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Advanced Financial Models
Example sheet 3 - Michaelmas 2015
Michael Tehranchi
Problem 1. Let f : [0, T ] → R be a continuous (non-random) function and W a Brownian
RT
motion. By considering the L2 construction of the stochastic integral, show that 0 f (t)dWt
RT
is a normal random variable with mean zero and variance 0 f (t)2 dt.
Solution 1. Let fn be the piece-wise constant left-continuous function
n
X
fn =
f (tni−1 )1(tni−1 ,tni ] .
i=1
where
tni
= iT /n. Now
T
Z
fn (t)dWt =
0
n
X
f (tni−1 )(Wtni − Wtni−1 )
i=1
is the sum of independent mean-zero normal random variables, and hence is mean-zero
normal with variance
Z T
n−1
X
n
2 n
n
f (ti−1 ) (ti − ti−1 ) →
f (t)2 dt,
0
i=0
RT
where the convergence is a standard result of Riemann integration. Since 0 fn (t)dWt →
RT
f (t)dWt in L2 by definition, we will be done if we appeal to the following standard fact:
0
Fact. Suppose Xn ∼ N (µn , σn2 ) and µn → µ and σn2 → σ 2 . If Xn → X in L2 then
X ∼ N (µ, σ 2 ).
Proof. Since Xn → X in L2 , then Xn → X in distribution. Let Fn and F be the distribution
functions of Xn and X, respectively. At the points of continuity of continuity of F we have
F (x) = lim Fn (x)
n
x − µn
= lim Φ
n
σn
x−µ
=Φ
σ
by the continuity of the standard normal distribution function Φ.
Problem 2. (Ornstein–Uhlenbeck process) Let W be a Brownian motion, and let
Z t
at
Xt = e x + b
ea(t−s) dWs
0
for some a, b, x ∈ R.
(a) Verify that (Xt )t≥0 satisfies the following stochastic differential equation:
dXt = aXt dt + b dWt ,
X0 = x.
(b) Show that
b2 2at
at
Xt ∼ N e x, (e − 1) .
2a
1
(c) What is the distribution of the random variable
RT
0
Xt dt?
Solution 2. (a) Since
Z t
−as
e
dWs
Xt = e
x+b
at
0
we can apply Itô’s formula
dXt = e
at
−at
be
Z t
−as
e
dWs aeat dt
dWt + x + b
0
= b dWt + aXt dt
(b) Since
t
Z
(ea(t−s) )2 ds =
0
e2at − 1
2a
this part follows from the previous problem.
(c) Method 1: Note that by rearranging the stochastic differential equation we have
Z T
1
Xt dt = (XT − x − bWT )
a
0
RT
and hence 0 Xt dt is normally distributed with mean (eaT −1)x/a. To compute the variance,
first note that
Z T
Z T
a(T −t)
e
, dWt
dWt
Cov(XT , WT ) = Cov b
0
Z
=b
0
T
ea(T −t) dt
0
b
= (eaT − 1)
a
by Itô’s isometry. Hence
Z T
1
Xt dt = 2 Var(XT ) − 2b Cov(XT , WT ) + b2 Var(WT )
Var
a
0
b2
= 3 (e2aT − 4eaT + 3 + 2aT ).
2a
Method 2:
Z T
Z T
Z TZ t
aT
Xt dt =
e x dt +
ea(t−s) b dWs dt
0
0
0
0
Z T
Z TZ T
=
eaT x dt +
ea(t−s) b dt dWs
0
0
s
Z T a(T −s)
eaT − 1
e
−1
=
x+
b dWs
a
a
0
RT
Hence 0 Xt dt is normally distributed with mean (eaT − 1)x/a and variance
Z
b2 T a(T −s)
b2 2aT
2
(e
− 4eaT + 3 + 2aT )
(e
−
1)
ds
=
a2 0
2a3
2
This calculation is useful in the study of the Vasicek interest rate model.
Problem 3. Let W be a Brownian motion. Show that if Yt = Wt3 − 3tWt then Y is a
martingale (1) by hand, and (2) by Itô’s formula.
