STK4510 - Mandatory Assignment II
Sindre Froyn
Exercise 3
We are working with two stocks with the dynamics:
n
o
p
dS1 (t) = µ1 S1 (t)dt + S1 (t) ρσ1 dB1 (t) + 1 − ρ2 σ1 dB2 (t)
dS2 (t) = µ2 S2 (t)dt + S2 (t)σ2 dB1 (t)
and the price of a spread option paying S2 (T ) − S1 (T )
(a)
+
at time T .
We begin by finding expressions for S1 (t) and S2 (t).
For S1p
(t) we simplify the notation by defining: S1 (t) = St , µ1 = µ, σ1 = σ
and γ = 1 − ρ2 . We have the dynamics
dSt = µSt dt + ρσSt dB1 + γσSt dB2 .
(1)
To solve this we make a guess which will provide some insight to the solution:
St = S0 exp {µt + ρσB1 + γσB2 } .
We find the dynamics for this process by using the multi dimensional Ito’s
formula:
f (t, x1 , x2 ) = S0 exp {µt + ρσx1 + γσx2 }
where
n
n
X ∂f
∂f
1 X ∂2f
df =
dBi +
dt.
dt +
∂t
∂xi
2 i=1 ∂x2i
i=1
In the last sum we don’t need to consider any cross double derivatives, and
we get dt because we are using Brownian motion, which means we have the
relation:
dt, i = j
dBi dBj =
0, i 6= j
1
1
df (t, B1, B2 ) = µSt dt + ρσSt dB1 + γσSt dB2 + ρ2 σ 2 St dt + γ 2 σ 2 St dt
2
2
h
i
1
1
dSt = St µ + ρ2 σ 2 + γ 2 σ 2 dt + ρσSt dB1 + γσSt dB2
2
2
1
Comparing with the dynamics we are supposed to get in (1), we see that we
have two excessive terms in the dt-term. These would be eliminated if we
included −1/2(ρ2 σ 2 + γ 2 σ 2 )t in our guess. We also note:
−
σ2 2
σ2
σ2
(ρ + γ 2 ) = − (ρ2 + 1 − ρ2 ) = − .
2
2
2
Hence, the correct solution for the SDE is, reintroducing the original notation:
p
σ12
2
t + ρσ1 B1 (t) + 1 − ρ σ1 B2 (t) .
S1 (t) = S1 (0) exp
µ1 −
2
For S2 (t) we have a normal Geometric Brownian motion, and by Ito’s formula
we can easily verify the solution:
σ22
S2 (t) = S2 (0) exp
µ2 −
t + σ2 B1 (t) .
2
Writing the price P at time 0 as an expectation of the price, is simply:
+ −rT
P = e E S2 (T ) − S1 (T )
We are going to write this under the risk neutral measure Q, and since we have
two stocks and two Brownian motions, we use Multi dimensional Girsanov’s
Theorem.
B1
W1
−1 (µ1 − r)t
+Σ
=
(µ2 − r)t
B2
W2
p
Recalling that γ = 1 − ρ2 :
1 0
ρσ1 γσ1
−1
σ2
=⇒ Σ = 1
Σ=
− γσρ 2
σ2
0
γσ1
Multiplying the matrix Σ−1 in the expression, and shifting terms gives us the
following expressions:
µ2 − r
dt
σ2
µs − r
µ1 − r
dt + ρ
dt
dB2 = dW2 −
γσ1
γσ2
dB1 = dW1 −
Now we return to the dynamics for S1 and S2 , and use the new Q-Brownian
motions instead, beginning with S2 :
µ2 − r dS2 = µ2 S2 dt + S2 σ2 dW1 −
dt = rS2 dt + σ2 S2 dW1
σ2
2
µ2 − r µ1 − r
µ2 − r dS1 = µ1 S1 dt + S1 ρσ1 dW1 −
dt + γσ1 dW2 −
dt + ρ
dt
σ2
γσ1
γσ2
µ
−
r
µ
−
r
2
1
+ r)dt + ρσ
dt
dt + γσ1 dW2 − (µ
=
µ1
S
1
1
1 dt + S1 ρσ1 dW1 − ρσ1 σ2
σ2
= rS1 dt + ρσ1 S1 dW1 + γσ1 S1 dW2
The dynamics under the Q measure have the same structure as under the
physical measure. We can solve the SDE’s under Q in the same manner, by
just replacing the appropriate variables. We write the solutions with T for
convenience.
σ12
S1 (T ) = S1 (0) exp rT − T + ρσ1 W1 (T ) + γσ1 W2 (T )
2
σ22
S2 (T ) = S2 (0) exp rT − T + σ2 W1 (T )
2
Now we can write the price of the spread option under Q.
