Hints/Solutions for Homework 4

MATH 8651 Fall 2015

Q1. Let X1, X2, . . . be i.i.d. exponential random variables with parameter λ > 0. Use Borel-Cantelli lemmasto show that

lim supn

Xn

log n=

1

λ.

[The p.d.f. of an exponential distribution with parameter λ > 0 is given by f(x) = λe−λx1(x>0).]

Solution: If X ∼ exp(λ), then P (X > x) =∫∞xλe−λtdt = e−λx for each x > 0. For any ε > 0,

∞∑n=1

P(Xn

logn >1+ελ

)=

∞∑n=1

e−(1+ε) logn =

∞∑n=1

n−(1+ε) <∞.

By the first Borel-Cantelli lemma, Xn

logn >1+ελ i.o. a.s. Hence, lim supn

Xn

logn ≤1+ελ a.s. for every ε > 0.

Hence lim supnXn

logn ≤1λ a.s. On the other hand,

∞∑n=1

P(Xn

logn >1−ελ

)=

∞∑n=1

e−(1−ε) logn =

∞∑n=1

n−(1−ε) =∞.

Since the events Xn

logn >1−ελ are independent, by the second Borel-Cantelli lemma, we have Xn

logn >1−ελ

i.o. a.s. Hence, lim supnXn

logn ≥1−ελ a.s. for every ε > 0. Hence lim supn

Xn

logn ≥1λ a.s. Combining the

above two inequalities, we obtainlim sup

n

Xn

logn = 1λ a.s.

Q2. Prove the following variant of the first Borel-Cantelli lemma.

If P(An)→ 0 and∑n=1 P(An ∩Acn+1) <∞ then P(An i.o.) = 0.

Solution: Let Xn = 1An . Note that if Xn = 1 occurs infinitely often, then either there are infinitelymany switches from 1 to 0 in the sequence (Xn)n≥1 or Xn = 1 for all but finitely many n. Writing inthis in terms of the events An, we obtain

{An i.o.} = {An ∩Acn+1 i.o.} ∪ lim inf An.

Thus, we have P(An i.o.) ≤ P(An ∩ Acn+1 i.o.) + P(lim inf An). By the first Borel-Cantelli lemma,P(An ∩Acn+1 i.o.) = 0 On the other hand, P (lim inf An) ≤ lim inf P(An) = 0.

Q3. Let X1, X2, . . . be i.i.d. with E|X1| <∞. Let Mn = max(X1, . . . , Xn). Prove that n−1Mn → 0 a.s.

Solution: For ε > 0,∑n

P(n−1|Xn| > ε) =∑n

P(|ε−1X1| > n) ≤∫ ∞0

P(|ε−1X1| > t)dt = E|ε−1X1| <∞,

where the last equality follows from an application of Fubini’s theorem. Hence, by Borel-Cantelli, wehave

lim supn−1|Xn| ≤ ε a.s.

This implies thatlim supn−1 max

1≤i≤n|Xi| ≤ ε a.s.

This is true for each ε > 0. Thus, lim supn−1 max1≤i≤n |Xi| = 0 a.s. and therefore limn−1Mn = 0 a.s.

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Q4. Suppose an unbiased coin is flipped independently infinitely many times. Let Tn be the minimumnumber of flips to get n heads. Show that Tn/n→ c a.s. and compute c.

Solution: We can write Tn = X1 + X2 + . . . + Xn, where Xi = the number of new tosses requiredafter the (i− 1)th head to get the ith head. Clearly, each Xi is i.i.d. with Geo( 1

2 ) distribution. Thus,

by SLLN, n−1Tna.s.→ EX1 = 2.

If you don’t want to assume that Xi is i.i.d. with Geo( 12 ) distribution (which, actually, requires a

proof), you can follow the argument of example 2.4.1 of Durrett (renewal theory example). Let Sk bethe number of heads in first k tosses. By SLLN, Sk

k →12 a.s. Note that Tn ≥ n, therefore Tn ↑ ∞

pointwise. So,STn

Tn→ 1

2 a.s.

Note that STn= n, so we can write

Tnn

=TnSTn

→ 2 a.s.

Q5. Durrett 2.3.6

Solution: Suppose, if possible, lim infnEXn < EX. So, there exists a subsequence {nk} such that

E[Xnk] ≤ E[X]− ε for each k for some ε > 0. Find a further subsequence {nkl} such that Xnkl

a.s.→ X.By Fatou’s lemma, we have that lim inf lE[Xnkl

] ≥ E[X], a contradiction.

Q6. Durrett 2.4.3

Solution: Let Wn+1 = |Xn+1|/|Xn|. Then, W1,W2, . . . are i.i.d. distributed according to the distanceof a random point on the unit disc from the origin. We have |Xn| =

∏ni=1Wi. By SLLN, n1− log |Xn| =

n−1∑ni=1 logWi → c = E[logW1] a.s. (provided E[| logW1|] < ∞). Let us now compute E[logW1].