Solution 3. By hand: Since Gaussian random variables have finite moments of all orders, Y
is integrable. Indeed, we have
E(|Yt |) ≤ E(|Wt3 |) + 3tE(|Wt |) = Ct3/2 < ∞
p
where C = 5 2/π. Therefore, using the indepedence of the increments of W we have
E(Yt |Fs ) =E(Wt3 − 3tWt |Fs )
=E[(Wt − Ws + Ws )3 − 3t(Wt − Ws + Ws )|Fs ]
=E[(Wt − Ws )3 ] + 3E[(Wt − Ws )2 ]Ws + 3E(Wt − Ws )Ws2 + Ws3
− 3tE(Wt − Ws ) − tWs
=0 + 3(t − s)Ws + 0 + Ws3 + 0 − tWs
=Ys
for 0 ≤ s < t.
By Itô’s rule:
dYt = d(Wt3 − 3tWt )
= (3Wt2 dWt + 3Wt dt) − 3(t dWt + Wt dt)
= 3(Wt2 − t)dWt
R
t 2
and hence Y is a local martingale. Recall that if E 0 αs ds < ∞ for all t ≥ 0 then
R
t
the process 0 αs dWs
is a martingale. Again, it’s clear that the integrand is square
t≥0
integrable in this case since Guassian random variables have finite moments of all orders.
But, just to be explicit,
Z t
Z t
Z t
2
2
4
2
2
E [3(Ws − s)] ds = 9
E(Ws − 2Ws s + s )ds = 9
2s2 ds = 6t3 < ∞
0
0
0
and hence (Yt )t≥0 is a martingale.
Problem 4. (Heat equation) Let W be a scalar Brownian motion, and let g : [0, T ] × R → R
satisfy the partial differential equation
∂g 1 ∂ 2 g
+
=0
∂t 2 ∂x2
with terminal condition
g(T, x) = G(x).
(a) Show that (g(t, Wt ))t∈[0,T ] is a local martingale.
(b) If the function g is bounded, deduce the formula
Z ∞
2
√
e−z /2
g(t, x) =
G(x + T − tz) √ dz.
2π
−∞
3
(c) Use Problem 2 to find explicitly the unique bounded solution to the PDE
∂h
∂h 1 ∂ 2 h
+x
+
=0
∂t
∂x 2 ∂x2
with terminal condition
h(T, x) = cos x.
Useful fact: If Z ∼ N (µ, σ ) then E(cos Z) = e−σ
2
2 /2
cos µ.
Solution 4. (a) By Itô’s formula:
∂g 1 ∂ 2 g
∂g
dg(t, Wt ) =
+
dt
+
dWt
∂t 2 ∂x2
∂x
∂g
=
dWt
∂x
and hence (g(t, Wt ))t∈[0,T ] is a local martingale.
(b) Recall a bounded local martingale is a true martingale. In particular, by the independence
of the increments of Brownian motion, we have
g(t, Wt ) = E[g(T, WT )|Ft ]
= E[G(WT )|Ft ]
= E[G(Wt + WT − Wt )|Ft ]
Z ∞
2
√
e−z /2
=
G(Wt + T − tz) √ dz
2π
−∞
since WT − Wt ∼ N (0, T − t). Since the above formula holds identically, we have the desired
integral representation of the solution of the heat equation.
(c) Let X be the Ornstein–Uhlenbeck process
dXt = Xt dt + dWt .
By Itô’s formula, the process (h(t, Xt ))t∈[0,T ] is a local martingale if and only if h is solution
to the PDE. Now if h is bounded, then (h(t, Xt ))t∈[0,T ] is a true martingale. Using problem
2, we have
h(t, Xt ) = E[cos XT |Ft ]
Z
T −t
= E cos e Xt +
T
T −t−s
e
dWs |Ft
t
= cos(eT −t Xt ) exp(−(e2(T −t) − 1)/4)
since the conditional distribution of XT given Ft is normal with mean eT −t Xt and variance
(e2(T −t) − 1)/2. In particular, the unique solution is
h(t, x) = cos(eT −t x) exp(−(e2(T −t) − 1)/4)
Problem 5. (Strictly local martingale) This is a technical exercise to exhibit a local martingale that is not a true martingale. Let W = (W 1 , W 2 , W 3 ) be a three-dimensional Brownian
motion and let u = (1, 0, 0). It is a fact that P(Wt 6= u for all t ≥ 0) = 1.
4
(a) Let Xt = |Wt − u|−1 . Use Itô’s formula and Lévy’s characterisation of Brownian motion
to show that
dXt = Xt2 dZt , X0 = 1
where Z is a Brownian motion. In particular, show that X is a positive local martingale.
