ST3180

UNIVERSITY OF WARWICK

Third Year EXAMINATIONS: SUMMER 2020

Probability Theory

Time allowed: 2 hours

Approved calculators may be used.

Full marks may be obtained by correctly answering 3 complete questions. Candi-

dates may attempt all questions. Marks will be awarded for the best 3 answers

only.

1

1: (a) State and prove the first and second Borel-Cantelli lemmas. [6]

For the rest of this question you may use, without proof, Kolmogorov’s zero one law.

(b) Let (Xn)n≥1 be a sequence of independent random variables on probability space (Ω,F ,P) taking

values in {1, 2, 3, . . .}. Suppose that P(Xn ≥ i) = 1/i for each n and each i ∈ {1, 2, 3, . . .}.

(i) Calculate

P(Xn ≥ nα i.o.)

for each fixed α > 0.

(ii) Show that the random variable

lim sup

n→∞

log(Xn)

log(n)

is almost surely constant, and find the value of this constant.

[8]

(b) Suppose (Xn)n≥1 is a sequence of independent identically distributed random variables such that

P(Xn = 1) = P(Xn = −1) = 1

2

for each n ∈ N .

Let Sn =

∑n

k=1Xk, and define

B− =

{

lim inf

n→∞ Sn = −∞

}

and B+ =

{

lim sup

n→∞

SN =∞

}

.

(i) Show that both B− and B+ belong to the tail σ-algebra of (Xn)n≥1, and that P(B+) =

P(B−) ∈ {0, 1}.

(ii) Using the Borel-Cantelli lemmas, show that for each k ≥ 1

lim sup

n→∞

(

Sn+k − Sn

)

= k a.s.

[Hint: Consider An = {Sn+k − Sn = k}.]

(iii) Deduce that P(Bc+ ∩Bc−) = 0 and hence P(B+) = P(B−) = 1.

[6]

[TOTAL: 20]

Continued...

2

2: (a) Suppose that (Xn)n≥1 is a sequence of random variables with |Xn| ≤ Y for each n, where Y ∈ L1

(i.e. E[|Y |] <∞). Show that {Xn}n≥1 is uniformly integrable. [4]

(b) Suppose that (Xn)n≥1 is a sequence of random variables converging to zero in probability. Fur-

ther, suppose that there exists a constant K ∈ (0,∞) such that E[|Xn|3] ≤ K for each n. Show

that Xn → 0 in L2.

(You may use results about uniform integrability so long as you state them clearly) [4]

(c) Suppose that (Xn)n≥1 is a sequence of independent normally distributed random variables, with

common mean µ and common variance 1. Let Zn = exp(X1 + . . .+Xn).

(i) Show that Zn → 0 in L2 if and only if µ < −1.

(Hint: If W is a standard normal random variable then E(eθW ) = eθ2/2.)

(ii) Show that Zn → 0 in probability if µ < 0.

[5]

(d) Let (Xn)n≥2 be a sequence of independent random variables such that for each n ∈ {2, 3 . . .}

P(Xn = n) = P(Xn = −n) = 1

2n log n

, and P(Xn = 0) = 1− 1

n log n

.

Let Sn = X2 + . . .+Xn. Show that

Sn

n → 0 in probability but not almost surely.

(Hint: You may use without proof that

∑

n

1

n logn =∞.) [7]

[TOTAL: 20]

Continued...

3

3: (a) Suppose (Xn)n≥1 is a sequence of random variables and X is a random variable all on probability

space (Ω,F ,P). Let µn, Fn be the law and distribution function of Xn respectively and µ, F be

the law and distribution function of X respectively.

(i) Suppose that µn → µ weakly; show that this implies Xn → X in distribution.

(ii) Suppose that Xn → X in distribution, and let (an)n≥1 be a sequence of real numbers such

that an → 0 as n→∞. Show that anXn converges in distribution to zero.

