1. All random variables in the following questions are defined on a probability space
(Ω,F , P ), and G denotes a σ-algebra contained in F .
(a) Let X and Y be random variables such that E|X| < ∞. Define precisely the conditional expectations E(X|G) and E(X|Y ). [7 marks] (b) Let X be a random variable with finite second moment, and consider R = X − E(X|G), the difference between the “true value” of X and the “predicted value” of X based on the “information” G. Compute ER and E(R|G). Show that ER2 = EX2 − EZ2, where Z = E(X|G). [7 marks] (c) Let X be a random variable with finite second moment, and let L2(G) denote the space of G-measurable random variables with finite second moment. Show that Z = E(X|G) minimises the mean square distance of X from L2(G), i.e., E(X − Z)2 = min{E(X − Y )2 : Y ∈ L2(G)}. [6 marks] (d) Let X1 and X2 be independent identically distributed random variables such that E|X1| = E|X2| <∞. Set X̄ = (X1 +X2)/2, and determine E(αX1 + (1− α)X2|X̄) as a function of X̄ for any constant α ∈ R. [Hint: You may use without proof that due to symmetry, E(X1|X̄) = E(X2|X̄).] [5 marks] 2. We want to model the evolution of the instantaneous interest rate by an Itô process r = (rt)t≥0 satisfying following conditions: (i) r0 = 4 and 3 ≤ rt ≤ 5 almost surely for all t ≥ 0; (ii) Ert = 4 for all t ≥ 0; (iii) E(|rt − 4|2) ≤ 1/200 for all t ≥ 0. The solution r of the stochastic differential equation drt = 100(4− rt) dt+ (|rt − 5||rt − 3|)1/2 dWt, r0 = 4 is suggested as a suitable model, where W = (Wt)t≥0 is a Wiener process. (a) State precisely a comparison theorem for SDEs, and applying it to suitable SDEs, show that property (i) holds for the solution r. [12 marks] (b) Prove that r satisfies property (ii). [6 marks] (c) Setting Yt := rt − 4 and using Itô’s formula, write an expression for the stochastic differential of e100tYt. Hence, estimating E(|e100tYt|2), or otherwise, deduce that r satisfies property (iii). [7 marks] 3. Consider the standard Black-Scholes market with bond price Bt = e rt and stock price St = S0 exp(αt+ σWt) at time t ∈ [0, T ], where W is a Wiener process, α is any constant, and S0 and σ are positive constants. Let f be a nonnegative function on R satisfying the polynomial growth condition, and denote by A0 and B0 the prices at time t = 0 of an American type option with pay-off process (f(St))t∈[0,T ] and a European type option with pay-off f(ST ) at maturity T , respectively. (a) Using appropriate formulas define precisely the prices A0 and B0, and hence show that A0 ≥ B0. [7 marks] (b) State precisely the Main Theorem on Pricing European Type Options, and hence show that the replicating strategy (ψ∗t , ϕ ∗ t )t∈[0,T ] for the European type option with pay-off f(ST ) is a hedging strategy for the American type option with pay-off process (f(St))t∈[0,T ] if and only if e−r(T−t)EQ (f(ST )|Ft) ≥ f(St) for every t ∈ [0, T ], where Q is the risk neutral probability measure. [7 marks] (c) Assume that f is a convex function such that f(0) = 0. Then show that λf(x) ≥ f(λx) for every x ∈ R and λ ∈ [0, 1], and hence using Part (b) prove that the replicating strategy (ψ∗t , ϕ ∗ t )t∈[0,T ] for the European type option with pay-off f(ST ) is a hedging strategy for the American type option with pay-off process (f(St))t∈[0,T ]. [7 marks] (d) Prove that if f is convex such that f(0) = 0 then A0 = B0. [4 marks] 4. Consider again the Black-Scholes market with bond and stock prices as in Question 3. We want to compute the price V at t = 0 of the European type option with payoff h := { L if maxt∈[0,T ] St ≥ K 0 otherwise } at expiry date T , where L > 0 and K > 0 are some constants.
(a) Using the main theorem on pricing European options, show that
V = Le−rTP
(
max
t∈[0,T ]
(Wt + at) ≥ b
)
,
where a := r
σ
− 1
2
σ, b := σ−1 ln(K/S0).
[7 marks]
(b) State precisely the Girsanov theorem.
[6 marks]
(c) Using Girsanov’s theorem show that
V = Le−rTE
(
1[maxt∈[0,T ]Wt≥b]e
aWT− 12a
2T
)
,
where a and b are the constants defined in Part (a).
