + σ T − tg − g /2 −
√
= s1 (t)
dg
s1 (t) 2π
2
2π
−d
Z ∞
√
s2 (t) 1
exp(−g 2 /2)
1
2
√ exp − (g − σ T − t) −
√
= s1 (t)
dg
s1 (t) 2π
2
2π
−d
Z ∞
Z ∞
√
1
1
exp(−g 2 /2)
2
√ exp − (g − σ T − t) dg − s1 (t)
√
dg
= s2 (t)
2
2π
2π
−d
−d
√
= s2 (t)(1 − N (−d − σ T − t)) − s1 (t)(1 − N (−d))
√
= s2 (t)N (d + σ T − t) − s1 (t)N (d),
where
ln(s2 (t)/s1 (t)) − σ2 (T − t)/2
√
,
σ T −t
and where N is the distribution function of the standard Gaussian distribution.
d=
Alternative solution:
We can also use both S1 and S2 as numeraires which gives that
S2 S1 ΠX (t) = S2 (t)EQ I (S2 (T ) > S1 (T ))|Ft − S1 (t)EQ I (S2 (T ) > S1 (T ))|Ft .
This can now be rewritten as
S2 (T )
S1 (T )
QS1
QS2
< 1 |Ft − S1 (t)E
I
> 1 |Ft .
ΠX (t) = S2 (t)E
I
S2 (T )
S1 (T )
Now S1 /S2 is a QS2 -martingale and S2 /S1 is a QS1 -martingale. By using the same type of argument and calculations as in the first solution we obtain that S1 (T )/S2 (T ) under QS2 has the
same distribution as
√
s1 (t)
1 2
exp − σ (T − t) + σ T − tG ,
s2 (t)
2
and that S2 (T )/S1 (T ) under QS1 has the same distribution as
√
1 2
s2 (t)
exp − σ (T − t) + σ T − tG ,
s1 (t)
2
where σ and G are as defined in the first solution. Straightforward calculation now give that
√
ΠX (t) = s2 (t)N (d + σ T − t) − s1 (t)N (d),
4
where d are defined as in the first solution.
Alternative solution 2:
We can also calculate the dynamics for S1 and S2 under both QS1 and QS2 . For this we need
to find two two-dimensional Girsanov kernels, where we also need to look at dynamics of the
bank-account to get the right Girsanov kernels. This is however as seen from the calculations
above an unnecessary detour. For the sake of completeness we supply the appropriate Girsanov
kernels:
g1S1 = −σ11 ,
g1S2 = −σ21 ,
g2S1 = −σ12
g2S2 = −σ22 .
By the same type of calculations as in the first two solutions we finally arrive at the same answer.
6.
(a) By using that Z (t) = S(t)/P(t, T ), for 0 ≤ t ≤ T , is a QT martingale and that volatilities do note change when we change measure, we see that we can calculate the volatility
function v(t, T ) using the diffusion part of the Q-dynamics for S(t)/P(t, T ). This gives
that
∂ S
∂ S
dW1 (t)
P(t, T )γ(T − t)dW1 (t) +
S(t)σdW2 (t)
Z (t)v(t, T )
=
dW2 (t)
∂P P
∂S P
= −Z (t)γ(T − t)dW1 (t) + Z (t)σdW2 (t)
dW1 (t)
= Z (t) −γ(T − t) σ
.
dW2 (t)
This gives that v(t, T ) = −γ(T − t) σ .
(b) Using that S(T ) = Z (T ) we can view the contract as written on the process Z instead of
S. So the price of the European call option can thus be calculated as
T
Π (t) = p(t, T )EQ
max(Z (T ) − K , 0)|Ft .
To calculate this we must find the distribution of Z (T ) under QT . Solving the SDE for Z
under QT gives that
1
Z (T ) = Z (t) exp −
2
Z
T
Z
2
|v(s, T )| ds +
t
T
v(s, T )dW
QT
(s) .
t
This now has the same distribution as
√
1
2
Z (t) exp − Σ(t, T ) (T − t) + Σ(t, T ) T − tG ,
2
where
sR
Σ(t, T ) =
T
t
γ2 (T − s)2 + σ2 ds
=
T −t
5
r
γ2 (T − t)3 /3 + σ2 (T − t)
,
T −t
and where G is standard Gaussian random variable. By exactly the same calculations as in
the derivation of the standard Black-Scholes formula but with σ replaced by Σ(t, T ) and
e−r(T −t) replace by p(t, T ) we obtain
√
Π (t) = p(t, T )Z (t)N (d + Σ(t, T ) T − t) − p(t, T )N (d )
√
= S(t)N (d + Σ(t, T ) T − t) − p(t, T )N (d),
where
d=
ln(Z (t)/K ) − Σ(t, T )2 (T − t)/2
ln(S(t)/K ) − ln(P(t, T )) − Σ(t, T )2 (T − t)/2
√
√
=
,
Σ(t, T ) T − t
Σ(t, T ) T − t
and where N is the distribution function of the standard Gaussian distribution.
To check that we get back the original formula if r(t) ≡ r and |v(t, T )| ≡ σ , we notice
that this imply that γ = 0 and p(t, T ) = exp(−r(T − t)). This gives that Σ(t, T ) ≡ σ
and − ln(p(t, T )) = r(T − t). Now plugging this into our price gives
√
Π (t) = S(t)N (d + σ T − t) − e−r(T −t) N (d),
where
ln(S(t)/K ) + r(T − t) − σ2 (T − t)/2
√
,
σ T −t
which we recognise as the ordinary Black-Scholes formula.
d=
6