Student
Number
Semester 1 Assessment, 2019
School of Mathematics and Statistics
MAST90059 Stochastic Calculus with Applications
Writing time: 3 hours
Reading time: 15 minutes
This is NOT an open book exam
This paper consists of 8 pages (including this page)
Authorised Materials
• Mobile phones, smart watches and internet or communication devices are forbidden.
• Calculators, tablet devices or computers must not be used.
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Instructions to Students
• You must NOT remove this question paper at the conclusion of the examination.
• This paper has 6 questions. Attempt as many questions, or parts of questions, as you
can. The number of marks allocated to each question is shown in the brackets after the
question statement. There are 80 total marks available for this examination. Working
and/or reasoning must be given to obtain full credit. Clarity, neatness and style count.
Instructions to Invigilators
• Students must NOT remove this question paper at the conclusion of the examination.
This paper must NOT be held in the Baillieu Library
Blank page (ignored in page numbering)
MAST90059 Semester 1, 2019
1. Let f : R → R be continuous with finite variation, and g : R → R be twice continuously
differentiable. Using the definition of Riemann-Stieltjes integral, show that for any t > 0,
g(f(t)) = g(f(0)) +
∫ t
0
g′(f(s))df(s).
[8 marks]
Sol: For a given partition {ti} of [0, t], Taylor’s theorem implies
g(f(t))− g(f(0)) =
∑
i
g(f(ti))− g(f(ti−1))
=
∑
i
g′(f(ti−1))(f(ti)− f(ti−1)) + 1
2
∑
i
g′′(f∗i )(f(ti)− f(ti−1))2,
where f∗i is in the interval with endpoints f(ti−1), f(ti). Since f is continuous,
sup
s∈[0,t]
|f(s)| <∞,
and, since g′′ is also continuous,
Kt = sup
s∈[0,t]
|g′′(f(s))| <∞.
Thus ∣∣∣∣∣∑
i
g′′(f∗i )(f(ti)− f(ti−1))2
∣∣∣∣∣ ≤ KtVf (t) supi |f(ti)− f(ti−1)| → 0,
as ‖{ti}‖ → 0.
From the definition of R-S integral:∫ t
0
g′(f(s))df(s) = lim
n→∞
∑
i
g′(f(tni−1))(f(t
n
i )− f(tni−1)),
where ({tni })n≥1 is any sequence of partitions of [0, t] with ‖{tni }‖ → 0 as n→∞.
Combining the above yields the result.
2. Let (Bt)t≥0 be a standard Brownian motion and Mt = max0≤s≤tBs.
(a) Derive the joint density function of Bt and Mt.
(b) Derive the density of Mt.
(c) Derive the joint density function of −Bt and Mt −Bt.
(d) For x < 0, find
lim
ε→0+
P(Bt ≤ x|Mt ≤ ε).
(e) Show that Xt =
∫ t
0 e
udBu exists for each t > 0, and derive and identify by name the
joint distribution of (Xs, Xt) for 0 < s < t.
(f) Find the stochastic differential of Xt = e
2Bt+3t2 .
(g) Find the stochastic exponential of B3t .
Page 2 of 8 pages
MAST90059 Semester 1, 2019
[27 marks]
Sol:
(a) (6 marks) Define Ty = inf{t : Bt = y}, then for y > x, y > 0,
P(Mt ≥ y,Bt ≤ x) = P(Ty ≤ t, Bt ≤ x) =
∫ t
0
P(Bt ≤ x|Ty = s)P(Ty ∈ ds)
=
∫ t
0
P(Bt ≥ 2y − x|Ty = s)P(Ty ∈ ds) (reflection principle)
= P(Bt ≥ 2y − x, Ty ≤ t)
= P(Bt ≥ 2y − x) since Bt ≥ 2y − x implies Ty ≤ t
= 1− Φ
(
2y − x√
t
)
,
where Φ is the cdf of N(0, 1). Thus,
fMt,Bt(y, x) = −
∂2
∂y∂x
P(Mt ≥ y,Bt ≤ x)
=
√
2
pi
(2y − x)
t3/2
e−
(2y−x)2
2t , y > x, y > 0.
(b) (2 marks) Integrating out x from the joint density:∫
fMt,Bt(y, x)dx =
√
2
pi
∫ y
−∞
(2y − x)
t3/2
e−
(2y−x)2
2t dx =
√
2
pit
e−
y2
2t , y > 0.
(c) (2 marks) Making the change of variable u = −x and v = y − x, we find the density
is the same as part (a).