[8]

(b) Let µX be the law of a random variable X. Show that the distribution of X is symmetric, i.e.

µX(−∞, x] = µX [−x,∞) for all x ∈ R, if and only if the characteristic function of X is real. [4]

(c) In the following, identify if the sequence of random variables (Yn)n≥1 converges weakly and if so

identify the limit. (You should explain your reasoning, and state clearly any results from lectures

that you use).

(i) Yn = max{U1, . . . , Un} where U1, U2, . . . are independent Uniform(−1, 1) random variables.

(ii) Yn = n(1 − max{U1, . . . , Un}) where U1, U2, . . . are independent Uniform(−1, 1) random

variables.

(iii) Yn =

√

3

n(U1 + . . .+Un) where U1, U2, . . . are independent Uniform(−1, 1) random variables.

(iv) Yn = nmin{U1, . . . , Un} where U1, U2, . . . are independent Uniform[0, 1] random variables.

(v) Yn is an exponential random variable with mean λn > 0 for each n, i.e. P(Yn > y) = e−y/λn ,

where λn → 0 as n→∞.

[8]

[TOTAL: 20]

Continued...

4

4: (a) Let X and Y and Z be integrable random variables on a probability space (Ω,F ,P), and let

G ⊆ F be a sub-σ-algebra. Show that if Y and Z are both versions of E[X | G] then Y = Z

almost surely. [3]

(b) Suppose that (Xn)n≥1 is a random process adapted to (Fn)n≥1. Define the hitting time τA of

(Xn)n≥1 on a Borel set A. Show that τA is a stopping time for this filtration. [3]

(c) Supose (Xn)n≥1 is a symmetric simple random walk on Z started from 0. Use the Optional

Stopping Theorem for Bounded Stopping Times to evaluate the probability (Xn)n≥1 hits b > 0

before −a < 0. [4]

(d) Let (Xn)n≥1 be a sequence of independent random variables on probability space (Ω,F ,P), with

E[Xn] = 1 for each n. Let F0 = {∅,Ω}, Fn = σ(X1, . . . , Xn) for n ∈ N, and

M0 = 1, and Mn =

n∏

k=1

Xk for n = 1, 2, . . . .

(i) Show that (Mn)n≥0 is a martingale with respect to (Fn)n≥0.

(ii) Now suppose ϕ(t) = E[etX1 ] <∞ for all t ∈ R. Let S0 = 0, Sn =

∑n

k=1Xk, and

Yn =

etSn

ϕ(t)n

, for n = 0, 1, 2, . . . .

Show, using part (i), that (Yn)n≥0 is a martingale with respect to (Fn)n≥0.

[4]

(c) Let (εn)n≥1 be independent random variables with

P(εn = +1) = p, P(εn = −1) = q, where 1/2 < p = 1− q < 1 .

Let Fn = σ(ε1, . . . , εn) and define Xn inductively by X0 = 1 and for n ≥ 1

Xn = Xn−1 + Vnεn .

Assume Vn is a Fn−1-measurable random variable for each n, and that Vn is strictly between 0

and Xn.

(i) Prove that if Xn > 0 then E[log(Xn+1/Xn) | Fn] = f(Vn+1/Xn) where

f(x) = p log(1 + x) + q log(1− x) .

(ii) Deduce that (log(Xn)− nα)n≥0 is a supermartingale, where

α = p log p+ q log q + log 2 .

[6]

[TOTAL: 20]

End.

5

ST3180

UNIVERSITY OF WARWICK

Third Year EXAMINATIONS: SUMMER 2020

Probability Theory

Time allowed: 2 hours

Approved calculators may be used.

Full marks may be obtained by correctly answering 3 complete questions. Candi-

dates may attempt all questions. Marks will be awarded for the best 3 answers

only.