(b) By directly evaluating the integral or otherwise, show that
E(Xt ) = 2Φ(t−1/2 ) − 1
for all t > 0, where Φ is the distribution function of a standard normal random variable.
Why does this imply that X is a strictly local martingale?
Solution 5. (a) Let f (x1 , x2 , x3 ) = ((x1 − 1)2 + x22 + x23 )−1/2 so that
∂f ∂f ∂f
,
,
= −[f (x1 , x2 , x3 )]3 (x1 − 1, x2 , x3 )
∂x1 ∂x2 ∂x3
∂ 2f
∂ 2f
∂ 2f
+
+
= −f 3 + 3f 5 (x1 − 1)2 + −f 3 + 3f 5 x22 − f 3 + 3f 5 x23
2
2
2
∂x1 ∂x2 ∂x3
= 0.
In particular, Itô’s formula yields
dXt = −Xt3 [(Wt1 − 1)dWt1 + Wt2 dWt2 + Wt3 dWt3 ].
Since X can be written as a stochastic integral of a three dimensional Brownian motion, it
is a local martingale. Now let Z be the local martingale such that Z0 = 0 and
dZt = −Xt [(Wt1 − 1)dWt1 + Wt2 dWt2 + Wt3 dWt3 ].
Since
dhZit = Xt2 [(Wt1 − 1)2 + (Wt2 )2 + (Wt3 )2 ]dt
= dt
by construction, the process Z is a Brownian motion by Lévy’s characterisation theorem.
(b) Switch to spherical coordinates:
ZZZ
2
2
2
e−x1 /2−x2 /2−x3 /2
−3/2
q√
dx1 dx2 dx3
E(Xt ) = (2π)
2
2
2
( tx1 − 1) + tx2 + tx3
Z ∞ Z π Z 2π
2
r2 sin θe−r /2
p
= (2π)−3/2
dφ dθ dr
√
r=0 θ=0 φ=0
tr2 − 2 t cos θ + 1
Z ∞Z π
2
r2 sin θe−r /2
−1/2
p
= (2π)
dθ dr
√
r=0 θ=0
tr2 − 2 t cos θ + 1
π
q
Z ∞
√
−1/2
−r2 /2
= (2πt)
re
tr2 − 2 t cos θ + 1 dr
θ=0
Zr=0
∞
√
2
= (2πt)−1/2
2(r1{r>t−1/2 } + tr2 1{r≤t−1/2 } )e−r /2 dr
r=0
Z
=2
0
t−1/2
−r2 /2
e
√
2π
dr
5
Note that E(Xt ) < X0 for all t > 0, so X is a strictly local martingale.
Problem 6. Consider a three asset market with prices given by
dBt
= 2 dt
Bt
(1)
dSt
(1)
= 3 dt + dWt
(1)
St
(2)
dSt
(2)
St
(2)
− 2 dWt
(1)
= 5 dt − 2 dWt
(2)
+ 4 dWt .
Construct an absolute arbitrage.
Solution 6. If the pure investment strategy is decomposed as (φ, π) a good choice for the
holding is stock is given by
!
2
1
, (2)
πt =
(1)
St St
but, of course, it is not unique. It remains to find the holding in the bank account φ. Note
that the wealth X evolves as
dXt = r(Xt − πt · St )dt + πt · dSt
= (2Xt + 5)dt
so the unique solution with X0 = 0 is
5
Xt = (e2t − 1).
2
Since Xt > 0 a.s. for t > 0, this is an arbitrage with the holding in the bank account given
by
φt =
X t − πt · S t
5 − 11e−2t
=
.
Bt
2B0
Problem 7. (Black–Scholes formula) Let X ∼ N (0, 1) be a standard normal random variable, and v and m be positive constants. Express the expectation
√
F (v, m) = E[(e−v/2+
in terms of Φ, the distribution function of X.
6
vX
− m)+ ]
Solution 7.