(d) (4 marks) From the definition of conditional probability,
P(Bt ≤ x|Mt ≤ ε) = P(Bt ≤ x,Mt ≤ ε)P(Mt ≤ ε)
=
∫ ε
0
∫ x
−∞ fMt,Bt(y, u)dudy∫ ε
0 fMt(y)dy
=
ε
∫ x
−∞ fMt,Bt(0, u)du+ o(ε)
εfMt(0) + o(ε)
,
where we have used Taylor’s theorem in the last line. Thus
lim
ε→0+
P(Bt ≤ x|Mt ≤ ε) =
∫ x
−∞
fMt,Bt(0, u)
fMt(0)
du
=
∫ x
−∞
√
2
pi
−u
t3/2
e−
u2
2t√
2
pit
du
=
∫ x
−∞
−u
t
e−
u2
2t du
= e−
x2
2t ,
where in the second equality we use part (b).
Page 3 of 8 pages
MAST90059 Semester 1, 2019
(e) (4 marks) Xt exists for all t by Itoˆ isometry since
∫ t
0 e
2udu = 12(e
2t − 1) < ∞. From
class, we know that (Xu)0≤u≤t is a centred Gaussian process with independent increments,
so (Xs, Xt) is bivariate normal with covariance
Cov(Xs, Xt) = Cov(Xs, Xs) + Cov(Xs, Xt −Xs) = 1
2
(e2s − 1).
(f) (4 marks) Using Itoˆ’s formula,
dXt = e
2Bt+3t2 ((2 + 6t)dt+ 2dBt) .
(g) (5 marks) In general, E(X)t = eXt−X0− 12 [X]t . So we need to find [B3]t. Using Itoˆ’s
formula, we have
d(B3)t = 3B
2
t dBt + 3Btdt,
and so
[B3]t = 9
∫ t
0
B4sds.
Thus
E(B3t ) = eB
3
t− 92
∫ t
0 B
4
sds.
3. Let (Bt)t≥0 be a standard Brownian motion and let (Xt)t≥0 satisfy Xt > 0 and
dXt =
1
2
(
1
Xt
− 1
)
dt+ dBt.
(a) Fix α > 0. Find an SDE for the process (Yt)t≥0 = (Xαt )t≥0 and write down the
generator of (Yt)t≥0.
(b) For c > 0, let Tc = inf{t > 0 : Xt = c}. Find a function g : (0,∞) → (0,∞) such
that for 0 < a < x < b
P(Ta < Tb|X0 = x) =
∫ b
x g(u)du∫ b
a g(u)du
.
(c) Find a stationary distribution for (Xt)t≥0 and identify it by name.
[18 marks]
Sol:
(a) (6 marks) Using Itoˆ’s formula:
dYt = d(X
α)t =
1
2
(
αXα−1t
(
1
Xt
− 1
)
+ α(α− 1)Xα−2t
)
dt+ αXα−1t dBt
=
α
2
(
αXα−2t −Xα−1t
)
dt+ αXα−1t dBt
=
α
2
(
αY
(α−2)/α
t − Y (α−1)/αt
)
dt+ αY
(α−1)/α
t dBt
The generator L is given by
Lf(y) =
α
2
(
αy(α−2)/α − y(α−1)/α
)
f ′(y) +
1
2
α2y2(α−1)/αf ′′(y).
Page 4 of 8 pages
MAST90059 Semester 1, 2019
(b) (6 marks) A scale function for (Xt)t≥0 is given by
S(x) =
∫ x
exp
{
−2
∫ u
µ(y)dy
}
du
=
∫ x
exp
{
−
∫ u
(y−1 − 1)dy
}
du
=
∫ x eu
u
du.
Since
P(Ta < Tb|X0 = x) = S(b)− S(x)
S(b)− S(a) ,
we can take g(u) = eu/u.
(c) (6 marks) If we can find a probability density pi satisfying the PDE
∂
∂y
(
µ(y)pi(y)− 1
2
∂
∂y
(σ(y)pi(y))
)
= 0,
then pi is the density of a stationary distribution for (Xt)t≥0. This is the same as(
y−1 − 1)pi(y)− pi′(y) = C,
for some constant C. Trying C = 0 (for the probability solution), we find
pi(y) = cye−y, y > 0.
Thus the stationary distribution is Gamma(2, 1) and c = 1.
4. Let (Bt)t≥0 be a standard Brownian motion, let (Xt)t≥0 satisfy X0 = x > 0 and
dXt = X
1/2
t dBt,
and define Mt(θ) = E[eθXt ].
(a) Assuming integrability as needed, show that Mt(θ) satisfies
∂
∂t
Mt(θ) =
θ2
2
M ′t(θ).
(b) Assuming there are constants a, b such that Mt(θ) = e
θ/(aθt+b), find a and b.
[10 marks]
Sol:
(a) (4 marks) Use Dynkin’s formula (or Itoˆ and integrate) to find
Mt(θ) = M0(θ) +
θ2
2
∫ t
0
E
[
Xse
θXs
]
ds
= M0(θ) +
θ2
2
∫ t
0
M ′s(θ)ds
Differentiating both sides with respect to t gives the result.
(b) (6 marks) Since Mt(θ) = e
θ/(aθt+b), we have M0(θ) = e
θ/b. But the initial condition
implies M0(θ) = e
θx, so b = 1/x.