1

1: (a) State and prove the first and second Borel-Cantelli lemmas. [6]

Answer: (Applied Bookwork:)

(Borel-Cantelli Lemmas) Let (An)n≥1 be a sequence of events on probability space (Ω,F ,P). Then,

(BC1) If

∑

n P(An) <∞ then P(lim supAn) = P(An i.o.) = 0,

(BC2) If (An)n≥1 are also independent then

∑

n P(An) = ∞ implies P(lim supAn) = P(An i.o.) = 1 (i.e.

the converse of (BC1) holds under independence).

(1 marks)

For (BC1) we assume that

∑

n P(An) < ∞. Let Gn =

⋃

m≥nAm, which is a decreasing sequence of sets

such that Gn ↘ G where G = lim supn→∞An. Fix k ∈ N and observe

P(lim sup

n→∞

An) = P

( ⋂

n≥1

Gn

) ≤ P(Gk) ≤∑

n≥k

P(An) ,

where the final inequality holds by σ-subadditivity. By assumption the right hand side converges to zero as

k →∞, which concludes the proof of (BC1).

(2 marks)

For (BC2) we assume that (An)n≥1 are independent and

∑

n P(An) =∞. The main idea of the proof is to

take compliments and use the standard bound 1− x ≤ e−x for x ∈ R. Fix m, r ∈ N, then by independence

P

( ⋂

m≤n

Acn

) ≤ P( ⋂

m≤n≤r

Acn

)

=

∏

m≤n≤r

P(Acn) =

∏

m≤n≤r

(

1− P(An)

)

,

Applying the bound 1− x ≤ e−x, we find

P

( ⋂

m≤n

Acn

) ≤ e−∑m≤n≤r P(An) → 0 as r →∞ ,

where convergence follows by assumption on

∑

n P(An). It follows that, since {Acn ev.} is a countable

union of null sets, that P(An i.o.) = 1− P(Acn ev.) = 1, as required.

(3 marks)

For the rest of this question you may use, without proof, Kolmogorov’s zero one law.

(b) Let (Xn)n≥1 be a sequence of independent random variables on probability space (Ω,F ,P) taking

values in {1, 2, 3, . . .}. Suppose that P(Xn ≥ i) = 1/i for each n and each i ∈ {1, 2, 3, . . .}.

(i) Calculate

P(Xn ≥ nα i.o.)

for each fixed α > 0.

(ii) Show that the random variable

lim sup

n→∞

log(Xn)

log(n)

is almost surely constant, and find the value of this constant.

[8]

Answer: (Unseen (seen similar example):)

(i) Apply (BC1) and (BC2) above with An = {Xn ≥ nα}. Observe

∑

n

P(An) =

∑

n

1

dnαe

{

=∞ if α ≤ 1 ,

<∞ if α > 1 .

2

So (since the Xn are independent) the Borel-Cantelli lemmas imply

P(Xn ≥ nα i.o.) =

{

1 if α ≤ 1 ,

0 if α > 1 .

(3marks)

(ii) Let Fn = σ(Xn), then (Fn)n≥1 is an independent sequence of σ-algebras. Let Ln = supm≥n logXmlogm

and L = limn→∞ Ln. Since Ln is σ

(⋃

m≥n Fm

)

-measurable we have L is T -measurable, in particular

L is almost surely constant by Kolmogorov’s zero-one law. Observe {L ≥ 1} ⊇ {Ln ≥ 1 i.o.} ⊇

{Xn ≥ n i.o.}. It follows that P(L ≥ 1) = 1. Now fix k ∈ N,

{L > 1 + 2

k

} ⊆ {L ≥ 1 + 2

k

} ⊆

⋂

m

{Ln > 1 + 2

k

− 1

m

i.o.} ⊆ {Ln > 1 + 1

k

i.o.} .