Let (U, V ) be a point chosen uniformly at random from the unit disc, i.e., (U, V ) has a bivariate density

f(u, v) = π−11{u2+v2≤1}. Since W1d=√U2 + V 2,

E[logW1] = E 12 log(U2 + V 2) =

∫ ∫12 log(u2 + v2)f(u, v)dudv

= π−1∫∫{u2+v2≤1}

12 log(u2 + v2)dudv

= π−1∫ 2π

0

∫ 1

0

log r · rdrdθ

= 2

∫ 1

0

log r · rdr = − 12 .

Since 0 ≤ W1 ≤ 1, | logW1| = − logW1 and thus by above calculation, E| logW1| = 12 < ∞, proving

the validity of SLLN.

Q7. Durrett 2.4.4

Solution: (i) The SLLN guarantees that n−1 logWn → c(p) := E log(ap+ (1− p)V1) a.s. (to show thefiniteness of E| log(ap+ (1− p)V1)|, use | log(ap+ (1− p)V1)| ≤ max(ap+ (1− p)V1, 1

(1−p)V1).)

(ii) By Theorem A.5.1 of Durrett, we can exchange differentiation and expectation. We obtain

c′(p) = E

[(a− V1

ap+ (1− p)V1

)]and c′′(p) = −E

[(a− V1

ap+ (1− p)V1

)2]

Since c′′ ≤ 0, c is concave.

(iii) In order to have the maximum of c inside (0, 1), we need to have c′(0) > 0 and c′(1) < 0, that is,E[V −11 ] > a−1 and E[V1] > a.

2

(iv) In this case, EV −11 = 58 and EV1 = 5

2 . Thus, when a ≤ 58 , the maximum of c occurs at p = 0 and

when a ≥ 52 , the maximum of c occurs at p = 1. In between, the maximum occurs for p ∈ (0, 1) such

that c′(p) = 0, that is,12

a−1ap+(1−p) + 1

2a−4

ap+4(1−p) = 0⇔ p = 5a−82(a−1)(4−a) .

Q8. Durrett 2.5.9

[Hint for Durrett 2.5.9: For m < j ≤ n if |Sm,j | > 2a and |Sj,n| ≤ a then |Sm,n| > a.]

Solution: Fix m < n. Let T := min{j ≥ m+ 1 : |Sm,j | > 2a}. Using the hint, it follows that

{|Sm,n| > a} ⊃⋃

m<j≤n

{T = j, |Sj,n| ≤ a}.

Note that the events {T = j, |Sj,n| ≤ a} are disjoint for m < j ≤ n and {T = j} ∈ σ(Xm+1, · · · , Xj) isindependent of {|Sj,n| ≤ a} ∈ σ(Xj+1, · · · , Xn). Therefore,

P(|Sm,n| > a) ≥n∑

j=m+1

P(T = j, |Sj,n| ≤ a) =

n∑j=m+1

P(T = j)P(|Sj,n| ≤ a)

≥ minm<k≤n

P(|Sk,n| ≤ a)

n∑j=m+1

P(T = j) = minm<k≤n

P(|Sk,n| ≤ a)P(m < T ≤ n)

= minm<k≤n

P(|Sk,n| ≤ a)P( maxm<j≤n

|Sm,j | > 2a).

Q9. Durrett 2.5.10

Solution: Let Sn converges in probability to S∞. Fix m ∈ N and a > 0. From the previous exercise,for any n > m,

P( maxm<j≤n

|Sj − Sm| > 2a) ≤ P(|Sn − Sm| > a)

minm<k≤nP(|Sn − Sk| ≤ a). (1)

Now P(|Sn − Sm| > a) ≤ P(|Sn − S∞| > a/2) + P(|Sm − S∞| > a/2) and taking n→∞ (and keepingm fixed) we have lim supnP(|Sn − Sm| > a) ≤ P(|Sm − S∞| > a/2). On the other hand,

minm<k≤n

P(|Sn − Sk| ≤ a) = 1− maxm<k≤n

P(|Sn − Sk| > a)

≥ 1− maxm<k≤n

(P(|Sk − S∞| > a/2) + P(|Sn − S∞| > a/2)

)≥ 1− 2 max

m<k≤nP(|Sk − S∞| > a/2)

Now, by taking n→∞ in (1), we obtain

P(supj≥m|Sj − Sm| > 2a) ≤ P(|Sm − S∞| > a/2)

1− 2 supk>mP(|Sk − S∞| > a/2).

Now taking m→∞, we have P(|Sm − S∞| > a/2)→ 0 and supk>mP(|Sk − S∞| > a/2) = 0. Hence,

lim supm

P(supj≥m|Sj − Sm| > 2a) = 0.

Finally, usingP( sup

j,`≥m|Sj − S`| > 4a) ≤ 2P(sup

j≥m|Sj − Sm| > 2a),

we see that Tm := supj,`≥m |Sj − S`|p→ 0 as m → ∞. Since Tm is decreasing, we have Tm

a.s.→ 0 andhence Sn is almost surely Cauchy and hence converges almost surely.

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