Now, aiming to use the PDE, we compute
∂
∂t
Mt(θ) =
−aθ2
(aθt+ b)2
Mt(θ),
θ2
2
M ′t(θ) =
1
2bθ
2
(aθt+ b)2
Mt(θ),
and equating implies a = −b/2 = −1/(2x).
Page 5 of 8 pages
MAST90059 Semester 1, 2019
5. A one period finite market model has riskless bond (βi)i=0,1 with βi = r
i for some r > 0,
and an asset price process (Si)i=0,1 with S1 ∈ {dS0, S0, uS0} for some d < 1 < u.
(a) Show that the market model has no arbitrage opportunities if and only if d < r < u.
(b) Determine for which values of r ∈ (d, u) the market model is complete.
[9 marks]
Sol:
(a) (7 marks) The first fundamental theorem says that there are no arbitrage opportuni-
ties if and only if there is an EMM Q such that (Si/βi)i=0,1 is a martingale under Q.
In this case, this is the same as saying
EQ[S1|S0] = rS0.
Letting pu, pn, pd be the probabilities of the three possible outcomes for S1, the
equation above becomes
S0(upu + (1− pu − pd) + dpd) = rS0,
and we solve
(u− 1)pu + (d− 1)pd = r − 1,
with 0 ≤ pu, pd and pu + pd < 1. The equation gives
pu =
r − 1
u− 1 −
d− 1
u− 1pd,
so that the inequality pu + pd < 1 becomes
r − 1
u− 1 +
u− d
u− 1pd < 1,
which is the same as
pd <
u− r
u− d,
which can only be satisfied for pd > 0 if r < u. Similarly, starting from
pd =
r − 1
d− 1 −
u− 1
d− 1pr,
gives
pu <
r − d
u− d,
which can only be satisfied if r > d. Thus we have shown that if r ≥ u or r ≤ d, then
there is no EMM and thus arbitrage opportunities exist. (Alternatively, an explicit
arbitrage if r ≤ d is to borrow money and buy stock, and if r ≥ u, shorting stock
and investing in bonds gives arbitrage).
Reversing the calculations above shows that if d < r < u, then
pu =
r − 1
u− 1 +
1− d
u− 1pd,
pn =
u− r
u− 1 −
u− d
u− 1pd,
gives an EMM for any 1−r1−d ∨ 0 < pd < u−ru−d .
Page 6 of 8 pages
MAST90059 Semester 1, 2019
(b) (2 marks) The market model is complete if and only if the EMM is unique. From
the previous part, there is always more than one EMM for the range of r, and so the
model is not complete.
6. The Black-Sholes model has riskless bond (ert)t≥0 for some r > 0 and risky asset (St)t≥0
given by dSt = µStdt+ σStdBt for some µ ∈ R, σ > 0 and (Bt)t≥0 a P-Brownian motion.
Find the price Ct at time 0 ≤ t ≤ T of the claim X that pays at time T :
X =
{
C ST ≥ K,
0 ST < K,
where C,K are positive parameters. Your answer should be in terms of St, µ, σ, r, C,K, T, t
and the CDF Φ of the standard normal distribution.
[8 marks]
Sol: The Black-Sholes model has EMM Q such that
ST = Ste
σ(WT−Wt)+(r−σ2/2)(T−t),
for (Wt)0≤t≤T a Q-Brownian motion. Using the general no arbitrage pricing formula, then
the Markov property of Brownian motion, we have
e−r(T−t)EQ[X|Ft] = Ce−r(T−t)Q(ST ≥ K|St)
= Ce−r(T−t)Q
(
WT −Wt ≥ log(K/St)− (r − σ
2/2)(T − t)
σ
∣∣St)
= Ce−r(T−t)
(
1− Φ
(
log(K/St)− (r − σ2/2)(T − t)
σ
√
T − t
))
= Ce−r(T−t)Φ
(
log(St/K) + (r − σ2/2)(T − t)
σ
√
T − t
)
.
Page 7 of 8 pages
MAST90059 Semester 1, 2019
Useful formulas and facts:
• If X ∼ N(0, 1), then, for an even positive integer p,
EXp = (p− 1)!! : s = (p− 1)× (p− 3)× · · · × 1.
• The moment generating function of X ∼ N(µ, σ2) is MX(t) = eµt+ 12σ2t2 .
• Feynman-Kac formula: For given bounded measurable functions r(x, t) and g(x) let
C(x, t) = E
(
e−
∫ T
t r(Xu,u)dug(XT )
∣∣∣Xt = x) , 0 ≤ t ≤ T,
where {Xt} is a diffusion with generator Lt. Assume there is a solution to
∂f
∂t
(x, t) + Ltf(x, t) = r(x, t)f(x, t), 0 ≤ t ≤ T ; f(x, T ) = g(x).
Then the solution is unique and is given by C(x, t).
End of Exam
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