By definition {Ln > 1 + 1k i.o.} ⊆ {Xn ≥ n1+1/k i.o}, and by part (i) P(Xn ≥ n1+1/k i.o) = 0. The

conclusion follows, since

{L > 1} =

⋃

k

{

L > 1 +

1

k

}

,

and a countable union of null sets is null. (5 marks)

(c) Suppose (Xn)n≥1 is a sequence of independent identically distributed random variables such that

P(Xn = 1) = P(Xn = −1) = 1

2

for each n ∈ N .

Let Sn =

∑n

k=1Xk, and define

B− =

{

lim inf

n→∞ Sn = −∞

}

and B+ =

{

lim sup

n→∞

SN =∞

}

.

(i) Show that both B− and B+ belong to the tail σ-algebra of (Xn)n≥1, and that P(B+) =

P(B−) ∈ {0, 1}.

(ii) Using the Borel-Cantelli lemmas, show that for each k ≥ 1

lim sup

n→∞

(

Sn+k − Sn

)

= k a.s.

[Hint: Consider An = {Sn+k − Sn = k}.]

(iii) Deduce that P(Bc+ ∩Bc−) = 0 and hence P(B+) = P(B−) = 1.

[6]

Answer: (Unseen (seen similar):)

(i) Notice that, for each N ∈ N fixed

lim inf

n→∞ Sn = −∞ if and only if lim infn→∞ (XN + . . . Xn) = −∞ ,

and hence B− is measurable with respect to σ(XN , XN+1, . . .) for each N ∈ N, and therefore B− is

in the tail σ-algebra. The same argument applies to B+. It follows from Kolmogorov’s 0-1 law that

P(B−),P(B+) ∈ {0, 1}. Finally, by symmetry (Xn)n≥1 and (−Xn)n≥1 have the same distribution and

hence so do (Sn) and (−Sn). It follows that

P(B+) = P(lim sup(−Sn) =∞) = P(− lim inf Sn =∞) = P(B−) .

(2 marks)

3

(ii) Fix k ≥ 1, consider An as defined in the hint. Then

P(An) = P(Xn+1 = 1, . . . , Xn+k = 1) =

1

2k

for each n ∈ N .

It follows that

∑

n≥1 P(Ank) =∞, since the {Ank}n≥1 are independent (BC2) implies P({Ank}n≥1 i.o.) =

1. Together with the fact that (Sn+k − Sn) ≤ k for each n we conclude that lim sup(Sn+k − Sn) = k

almost surely. (2 marks)

(iii) Since

(Bc+ ∩Bc−) ⊆ {(Sn) bounded above and below} ⊆

∞⋃

k=1

{

lim sup(Sn+k − Sn) = k

}c

,

and a countable union of null sets is null, we have P(Bc+ ∩Bc−) = 0. Combining with part (i) we get

P(B+) = P(B−) = 1 as required. (2 marks)

[TOTAL: 20]

Continued...

4

2: (a) Suppose that (Xn)n≥1 is a sequence of random variables with |Xn| ≤ Y for each n, where Y ∈ L1

(i.e. E[|Y |] <∞). Show that {Xn}n≥1 is uniformly integrable. [4]

Answer: (Applied Bookwork) Observer, by assumptions E(|Xn|1|Xn|≥K) ≤ E(|Y |1|Xn|≥K) ≤ E(|Y |1|Y |≥K)

and the right hand side tends to zero (uniformly in n), by dominated convergence, since Y ∈ L1. For full

marks it must be made clear that convergence is uniform in n. (4 marks)

(b) Suppose that (Xn)n≥1 is a sequence of random variables converging to zero in probability. Fur-

ther, suppose that there exists a constant K ∈ (0,∞) such that E[|Xn|3] ≤ K for each n. Show

that Xn → 0 in L2.