σ22
−rT
P = e EQ S2 (0) exp rT − T + σ2 W1 (T ) −
2
+ σ12
S1 (0) exp rT − T + ρσ1 W1 (T ) + γσ1 W2 (T )
2
We factor out erT which cancels out the discounting factor, and S2 (0) from
both terms.
2
σ2
P = S2 (0)EQ exp − T + σ2 W1 (T ) −
2
2
+ S1 (0)
σ1
exp − T + ρσ1 W1 (T ) + γσ1 W2 (T )
S2 (0)
2
Finally we factor out the remaining exponential in the first term. This
must remain within the expectation as it contains a Brownian component.
σ2 σ2
S1 (0)
exp (− 1 + 2 )T + (ρσ1 − σ2 )W1 (T )
(2)
P = S2 (0)EQ 1 −
S2 (0)
2
2
+
2
p
σ2
2
+ 1 + ρ σ1 W2 (T )
(3)
× exp − T + σ2 W1 (T )
2
We can factor out of the max-function, because if the difference S2 (T )−S1 (T )
is positive, we have only reformulated the expression, and if the difference is
0 or negative, the max-function becomes 0 and so does the entire expression.
3
(b)
e Brownian motion.
Using Girsanov’s theorem, we introduce a Q
f1 (t) = −σ2 dt + dW1 (t) ⇐⇒ dW1 (t) = σ2 dt + dW
f1 (t)
dW
Writing in integral form up to T , we can calculate the deterministic part.
Z T
f1 (T ) =⇒ W1 (T ) = σ2 T + W
f1 (T )
W1 (T ) =
σ2 dt + W
0
We perform som intermediate calculations. For the expression in the
exponential in (2):
f1 (T ) =⇒
(ρσ1 − σ2 )W1 (T ) = (ρσ1 σ2 − σ22 )T + (ρσ1 − σ2 )W
σ12 σ22
σ2
σ2
f1 (T )
+ )T + (ρσ1 − σ2 )W1 (T ) = (− 1 + ρσ1 σ2 − 2 )T + (ρσ1 − σ2 )W
2
2
2
2
For the exponent in (3), we observe that at time T , W1 (T ) ∼ N (0, T ), or
2
µW = 0 and σW
= T , and by independence
1 2
1 2 2
σ T .
E[exp(σ2 W1 (T ))] = exp µW σ2 + σW σ2 = exp
2
2 2
(−
This cancels against the first term in the exponent from (3), so we get
exp(0) = 1 and this factor vanishes, and we are left with:
S1 (0)
1
f1 (T )
P = S2 (0)EQe 1 −
exp − (σ12 − 2ρσ1 σ2 + σ22 )T + (ρσ1 − σ2 )W
S2 (0)
2
+ p
+ 1 + ρ2 σ1 W2 (T )
(c)
We define: a = (ρσ1 − σ2 ) and b = σ1
f1 (T ) + bW2 (T ) with:
distribution aW
p
h
i
f
E aW1 (T ) + bW2 (T ) = 0,
1 − ρ2 . We have the centered normal
h
i2
f
E aW1 (T ) + bW2 (T ) = 0.
We calculate the variance:
i
h
i2
h
i
h
f1 (T ) + bW2 (T ) .
f1 (T ) + bW2 (T ) = E (aW
f1 (T ) + bW2 (T ))2 − E aW
Var aW
4
The last term is simply zero, and we get:
h
i
h
i
h
i
h
i
f1 (T )+bW2 (T ) = a2 E W
f12 (T ) +ab E W
f1 (T )W2 (T ) +b2 E W22 (T ) .
Var aW
| {z }
|
{z
}
| {z }
=0
=T
=T
2
2
2
2
2
2
2
ρ2
σ
ρ2
σ
= (a2 +b2 )T = 1 +σ2 −2ρσ1 σ2 +σ1 −
1 T = (σ1 +σ2 −2ρσ1 σ2 )T := σ T.
To calculate the price P we observe that we have the form of a put option
with strike price K = 1 and stock price ST := S1 (T )/S2(T ). By Black-Scholes
formula, the price of a standard put option E[(K − ST )+ ] at time t = 0 is
given by, using put/call parity:
P0p = S0 − Ke−rT − P0c = S0 − Ke−rT − S0 Φ(u1 ) + Ke−rT Φ(u2 ),
where P0c is the call option price. Replacing with our parameters and
multiplying with S2 (0) and u1 and u2 as given as in theorem 4.3 with σ
as defined above, we get:
S1 (0)
S1 (0)
−rT
−rT
P = S2 (0)
−e
−
Φ(u1 ) + e Φ(u2 )
S2 (0)
S2 (0)
S1 (0) −rT
1 − Φ(u1 ) − e
1 − Φ(u2 )
= S2 (0)
S2 (0)
−rT
1 − Φ(u2 )
= S1 (0) 1 − Φ(u1 ) − S2 (0)e
= S1 (0)Φ(−u1 ) − S2 (0)e−rT Φ(−u2 ).
5