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
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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 Ito 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 Ito ’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 Ito ’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 Ito ’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
