MATH 503 - PROBABILITY
Homework Assignment 5 solutions.
Note, that for some problems I may only give a sketch. You should fill in the details on your
own.
2.3.4 Since
X
X
P(ln / log2 n > 1 + ε) =
n−1−ε < ∞,
n
n
the first Borel-Cantelli lemma implies that
lim sup ln / log2 n ≤ 1 + ε a.s. ∀ε > 0.
n→∞
We have
P(ln / log2 n > 1 − ε i.o.) ≥ P(lnk / log2 nk > 1 − ε i.o.),
where nk = k 1+δ and 1 + δ < 1/(1 − ε). Since for k large enough nk < nk+1 − log2 nk+1 ,
we have that the events lnk / log2 nk > 1 − ε are independent (they depend on distinct
sections of the sequence Xn ), and since
X
P(lnk / log2 nk > 1 − ε) = ∞,
k
by the second Borel-Cantelli lemma we have
P(ln / log2 n > 1 − ε i.o.) ≥ P(lnk / log2 nk > 1 − ε i.o.) = 1,
so lim sup ln / log2 n ≥ 1 a.s.. For lim inf it’s obvious it is nonnegative. To show it is
zero, note that P(ln = 0 i.o.) = 1.
2.3.5 Let δ > 0 and ε > 0 be arbitrary. Since limk→∞ P(|X| > k) = 0, there exists a positive
m such that P(|X| > m) ≤ δ/3. Since limn P(|Xn − X| > 1) = 0, there exists an n1
such that for all n > n1 we have P(|Xn − X| < 1) > 1 − δ/3. We have
P(|f (Xn ) − f (X)| > ε) ≤ P(|f (Xn ) − f (X)| > ε, |X| ≤ m, |Xn − X| < 1)
+ P(|X| > m) + P(|Xn − X| ≥ 1)
≤ P(|f (Xn ) − f (X)| > ε, |X| ≤ m, |Xn − X| < 1) + 2δ/3.
Now, if |X| < m and |Xn − X| < 1, then |Xn | < m + 1. Since f is continuous, it is
uniformly continuous on [−(m + 1), m + 1], so ∃γ > 0 : ∀x, y ∈ [−m + 1, m + 1], |f (x) −
f (y)| > ε ⇒ |x−y| > γ. Thus P(|f (Xn )−f (X)| > ε) ≤ P(|Xn −X| > γ)+2δ/3, ∀n > n1 .
Since Xn → X in probability, the first term on the right hand side goes to zero as n → ∞.
Since this holds for all δ > 0, we are done.
2.3.18 For any ε > 0 we have
X
X
X
P(Xn / ln n > 1 + ε) =
e−(1+ε) ln n =
n−(1+ε) < ∞,
so P(Xn / ln n > 1+ε i.o.) = 0. Similarly P(Xn / ln n > 1−ε i.o.) = 1, so lim sup Xn / ln n =
1 a.s..
1
2
MATH 503 - PROBABILITY
For Mn we can show that P(|Mn / ln n − 1| > ε i.o.) = 0 for any ∀ε > 0 as follows.
We have
X
X
P(|Mn / ln n − 1| > ε) =
P(Mn − ln n > ε ln n) + P(Mn − ln n < −ε ln n).
For the first term we have
P(Mn − ln n > ε ln n) = 1 − P(Mn ≤ (1 + ε) ln n)
= 1 − P(Xm ≤ (1 + ε) ln n, ∀m ≤ n)
= 1 − P(X ≤ (1 + ε) ln n)n
= 1 − (1 − e−(1+ε) ln n )n
= 1 − (1 − e−(1+ε) ln n )n
= 1 − (1 − n−(1+ε) )n ,
which has leading order n−ε . Similarly
P(Mn − ln n < −ε ln n) = (1 − n−(1−ε) )n ,
ε
which has leading order e−n . Combining the two estimates, we have P(|MnP
/ ln n−1| > ε)
has order n−ε . We would like to apply the first Borel-Cantelli lemma, but 1/n−ε = ∞.
However since Mn ≤ Mn+1 we can apply the technique used in e.g. 2.3.2.
2.4.1 Take Ω = [0, 1] with the Borel measure, Xn = 1An , where k ∈ An iff
{
n
X
1/m} < k < {
m=1
n+1
X
1/m}
m=1
or
{
n+1
X
1/m} < {
m=1
n
X
1/m} < k
m=1
or
k<{
n+1
X
m=1
1/m} < {
n
X
1/m},
m=1
where {r} is the fractional part of a real number r, and N (n)(ω) is the n’th smallest
integer m such that Xm (ω) = 1.
P
P
You can think about this as follows. Let Bb be the interval ( nm=1 1/m, n+1
m=1 1/m) ⊂
R. Then An is the image of Bn when you wrap the positive part of the real line around
a circle of circumference 1 and N (n)(w) is the index of the n’th layer over ω.
2.4.2 Let N (t) be the largest integer such that X1 +Y1 +· · ·+XN (t) +YN (t) < t. From Theorem
2.4.6 and the remark after it, it follows that
N (t) → ∞ a.s.,
N (t)
X
Xi /N (t) → E(X) a.s.,
i=1
and
N (t)
X
i=1
(Xi + Yi )/N (t) → E(X) + E(Y ) a.s. .
MATH 503 - PROBABILITY
3
Since
0 < R(t) −
N (t)
X
Xi < XN (t)+1
i=1
and
XN (t) /N (t) → 0 a.s.,
it follows that
R(t)/N (t) → E(X) a.s. .
Similarly
0<t−
so
N (t)
X
(Xi + Yi ) < XN (t)+1 + YN (t)+1
i=1
t/N (t) → E(X) + E(Y ) a.s. .
Combining the last two conclusions gives the desired result.
2.4.3 Let Yn = |Xn |/|Xn−1 | for n ≥ 1. Then |Xn | = Y1 . . . Yn , and the Y ’s are i.i.d. with P(Y ≤
r) = r2 for 0 ≤ r ≤ 1 and Y ≥ 0. Apply the Strong LLN numbers to ln Y1 , ln Y2 , . . . .
c = E ln Y .
2.4.4 (i) The Strong LLN implies
1/n ln Wn → c(p) = E ln(ap + (1 − p)V ) a.s. .
For this we need E ln(ap + (1 − p)V ) to exist, which follows from
ln(ap + (1 − p)V ) ≤ max{a, V, 1/V }
and E V 2 , E V −2 < ∞.
a−V
(ii) We have c0 (p) = E ap+(1−p)V
. If a = V (ω), then the integrand is constant in p. If
1
0
a 6= V (ω), then c (p) = E p+V (ω/(a−V (ω)) , which is decreasing in p since the denominator
cannot be zero. Thus c0 (p) is decreasing in p so c(p) is concave. The only remaining
thing left to do is to justify differentiation under expectation, for which you can use
Theorem A.5.1.
(iii) Looking at c0 (0) and c0 (1) it follows that the optimal p is in (0, 1) iff a − E V and
E 1/V − 1/a have opposite signs.
(iv) Solve E(a − V )/(ap + (1 − p)V ) = 0.
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