(You may use results about uniform integrability so long as you state them clearly) [4]

Answer: (Applied Bookwork) We know from lectures that if (Yn)n≥1 is a sequence of random variables

bounded in Lp for p > 1 then {Yn}n≥1 are uniformly integrable. Now let Yn = X2n for each n, this forms a

sequence which is bounded in L3/2 and hence is U.I.. Also from lectures, a U.I. sequence which converges

in probability converges in L1. This implies (Xn)n≥1 converges in L2. (Alternative proofs, for example

using law of total expectation are also acceptable). (4 marks)

(c) Suppose that (Xn)n≥1 is a sequence of independent normally distributed random variables, with

common mean µ and common variance 1. Let Zn = exp(X1 + . . .+Xn).

(i) Show that Zn → 0 in L2 if and only if µ < −1.

(Hint: If W is a standard normal random variable then E(eθW ) = eθ2/2.)

(ii) Show that Zn → 0 in probability if µ < 0.

[5]

Answer: (Unseen example) (i) Consider

E(Z2n) = E[exp(2(X1 + . . .+Xn))] =

(

E[exp(2(X1 − µ))]

)n

e2nµ ,

and apply the hint.

(ii) Either by considering E[( 1n (X1 + . . . + Xn))] = 1/n → 0 as n → ∞ (convergence in L2 implies in

prob.), or by the Weak Law of Large Numbers, 1n (X1 + . . . + Xn) to µ in probability as n → ∞. Hence

P[X1 + . . .+Xn ≥ n(µ+ε)]→ 0 for each ε > 0. Fix ε > 0 such that µ+ε < 0 (possible since µ < 0). Now

P[Zn ≥ en(µ+ε)]→ 0 and en(µ+ε) → 0 as n→∞, which implies convergence in probability. (5 marks)

(d) Let (Xn)n≥2 be a sequence of independent random variables such that for each n ∈ {2, 3 . . .}

P(Xn = n) = P(Xn = −n) = 1

2n log n

, and P(Xn = 0) = 1− 1

n log n

.

Let Sn = X2 + . . .+Xn. Show that

Sn

n → 0 in probability but not almost surely.

(Hint: You may use without proof that

∑

n

1

n logn =∞.) [7]

Answer: (Unseen example) Fix ε > 0 and consider P(|Sn/n| > ε), since Sn is mean zero we can apply

Chebyshev’s inequality. Hence consider the variance of Sn,

Var(Sn) =

n∑

k=2

Var(Xk) =

n∑

k=2

E(X2k) =

n∑

k=2

k

log k

≤ (n− 2)n

log n

+

2

log 2

,

where in the final line we used x 7→ x/ log x is increasing for x ≥ 3 > e. So by Chebyshev’s inequality

P(|Sn/n| > ε) ≤ E(S

2

n)

ε2n2

≤ 1

ε2 log n

+

2

ε2n2 log 2

→ 0 ,

5

as n→∞, as required. For (lack of) almost sure convergence, consider An = {Xn = n} and Bn = {Xn =

−n}, then ∑

n≥2

P(An) =

∑

n≥2

P(Bn) =

∑

n≥2

1

n log n

=∞ ,

so by (BC2) P(An i.o.) = P(Bn i.o.) = 1, so in particular Sn can not converge almost surely. (7 marks)

[TOTAL: 20]

Continued...

6

3: (a) Suppose (Xn)n≥1 is a sequence of random variables and X is a random variable all on probability

space (Ω,F ,P). Let µn, Fn be the law and distribution function of Xn respectively and µ, F be

the law and distribution function of X respectively.

(i) Suppose that µn → µ weakly; show that this implies Xn → X in distribution.

(ii) Suppose that Xn → X in distribution, and let (an)n≥1 be a sequence of real numbers such

that an → 0 as n→∞. Show that anXn converges in distribution to zero.

[8]

Answer:

(i) (Applied Bookwork) Fix x ∈ R a continuity point of F , and δ > 0. Define hx ∈ Cb(R) by

hx(y) =

1 if y ≤ x ,

1− y−xδ if y ∈ (x, x+ δ) ,

0 if y ≥ x+ δ .