√
−v/2+ vX
E[(e
+
Z
∞
− m) ] =
(e
√
−v/2+ vx
−∞
Z ∞
−x2 /2
+e
− m) √
2π
dx
2
e−x /2
√ dx
(e
−
m)
=
√ √
2π
log m/ v+ v/2
√
Z ∞
Z ∞
2
2
e−v/2+ vx−x /2
e−x /2
√
√ dx
dx − m
=
√ √
√ √
2π
2π
log m/ v+ v/2
log m/ v+ v/2
Z − log m/√v+√v/2 −s2 /2
Z − log m/√v−√v/2 −t2 /2
e
e
√ ds − m
√ dt
=
2π
2π
−∞
−∞
√ √ log m
log m
v
v
−m Φ − √ −
= Φ − √ +
2
2
v
v
√
−v/2+ vx
Problem 8. Consider a Black–Scholes market with two assets with dynamics given by
dBt = Bt r dt
dSt = St (µ dt + σdWt )
Find a replicating strategy H and the associated wealth process for a claim with payout
(1) ξT = STp for some p ∈ R
(2) ξT = (log ST )2
RT
(3) ξT = 0 Ss ds
Show that your answer to part (3) is unchanged if we only assume that S is a positive Itô
process.
2
Solution 8. Note that ST = St e(r−σ /2)(T −t)+σ(ŴT −Ŵt ) where Ŵt = Wt + (µ − r)t/σ defines a
Brownian motion for the equivalent martingale measure Q.
(1) We could solve the Black–Scholes PDE with boundary condition V (T, S) = S p , but
it is easier to compute conditional expections.
ξt = EQ [e−r(T −t) STp |Ft ] = Stp e(p−1)(r+pσ
2 /2)(T −t)
so that
V (t, S) = S p e(p−1)(r+pσ
2 /2)(T −t)
.
The hedging portfolio is
πt =
∂V
2
(t, St ) = pStp−1 e(p−1)(r+pσ /2)(T −t) .
∂S
(2) Similarly, since
ξt = EQ [e−r(T −t) (log ST )2 |Ft ] = e−r(T −t) {[log St + (r − σ 2 /2)(T − t)]2 + σ 2 (T − t)}
we have
πt = 2e−r(T −t) [log St + (r − σ 2 /2)(T − t)]/St .
7
(3) Finally,
Z
−r(T −t)
ξt = E e
Q
T
−r(T −t)
Z
t
Ss ds|Ft = e
0
Ss ds + St
0
1 − e−r(T −t)
.
r
Unfortunately, the payout is not of the form g(ST ) so our theorem for calculating the
replication portfolio doesn’t apply. Still, let
πt =
1 − e−r(T −t)
r
and
Z
ξt − πt St
1 −rT t
Ss ds.
φt =
=
e
Bt
B0
0
Notice that φt Bt + πt St = ξt by construction and that by a calculation
dξt = φt dBt + πt dSt ,
so (φ, π) is a self-financing replication strategy. To check the self-financing condition
we did not need to use the specific dynamics of S. Indeed, we need only assume that
we can apply Itô’s formula.
Problem 9. (Hull–White extension of Black–Scholes) Consider a market with zero interest
rate r = 0 and with a stock price modelled as
dSt = St (µ dt + σ(t)dWt )
for a given non-random function σ : [0, ∞) → (0, ∞).
(a) Compute the replication cost of a European call option.
(b) A call option is said to be at-the-money if its strike K equals the initial price S0 of the
underlying asset. Explain how you could use your answer to part (a) and quoted time-0
prices of at-the-money calls of different maturities T1 < · · · < Tn to estimate the function σ.
Solution 9. (a) The replication cost is given by
C(T, K) = EQ [(ST − S0 )+ ]
for each maturity
T where Q is the unique equivalent martingale measure with density
Rt
process E(− 0 µ/σ(s)ds). But, note that d log St = − 21 σ(t)2 dt + σ(t)dŴt by Itô’s formula,
where dŴt = dWt + µ/σ(t)dt defines a Q-Browian motion. Since σ is not random, we can
conclude from Problem 1 that
Z
Z t
1 t
2
2
σ(s) ds,
σ(s) ds
log St ∼ N log S0 −
2 0
0
Hence, the local volatility model prices the call with maturity T as
Z T
2
C(T, K) = S0 F
σ(s) ds, K/S0
0
where F is the function from Problem ??.
(b) Note that for K = S0 we have
( " Z
)
1/2 #
t
1
C(T, S0 ) = S0 2Φ
σ(s)2 ds
−1
2
0
8
Note that the at-the-money call prices are increasing in T , assuming that the calls are priced
to be their replication cost.
Hence, we need only choose σ in such a way that
2
Z Ti
C(Ti , S0 ) 1
2
−1
+
σ(s) ds = 2Φ
2S0
2
0
for each i.