[7 marks]
(d) Denote the event [maxt≤T Wt ≥ b,WT ≤ x] byBx. Knowing that, due to the reflection
principle for the Wiener process W ,
P (Bx) =
{
P (WT ≥ 2b− x) if x < b
P (WT ≥ b)− P (WT ≥ x) if x ≥ b
}
,
find a function g such that
P (Bx) =
∫ x
−∞
g(y) dy for every x ∈ (−∞,∞).
[3 marks]
(e) Deduce from Parts (c) and (d) that
V = C
∫ ∞
b
(
ea(2b−y) + eay
)
e−
y2
2T dy,
with C := Le−(r+
1
2
a2)T/
√
2πT , a = r
σ
− 1
2
σ and b := σ−1 ln(K/S0).
You may use without proof the following fact:
If A ∈ F is an event and X is a random variable such that
P (A ∩ [X ≤ x]) =
∫ x
−∞
g(y) dy, for all x ∈ (−∞,∞)
with a function g, then for every non-negative function f
E
(
1Af(X)
)
=
∫ ∞
−∞
f(x)g(x) dx.
[2 marks]
END PAPER
Solution
1. (a) (i) Z := E(X|G) is a G-measurable random variable such that E(Z1G) =
E(X1G) for every G ∈ G. (ii) E(X|Y ) = E(X|σ(Y )) where σ(Y ) is the σ-
algebra generated by Y . [7
marks]
(b) (i) ER = E(X − E(X|G)) = EX − EE(X|G) = EX − EX = 0,
(ii) E(R|G) = E(X − E(X|G|G) = E(X|G)− E(X|G) = 0
(iii) Notice that E(XZ) = EE(XZ|G) = EZ2. Hence
ER2 = E(X − Z)2 = EX2 − 2E(XZ) + EZ2 = EX2 − EZ2.
[7 marks]
(c)
E(X−Y )2 = E(X−Z+Z−Y )2 = E(X−Z)2+2E{(X−Z)(Z−Y )}+E(Z−Y )2.
Notice that E{(X−Z)(Z−Y )} = EE ((X − Z)(Z − Y )|G) = E ((Z − Y )E (X − Z|G)) =
0. Consequently,
E(X − Y )2 = E(X − Z + Z − Y )2 = E(X − Z)2 + E(Z − Y )2 ≥ E(X − Z)2,
which proves the statement since Z ∈ L2(G). [6 marks]
(d) By symmetry E(X1|X̄) = E(X2|X̄). Hence
E(Xi|X̄) =
1
2
{E(X1|X̄) + (X2|X̄)} = E(X̄|X̄) = X̄ for i = 1, 2.
Consequently,
E(αX1 + (1−α)X2|X̄) = αE(X1|X̄) + (1−α)E(X2|X̄) = αX̄+ (1−α)X̄ = X̄.
[5 marks]
2. (a) Comparison Theorem: Consider the SDEs
dXt = b(t,Xt) dt+ σ(t,Xt) dWt, X0 = ξ
dYt = B(t, Yt) dt+ σ(t, Yt) dWt, Y0 = η,
such that the conditions of the existence and uniqueness theorem for both
equations hold. Assume moreover that
ξ ≤ η (a.s.),
and
b(t, x) ≤ B(t, x)
for all t ∈ [0, T ], x ∈ (−∞,∞). Then almost surely Xt ≤ Yt for all t ∈ [0, T ].
Applying this theorem with
ξ := 3, b(t, x) := 400(3− x), σ(t, x) :=
√
|x− 5||x− 3|
and
η := 4, B(t, x) := 400(4− x),
we get that
rt = Yt ≥ Xt = 3.
Applying the comparison theorem with
η := 5, B(t, x) := 400(5− x), σ(t, x) :=
√
|x− 5||x− 3|,
and
ξ := 4, b(t, x) := 400(4− x),
we get that almost surely
rt = Xt ≤ Yt = 5.
[12 marks]
(b) By definition r satisfies the equation
rt = 4 +
∫ t
0
100(4− rs) ds+
∫ t
0
√
|rs − 5||rs − 3| dWs.
Taking expectation in both sides, for mt := Ert we get
mt = 4 +
∫ t
0
400(4−ms) ds,
by noticing that
E
∫ t
0
√
|rs − 5||rs − 3| dWs = 0,
since
E
∫ t
0
|rs − 5||rs − 3| ds ≤ t <∞.
Hence clearly Ert = mt = 4 for all t ≥ 0.