Then by assumption µn(hx)→ µ(hx) as n→∞, and by construction of hx we have

Fn(x) ≤ µn(hx) and µ(hx) ≤ F (x+ δ) ,

which implies that

lim sup

n→∞

Fn(x) ≤ lim sup

n→∞

µn(hx) = µ(hx) ≤ F (x+ δ) .

Now take the limit δ → 0 and use continuity of F at x to get lim supn→∞ Fn(x) ≤ F (x). We use the

same trick to get the desired upper bound on F (x). That is, define gx ∈ Cb(R) by

gx(y) =

1 if y ≤ x− δ ,

1− y−(x−δ)δ if y ∈ (x− δ, x) ,

0 if y ≥ x .

Then by the same argument,

lim inf

n→∞ Fn(x) ≥ lim infn→∞ µn(gx) = µ(gx) ≥ F (x− δ) .

Now take the limit δ → 0 and use continuity of F at x to get lim infn→∞ Fn(x) ≥ F (x). It follows

that Fn

d−→ F . (4 marks)

(ii) (Unseen) Let Yn = anXn. Fix ε > 0 and u a continuity point of F such that F (u) > 1− ε. If x > 0

then for all n sufficiently large x/an > u, and |Fn(u)− F (u)| < ε. It follows that

FYn(x) = P(anXn ≤ x) = P(Xn ≤ x/an) ≥ Fn(u) > 1− 2ε .

Thus limn→∞ FYn(x) = 1. By the same argument for x < 0 limn→∞ FYn(x) = 0.

(4 marks)

(b) Let µX be the law of a random variable X. Show that the distribution of X is symmetric, i.e.

µX(−∞, x] = µX [−x,∞) for all x ∈ R, if and only if the characteristic function of X is real. [4]

Answer: (Similar to exercise) ϕX(t) = E[eitX ] = E[ei(−t)(−X)] = E[eit(−X)] = ϕ−X(t). So by Le´vy’s in-

version formula (i.e. one-to-one correspondence between probability measures on (R,B) and characteristic

functions) X has the same law as −X if and only if ϕX(t) is real. (4 marks)

(c) In the following, identify if the sequence of random variables (Yn)n≥1 converges weakly and if so

identify the limit. (You should explain your reasoning, and state clearly any results from lectures

that you use).

(i) Yn = max{U1, . . . , Un} where U1, U2, . . . are independent Uniform(−1, 1) random variables.

7

(ii) Yn = n(1 − max{U1, . . . , Un}) where U1, U2, . . . are independent Uniform(−1, 1) random

variables.

(iii) Yn =

√

3

n(U1 + . . .+Un) where U1, U2, . . . are independent Uniform(−1, 1) random variables.

(iv) Yn = nmin{U1, . . . , Un} where U1, U2, . . . are independent Uniform[0, 1] random variables.

(v) Yn is an exponential random variable with mean λn > 0 for each n, i.e. P(Yn > y) = e−y/λn ,

where λn → 0 as n→∞.

[8]

Answer: (Unseen example)

(i) Weak convergence to 1. Check (equivalently) convergence in distribution. For u ≥ 1, P(Yn ≤ u) = 1,

if u ≤ −1 then P(Yn ≤ u) = 0 and for u ∈ (−1, 1), P(Yn ≤ u) = ((1 + u)/2)n → 0.

(ii) Weak convergence to an Exp(1/2) random variable. Check (equivalently) convergence in distribution.

For u > 0, P(Yn ≤ u) = 1 − P(max{U1, . . . , Un} < 1 − u/n) = 1 − P(X1 < 1 − u/n)n = 1 − ((2 −

u/n)/2)n → 1− e−u/2.

(iii) Weak convergence to standard normal. This follows from the Central Limit Theorem, which states

if X1, X2, . . . is a sequence of independent identically distributed random variables with mean µ and

variance σ, then

1√

nσ2

(X1 + . . .+Xn − nµ) weakly−−−−→ N (0, 1) .