Problem 10. (variance swap) Consider a market a stock with price S and interest rate
r = 0. A variance swap is a European contingent claim with payout
2
N X
Stn
.
ξT =
log
Stn−1
n=1
where 0 ≤ t0 < · · · < tN = T are fixed non-random dates. The goal of this exercise is to
show that the variance swap can be replicated approximately.
(a) Let S be a positive Itô process. Confirm the identity
Z T
dSt 1
log(ST /S0 ) =
− hlog SiT .
St
2
0
(b) Confirm the identity
Z
log x = x − 1 −
0
1
(k − x)+
dk −
k2
Z
∞
1
(x − k)+
dk.
k2
(c) Explain how to replicate the variance swap by trading in the stock, cash, and a family of
call and put options of different strikes but all with maturity date T . Show that the number
of shares of the stock varies dynamically but the portfolio of calls and puts is static. To what
extent is this replication strategy independent of the details of the model for S?
Solution 10. (a) Note that by Itô’s formula:
d log St =
dSt dhSit
.
−
St
2St2
and hence
dhlog Sit =
dhSit
.
St2
The conclusion follows.
(b) Since (k − x)+ ≥ 0 only if k ≥ x we have
Z 1
Z 1 (k − x)+
1
x
dk =
−
dk
k2
k k2
0
x∧1
= − log(x ∧ 1) − (1 − x)+
Similarly,
Z
1
∞
Z x∨1 (x − k)+
x
1
dk =
−
dk
k2
k2 k
1
= (x − 1)+ − log(x ∨ 1)
9
(c) By the construction of quadratic variation in lecture, we have
ξT ≈ hlog SiT
Z T
dSt
− 2 log(ST /S0 )
=2
St
0
Z T
Z 1
Z ∞
dSt
PT (T, K)
CT (T, K)
=2
− 2ST + 2(log S0 + 1) + 2
dK
+
2
dK
St
K2
K2
0
0
1
where PT (T, K) = (K − ST )+ is the payout of a put option and CT (T, K) = (ST − K)+ is
the payout of a call. Therefore, a replication strategy is to hold the static position K22 dK
puts of strike K ≤ 1 and K22 dK calls of strike K > 1, and the dynamic position of S2t − 2
shares of the stock at time t.
This replication strategy only requires that S is positive and that Itô’s formula applies, but
is otherwise model-independent. (But don’t forget about notions of admissibility, and even
more fundamentally, that we need to be in a situation where we can safely ignore bid-ask
spread, price impact, transaction costs, etc.)
Problem 11. (strictly local martingale) Consider a market with zero interest rate r = 0
and stock price with dynamics
dSt = St2 dWt .
Consider a European claim with payout ξT = ST .
(a) Show that there exists a trading strategy which replicates the claim with corresponding
wealth ξt = V (t, St ) where
1
√
V (t, S) = S 2Φ
−1 .
S T −t
(b) Consider the strategy of buying S0 claims and selling ξ0 shares. The time 0 wealth is
V0 = 0 and the time T wealth is VT = (S0 −ξ0 )ST > 0. Is this strategy an absolute arbitrage?
Solution 11. (a) It is straight-forward, if a bit tedious, to verify
1 ∂ 2V
∂V
+ S4 2 = 0
∂t
2 ∂S
and limt↑T V (t, S) = S. The replication strategy is given by πt = ∂V
(t, St ) as usual.
∂S
(b) It depends on what we assume about the credit limit process L. For instance, this
candidate arbitrage is not admissible. Indeed, let Vt = S0 ξt − ξ0 St . Note that V is a local
martingale, as it is the linear combination of two local martingales. Now if the strategy
were admissible, the wealth V would be bounded from below and hence a supermartingale.
Therefore we would have to conclude
0 = V0 ≥ E(VT ) = E(S0 ξT − ξ0 ST ) = (S0 − ξ0 )E(ST )
But this contradicts S0 > ξ0 . So in this market it is impossible to lock in the sure future
profit at zero initial cost, because doing so leaves open the possibility that the wealth is goes
negative between times t = 0 and t = T .
Note, however, that there is an admissible arbitrage relative to the asset with price S.
Indeed, the strategy of holding one share of the claim is a relative arbitrage, since the initial
discounted wealth is ξ0 /S0 < 1 and the terminal discounted wealth is ξT /ST = 1. This is
10
only a relative arbitrage since you do not short S, and hence must start with positive initial
wealth.
11