MASx52: Assignment 5
Solutions and discussion are written in blue. A sample mark scheme, with a total of 35 marks, is
given in red, with each mark placed after the statement/deduction for which the mark would be
given. As usual, mathematically correct solutions that follow a different method would be marked
analogously.
1.
(a) Within the Black-Scholes model, use the risk neutral valuation formula
F (t, St ) = e−r(T −t) EQ [Φ(ST ) | Ft ]
to show that price at time t of the contingent claim Φ(ST ) = 3ST + 5 is given by
F (t, St ) = 3St + 5e−r(T −t) .
(b) Describe a portfolio strategy that replicates Φ(ST ) during time [0, T ].
Solution.
(a) Using the explicit formula for geometric Brownian motion (see the formula sheet) we
obtain
h
i
1 2
e−r(T −t) EQ [3ST + 5 | Ft ] = e−r(T −t) EQ 3St e(r− 2 σ )(T −t)+σ(BT −Bt ) + 5 | Ft
h
i
1 2
= e−r(T −t) 3St e(r− 2 σ )(T −t) EQ eσ(BT −Bt ) | Ft + 5
−r(T −t)
Q σ
=e
3St E [e (BT − Bt )] + 5
1
(r− 2 σ 2 )(T −t)+ 12 σ 2 (T −t)
−r(T −t)
=e
3St e
+5
= e−r(T −t) 3St er(T −t) + 5
= 3St + 5e−r(T −t)
[4] Here, we use that St is Ft measurable,[1] and that Z = σ(BT − Bt ) ∼ N (0, σ 2 (T − t)
is independent of Ft . [1] We use the formula sheet to provide an explicit formula for
E[eZ ].
(b) At time 0, we buy three units of stock [1] and 5e−rT in cash. [1] It’s value at time t is
then
3St + 5e−rT ert = Φ(ST ).
Therefore, this portfolio replicates Φ(ST ) for all t ∈ [0, T ].
2.
(a) Let α ∈ R, σ > 0 and St be an Ito process satisfying dSt = αSt dt + σSt dBt . Let
Yt = St3 . Show that
dYt = 3α + 3σ 2 Yt dt + 3σYt dBt
Deduce that Yt is a geometric Brownian motion, and write down its drift and volatility.
(b) Within the Black-Scholes model, find the price F (t, St ) at time t ∈ [0, T ] of the contingent claim Φ(ST ) = ST3 .
Solution.
1
(a) By Ito’s formula,
1 2 2
2
dYt = (0) + αSt (3St ) + σ St (6St ) dt + σSt (3St2 )dBt
2
2
= 3α + 3σ Yt dt + 3σYt dBt .
[5] So, Yt is a geometric Brownian motion with drift 3α + 3σ 2 [1] and volatility 3σ. [1]
(b) Using the explicit formula for geometric Brownian motion (see the formula sheet) with
drift 3α + 3σ 2 and volatility 3σ, we have that
YT = Yt exp 3α + 3σ 2 − 92 σ 2 (T − t) + 3σ(BT − Bt )
= Yt exp 3α − 32 σ 2 (T − t) + 3σ(BT − Bt ) .
[2] Note that in the risk neutral world Q we have α = r. [1] Therefore, using the risk
neutral valuation formula (see the question, or the formula sheet), the arbitrage free
price of the contingent claim YT = Φ(ST ) = ST3 at time t is
e−r(T −t) EQ [YT | Ft ] = e−r(T −t) EQ St3 exp 3α − 32 σ 2 (T − t) + 3σ(BT − Bt ) | Ft
h
i
3 2
= e−r(T −t) St3 e(3r− 2 σ )(T −t) EQ e3σ(BT −Bt ) | Ft
i
h
3 2
= e−r(T −t) St3 e(3r− 2 σ )(T −t) EQ e3σ(BT −Bt )
3
= e−r(T −t) St3 e(3r− 2 σ
= St3 e2r(T −t)+3σ
2 )(T −t)
2 (T −t)
9
e2σ
2 (T −t)
.
[3] Here, we use that St is Ft measurable. [1] We then use the properties of Brownian motion to tell us that 3σ(BT − Bt ) is independent of Ft [1] with distribution
N (0, (3σ)2 (T − t)), followed by the formula sheet to explicitly evaluate EQ e3σ(BT −Bt ) .
[1]
3. Let Xt be an Ito process satisfying dXt = Xt2 dBt , and let F (t, x) be a solution of the partial
differential equation
∂F
1 ∂2F
(t, x) + x4 2 (t, x) = 0
∂t
2 ∂x
with the boundary condition F (T, x) = x. Use Ito’s formula to find dF (t, Xt ) and hence
show that F (t, x) = Et,x [XT ].
Solution. Using Ito’s formula, we have
∂F
1 2 2 ∂2F
2 ∂F
dF (t, St ) =
+ (0) + (Xt )
dt + Xt
dBt
∂t
2
∂x2
∂x
∂F
= Xt2
dBt
∂x
[6] Writing in integral form, over [t, T ], we obtain
Z
F (T, ST ) = F (t, St ) +
t
T
Xu2
∂F
dBu .
∂x
[1] Taking expectations Et,x (which denotes that X runs during time [t, T ] and has initial
state Xt = x), we obtain
Et,x [F (T, XT )] = Et,x [F (t, Xt )] + 0
2
because Ito integrals are martingales. [1] On the right hand side, since Et,x specifies that
Xt = x, we have F (t, Xt ) = F (t, x), which is deterministic [1]. On the left hand side,
F (T, XT ) = Φ(XT ) = XT [1], so we obtain
Et,x [Φ(XT )] = F (t, x)
as required. [1]
3