Here µ = 0 and σ2 = E[X21 ] = 23 .

(iv)

P(nmin{U1, . . . , Un} ≤ x) = 1− P

(

U1 >

x

n

, . . . , Un >

x

n

)

= 1−

(

1− x

n

)n

→ 1− e−x ,

as n→∞. So Yn converges in distribution to an Exp(1) random variable.

(v) Fix f ∈ Cb(R) then

E f(Yn) =

∫

R

f(x)

1

λn

e−x/λndx =

∫

R

f(λny)e

−ydy → f(0) ,

as n→∞, where the final limit follows from the dominated convergence theorem since the integrand

is dominated by |f(λny)e−y| ≤ (sup |f |)e−y and converges to f(0)e−y. Hence Yn converges weakly to

0.

(8 marks)

[TOTAL: 20]

Continued...

8

4: (a) Let X and Y and Z be integrable random variables on a probability space (Ω,F ,P), and let

G ⊆ F be a sub-σ-algebra. Show that if Y and Z are both versions of E[X | G] then Y = Z

almost surely. [3]

Answer: (Applied bookwork:) Suppose Z1 and Z2 are both versions of E[X | G], let Aε = {Z2−Z1 > ε} ∈

G (because Z1 and Z2 are both G-measurable so Z1 − Z2 is). Therefore, by (3.), E[X ; Aε] = E[Z1 ; Aε] =

E[Z2 ; Aε]. So by linearity of expectation, and construction of Aε, we have 0 = E[Z1−Z2] ≥ εP(Aε), hence

P(Z2 − Z1 > ε) = 0. By the same argument (symmetry) P(Z1 − Z2 > ε) = 0, and since a union of null

events is null P(|Z1 − Z2| > ε) = 0. Finally, by taking compliments and applying monotone convergence

of measures

P(Z1 = Z2) = P

( ⋂

n≥1

{|Z1 − Z2| ≤ 1

n

})

= lim

n→∞P

(|Z1 − Z2| ≤ 1/n) = 1.

(3 mark)

(b) Suppose that (Xn)n≥1 is a random process adapted to (Fn)n≥1. Define the hitting time τA of

(Xn)n≥1 on a Borel set A. Show that τA is a stopping time for this filtration. [3]

Answer: (Applied bookwork:) (Fn)n≥1 is a filtration on (Ω,F ,P) if for all n ∈ N, Fn ⊆ Fn+1 ⊆ F . The

hitting time is defined by τA = inf{n ∈ N : Xn ∈ A}. τA is a stopping time if and only if {τA ≤ n} ∈ Fn

for each n ∈ N. This holds because

{τA ≤ n} =

⋃

k≤n

{Xk ∈ A}

and since (Xn)n≥1 is adapted to (Fn)n≥1 we have {Xk ∈ A} ∈ Fk ⊆ Fn. Hence {τA ≤ n} ∈ Fn.

(3 marks)

(c) Supose (Xn)n≥1 is a symmetric simple random walk on Z started from 0. Use the Optional

Stopping Theorem for Bounded Stopping Times to evaluate the probability (Xn)n≥1 hits b > 0

before −a < 0. [4]

Answer: (Seen in Exercises:)

Let T = τ{−a,b}, then T is not bounded, but it is almost surely finite by a Borel-Cantelli argument, and

Tk = T ∧ k is bounded for each k ∈ N. Hence, by dominated convergence, E[XT ] = limk→∞ E[XTk ], and

E[XTk ] = 0 by OST. Let p = P(XT = b) = 1 − P(XT = −a), so that E[XT ] = p b − a(1 − p) and solving

gives p = a/(b+ a).

