MATH 5050: Homework 4 (Due Monday, April 18)
Problem 1 Consider the heat equation
∂u
(t, x) = ∆u(t, x) + 3t for t > 0 and all x,
∂t
2
with initial heat profile u(0, x) = e−x .
2
(a) According to the lectures, the solution is u(t, x) = E[e−(x+σB(t)+µt) ] where B(t) is
standard Brownian motion, and σ and µ are chosen appropriately, based on the coefficients
of the above heat equation. What are the values of σ and µ?
(b) Plot u(0, x).
(c) Let t = 0.1. What is the distribution of B(0.1)? Numerically compute and then plot
u(0.1, x), for x ∈ [−20, 20].
Hint: You do not need to generate a whole path of Brownian motion. Just need to
generate many (say 10, 000) copies of the variable B = B(0.1). Then let x=-20:0.01:20
2
and calculate the average of e−(x+σB+0.1µ) to get the vector of u(0.1, x). [Make sure the
average is over the samples, not over the x’s!]
(d) Repeat the above for t = 1 and t = 10. Do plots look like what you expect? Explain.
Problem 2 Let B(t) be standard Brownian motion. Let Y (t) take the value 1 if 0 ≤ t < 2
and the value 3B(2) if t ≥ 2. Prove that
Z t
M (t) =
Y (s) dB(s)
0
is a martingale relative to the filtration given by Ft = {B(s) : s ≤ t}.
Hint: M (t) can be easily computed. Consider the cases t < 2 and t ≥ 2 separately. After
that you need to prove that for s < t we have E[M (t)|Fs ] = M (s). For this, consider the
cases s < t < 2, 2 ≤ s < t, and s < 2 ≤ t separately.
Problem 3 Let B(t) be standard Brownian motion.
Rt
(a) Calculate numerically 0 B(s) dB(s) for 0 ≤ t ≤ 10. Use the leftmost points of the
partition intervals and then the rightmost points. Plot both solutions (using the same copy
of Brownian motion, not two independent copies).
(b) Based on the plots, which of the two is the right one and why?
Rt
(c) Calculate 0 B(s) dB(s) explicitly (not numerically).
Hint: Use Ito’s formula:
Z
f (B(t)) = f (B(0)) +
t
f 0 (B(s)) dB(s) +
0
1
2
Z
t
f 00 (B(s)) ds
0
with f (x) = x2 .
(d) If you use the same Brownian motion for the plots in (a) as well as to plot the explicit
solution you got in (c), then the latter should match one of the former two. Is this the
case? Does this agree with your answer to (b)?
1
2
Problem 4 Let B(t) be standard Brownian motion. Let λ be a positive constant and let
X(t) = eλB(t) .
(a) Apply Ito’s formula from Problem 3(c) to the function f (x) = eλx to prove that X(t)
solves
Z
Z t
λ2 t
X(t) = 1 +
X(s) ds + λ
X(s) dB(s).
2 0
0
(b) Let A(t) = E[X(t)]. Observe that A(0) = 1. Using the fact that a stochastic integral
2 Rt
is a martingale prove that A(t) = 1 + λ2 0 A(s) ds.
(c) The equation in (b) says that A0 (t)/A(t) =
1) to find the formula for A(t).
λ2
2 .
Integrate both sides (and use A(0) =
(d) What is the distribution of B(t)? Can you compute E[eλB(t) ] directly? Does your
answer agree with (c)?
Hint: Recall moment generating functions!
Recall the following variant of Ito’s formula: Let dZ(t) = X(t)dt + Y (t)dB(t), let f (t, x)
have two continuous derivatives in x and one continuous derivative in t, and let U (t) =
f (t, Z(t)). Let f˙ (notice the dot!) denote the derivative in the t variable and f 0 and f 00
the derivatives in the x variable. Then a Taylor expansion (and ignoring terms that have
something smaller than ds) gives
Z t
Z
Z t
1 t 00
0
˙
f (s, Z(s)) (dZ(s))2 .
U (t) − U (0) =
f (s, Z(s)) ds +
f (s, Z(s)) dZ(s) +
2
0
0
0
Replacing dZ(s) by X(s)ds + Y (s)dB(s) and ignoring again terms that have something
smaller than ds (like (ds)2 or ds dB(s)), then recalling that (dB(s))2 = ds we get
Z t
Z th
i
1
f˙(s, Z(s))+f 0 (s, Z(s))X(s)+ f 00 (s, Z(s))(Y (s))2 ds + f 0 (s, Z(s))Y (s) dB(s) .
U (t)−U (0) =
2
0
0
2 t/2
Problem 5 Let B(t) be a standard Brownian motion. Prove that M (t) = eλB(t)−a
martingale in the filtration of Brownian motion Ft = {B(s) : s ≤ t}.
2
is a
Hint: Apply Ito’s formula to Z(t) = B(t) and f (t, x) = eλx−λ t/2 to prove that M (t)
is equal to a stochastic integral. An alternate way would be by computing E[M (t) | B(s)]
using moment generating functions. It is a good exercise that way too, but it does not let
you practice your Ito’s formula skills! So maybe it is best to do it both ways!