[7 marks]
(c) By Itô’s formula for Yt := rt − 4 we have
d(e100tYt) = e
100t
√
|Y 2t − 1| dWt, Y0 = 0,
which means
e100tYt =
∫ t
0
e100s
√
|Y 2s − 1| dWs.
Hence by Itô’s identity
E|e100tYt|2 = E
∫ t
0
e200s|Y 2s − 1| ds ≤
1
200
(e200t − 1) ≤ 1
200
e200t,
by taking into account 0 ≤ Y 2s ≤ 1, which gives
E|rt − 4|2 ≤ 1200 .
[6 marks]
(3) (a) The payoff is h = L1{maxt≤T St≥K}. By the Main Theorem on Pricing European
Type Options
V = e−rTEQh = Le
−rTEQ1{maxt≤T St≥K} = Le
−rTQ
(
max
t≤T
St ≥ K
)
,
where Q is the risk neutral probability measure. Since St = S0 exp(σW̃t + (r −
1
2
σ2)t) with a Wiener process (W̃t)t∈[0,T ] under Q, and exp(x) is increasing in x,[
max
t≤T
St ≥ K
]
=
[
max
t≤T
(σW̃t + (r −
1
2
σ2)t) ≥ ln K
S0
]
=
[
max
t≤T
(W̃t + at) ≥ σ−1 ln
K
S0
]
=
[
max
t≤T
(W̃t + at) ≥ b
]
with a := r
σ
− 1
2
σ, b := σ−1 ln(K/S0). Hence
V = Le−rTQ
(
max
t∈[0,T ]
(W̃t + at) ≥ b
)
= Le−rTP
(
max
t∈[0,T ]
(Wt + at) ≥ b
)
.
[7 marks]
(b) Girsanov’s theorem: Let W = (Wt)t∈[0,T ] be a Wiener with respect to (Ft)t≥0 on
a probability space (Ω,F , P ) with a filtration (Ft)t≥0. Consider the process
Xt := Wt +
∫ t
0
bs ds, t ∈ [0, T ],
where (bt)t∈[0,T ] ∈ S([0, T ]). Set
γT = exp
(
−
∫ T
0
bt dWt −
1
2
∫ T
0
b2t dt
)
,
and define the measure Q by dQ = γT dP . Assume that EγT = 1. Then Q is a
probability measure, and under Q the process (Xt)t∈[0,T ] is a Wiener martingale
with respect to (Ft)t≥0.
[6 marks]
(c) Set γ := exp(−aWT − 12a
2T ) and define the measure Q by dQ = γ dP . Then
Eγ = e−
1
2
a2TEe−aWT = 1.
Hence by Girsanov’s theoremQ is a probability measure and underQ the process
Vt := Wt + at, t ∈ [0, T ]
is a Wiener process. Therefore
P
(
max
t∈[0,T ]
(Wt + at) ≥ b
)
= P
(
max
t∈[0,T ]
Vt ≥ b
)
= EQ(1maxt∈[0,T ] Vt≥bγ
−1).
Notice that γ−1 = exp(aWT +
1
2
a2T ) = exp(aVT − 12a
2T ). Hence
P
(
max
t∈[0,T ]
(Wt+at) ≥ b
)
= EQ(1maxt≤T Vt≥be
aVT− 12a
2T ) = E(1maxt≤T Wt≥be
aWT− 12a
2T ).
Consequently,
V = Le−rTE
(
1maxt≤T Wt≥be
aWT− 12a
2T
)
.
[7 marks]
(d) Since
d
dx
P (Bx) =
1√
2πT
e−
(2b−x)2
2T if x < b
1√
2πT
e−
x2
2T if x ≥ b
=: g(x)
is nonnegative and continuous at every x ∈ R such that
∫∞
∞ g(x) dx <∞,
P (Bx) =
∫ x
−∞
g(y) dy for all x ∈ (∞,∞).
[3 marks]
(i) From (c) and (d)
V =
Le−rT
√
2πT
∫ ∞
−∞
eax−
1
2
a2Tg(x) dx = C
{∫ b
−∞
eaxe−
(2b−x)2
2T dx+
∫ ∞
b
eaxe−
x2
2T dx
}
with C := Le
−rT
√
2πT
e−
1
2
a2T = Le
−(r+1
2
a2)T
√
2πT
. By the change of variable z := 2b− x∫ b
−∞
eaxe−
(2b−x)2
2T dx =
∫ ∞
b
ea(2b−z)e−
z2
2T dz.
Thus
V = C
∫ ∞
b
(
ea(2b−x) + eax
)
e−
x2
2T dx.
[2 marks]