(4 marks)

(d) Let (Xn)n≥1 be a sequence of independent random variables on probability space (Ω,F ,P), with

E[Xn] = 1 for each n. Let F0 = {∅,Ω}, Fn = σ(X1, . . . , Xn) for n ∈ N, and

M0 = 1, and Mn =

n∏

k=1

Xk for n = 1, 2, . . . .

(i) Show that (Mn)n≥0 is a martingale with respect to (Fn)n≥0.

(ii) Now suppose ϕ(t) = E[etX1 ] <∞ for all t ∈ R. Let S0 = 0, Sn =

∑n

k=1Xk, and

Yn =

etSn

ϕ(t)n

, for n = 0, 1, 2, . . . .

Show, using part (i), that (Yn)n≥0 is a martingale with respect to (Fn)n≥0.

[4]

Answer: (Unseen example (seen similar):)

(i) Fix n ∈ N. Firstly, by independence and integrability of the (Xn)n≥1

E[|Mn|] =

n∏

k=1

E[|Xk|] <∞ ,

9

and hence Mn is integrable. Since Mn depends only on X1, . . . , Xn it is Fn-measurable. Since Mn+1 =

Xn+1Mn and Mn is Fn-measurable, we have

E[Mn+1 | Fn] = E[Xn+1Mn | Fn] = MnE[Xn+1 | Fn] = MnE[Xn+1] = Mn a.s. ,

where we used ‘taking out what is known’ in the second equality and independence in the third.

(ii) Using part (i), since etSn+1 = etXn+1etSn , we have

Y0 = 1, , and Yn =

n∏

k=1

etXk

ϕ(t)

for n = 1, 2, . . . ..

Also E[etXk/ϕ(t)] = 1 and (etXn/ϕ(t))n≥1 is and independent sequence. Hence the result follows

from part (i) with etXn/ϕ(t) in place of Xn.

(4 marks)

(c) Let (εn)n≥1 be independent random variables with

P(εn = +1) = p, P(εn = −1) = q, where 1/2 < p = 1− q < 1 .

Let Fn = σ(ε1, . . . , εn) and define Xn inductively by X0 = 1 and for n ≥ 1

Xn = Xn−1 + Vnεn .

Assume Vn is a Fn−1-measurable random variable for each n, and that Vn is strictly between 0

and Xn.

(i) Prove that if Xn > 0 then E[log(Xn+1/Xn) | Fn] = f(Vn+1/Xn) where

f(x) = p log(1 + x) + q log(1− x) .

(ii) Deduce that (log(Xn)− nα)n≥0 is a supermartingale, where

α = p log p+ q log q + log 2 .

[6]

Answer: (Unseen example:)

(i) Observe

log

(

Xn+1

Xn

)

= log

(

Xn + Vn+1

Xn

)

1εn+1=+1 + log

(

Xn − Vn+1

Xn

)

1εn+1=−1 .

Also E[1εn+1=+1 | Fn] = E[1εn+1=−1] = p and E[1εn+1=+1 | Fn] = q. Taking expectation conditioned

on Fn, and ‘taking out what is known’ we get

E[log(Xn+1/Xn) | Fn] = p log

(

1 +

Vn+1

Xn

)

+ q log

(

1− Vn+1

Xn

)

= f

(

Vn+1

Xn

)

.

(3 marks)

(ii) By part (i),

E[log(Xn+1)− (n+ 1)α | Fn] = log(Xn)− nα+ f

(

Vn+1

Xn

)

− α ,

so we have to show f(Vn+1/Xn)− α ≤ 0. Notice f is differentiable, so to find the max of f on [0, 1]

consider the derivative. For each x ∈ (0, 1)

f ′(x) =

p

x+ 1

− q

1− x ,

so f is increasing on (0, p− q] and decreasing on [p− q, 1], hence attains its maximum at

f(p− q) = p log(2p) + q log(2q) = α .

It follows that (log(Xn)− nα)n≥0 is a supermartingale.

(3 marks)

10

[TOTAL: 20]

End.

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