Problem Set 3 - Department of Mathematics, Hong Kong ...makchen/MAFS5020/ProbsSols3.pdf · Problem...
Transcript of Problem Set 3 - Department of Mathematics, Hong Kong ...makchen/MAFS5020/ProbsSols3.pdf · Problem...
Problem Set 3
Sn =∑n
i=1Xi throughout.
1. Suppose Sn/n→ 0 in probability. Show that Xn/n→ 0 in probability.
Solution. |Xn|/n = |Sn − Sn−1|/n ≤ |Sn|/n+ |Sn−1|/n→ 0 in probability.
2. Suppose X1, ..., Xn, ... are iid Cauchy r.v.s. (i.e., their common density is 1/[π(1 +x2)]). Suppose bn = cnn and cn ↑ ∞. Show that Sn/bn → 0 in probability. (Hint:verify the two conditions of WLLN.)
Solution. For condition (i),
n∑i=1
P (|Xi| > bn) = nP (|X1| > bn) = 2n
∫ ∞bn
1
π(1 + x2)dx
≤ 2n
π
∫ ∞bn
1
x2dx =
2n
πbn=
2
πcn↓ 0.
For condition (ii),
n∑i=1
E(Y 2i )/b2n = nE(X2
11{|X1|≤bn})/b2n =
2n
b2n
∫ bn
0
x2
π(1 + x2)dx ≤ 2nbn
πb2n=
2
πcn↓ 0.
Moreover, an = nE(Yj) = 0 as Xi are symmetric. Hence, Sn/bn → 0 in probability.
3. Suppose X1, ..., Xn, ... are independent with mean 0 and variance 1. Show that∑∞n=1Xn/n
p converges a.e. for all p > 1/2.
Solution.∑∞
n=1 var(Xn/np) =
∑∞n=1 1/n2p <∞. Then,
∑∞n=1Xn/n
p <∞, a.e.
4. If ψ1, ..., ψn are characteristic functions, then∏n
i=1 ψi is also a characteristic function.
Solution. Let X1, ..., Xn be r.v.s such that they are independent of each other withcharacteristic functions ψ1, ..., ψn, respectively. Then, the r.v. X1 + · · · + Xn hascharacteristic function
∏nj=1 ψj, since
E(exp{it(X1 + · · ·+Xn)}) =n∏j=1
E(exp(itXj)) =n∏j=1
ψj(t).
5. Suppose X1, ..., Xn, ... are independent with mean 0 variance 1. Suppose P (|Xn| ≤2) = 1 for all n ≥ 1. Show that Sn/
√n → N(0, 1) using characteristic functions.
(Hint: mimic the proof of the CLT in 6.2, but now {Xn} are not iid).
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Solution. All Xn, n ≥ 1 are bounded by 2, the following approx holds by the Taylorexpansion.
E(exp(itSn/√n)) =
n∏j=1
E(exp(itXj/√n) ≈
n∏j=1
E(
1 + itXj/√n− .5t2X2
j /n)
=n∏j=1
(1− .5t2/n)→ exp{−t2/2},
as n→∞. It implies Sn/√n→ N(0, 1).
Exercises
1. Suppose Xn, n = 1, 2, ... are independent such that P (Xn = cn) = P (Xn = −cn) =1/2 where 0 < cn are constants. Show that Sn/bn → 0 in L2 and in probability if∑n
i=1 c2i /b
2n → 0.
Solution.
E(S2n/b
2n) =
n∑j=1
E(X2j )/b2n =
n∑j=1
c2j/b2n → 0.
This is the L2 convergence to 0, which also implies convergence in probability.
2. Following Exercise 1, raise a counter example to show that Sn/∑n
j=1 cj 9 0 inprobability.
Solution. Suppose cn = n2n. Then |Sn−1| ≤ (n− 1)(n− 1)2n−2 ≤ n2n−1. Therefore|Sn−1|/cn ≤ 1/n → 0. Likewise,
∑n−1j=1 cj/cn → 0 and
∑nj=1 cj/cn → 1. |Sn|/cn ≥
−|Sn−1|/cn + |Xn|/cn ≥ 1− 1/n→ 1. As a result, Sn/∑n
j=1 cj 9 0.
3. Suppose X1, ..., Xn are independent with mean 0 and variance 1. Apply Kol-mogorov’s inequality to show
E( max1≤j≤n
|Sj|) ≤ 1 + n.
(Hint: E(|X|) =∫∞
0P (|X| > t)dt)
Solution.
E( max1≤j≤n
|Sj|) =
∫ ∞0
P ( max1≤j≤n
|Sj| > t)dt ≤ 1 +
∫ ∞1
P ( max1≤j≤n
|Sj| > t)dt
≤ 1 +
∫ ∞1
(var(Sn)/t2)dt = 1 + n
∫ ∞1
1/t2dt = 1 + n.
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4. Suppose X1, ..., Xn, ... are i.i.d. with mean 1. Show that max1≤k≤n |Xk|/|Sn| → 0a.e.. (Hint: max1≤k≤n |Xk|/n ≤ max1≤k≤n |Xk|/k and |Xn|/n→ 0 a.e..)
Solution. By the SLLN, Sn/n→ 1 a.e. It suffices to show max1≤k≤n |Xk|/n→ 0 a.e..First, since Xi has finite mean, we know E|X| < ∞ and therefore
∑∞i=1 P (|Xi| >
i/ε) <∞. It follows from the Borel-Cantelli lemma that, with probability 1, Xn/n ≤ε for all large n. In other words, Xn/n→ 0 a.e.. Write
max1≤k≤n
|Xk|/n ≤ max1≤k≤M
|Xk|/n+ maxM≤k≤n
|Xk|/n
≤ max1≤k≤M
|Xk|/n maxM≤k≤n
|Xk|/k ≤ max1≤k≤M
|Xk|/n+ maxM≤k≤∞
|Xk|/k
→ maxM≤k≤∞
|Xk|/k a.e., by letting n→∞
→ 0, a.e., by letting M →∞.
5. Suppose X1, ...Xn, ... are independent with mean 0 and variance 1. Suppose cn areconstants such that cn/n
p ↓→ 0 for some 0 < p < 1/2. Show that∑n
j=1 cjXj/n→ 0a.e.
Solution.∞∑n=1
var(cnXn)/n2 =∞∑n=1
c2n/n2 ≤
∞∑n=1
( cnnp
)2 1
n2−2p<∞.
By Kolmogorov’s criterion of SLLN,∑n
j=1 cjXj/n→ 0, a.e..
6. Suppose X1, ..., Xn, ... are iid with common density f(x) = cg(x) with g(x) = x−α
if x > 1, 1 if |x| ≤ 1 and (−x)−β if x < −1. c > 0 is some constant and α > 1and β > 2. Find the a.e. limit of Sn/n in terms of α and β. What condition onα and β would ensure
∑∞n=1Xn/n < ∞ a.e.? (Hint: verify the three conditions in
Kolmogorov’s three series theorem; first make sure that α = β (why?)).
Solution.
E(Xn/n1{|Xn|/n≤1}) = (c/n)[
∫ n
1
xx−αdx+
∫ −1
−nx(−x)−βdx+
∫ 1
−1
xdx]
= (c/n)(n−α+2 − 1
−α + 2− n−β+2 − 1
−β + 2
)(If α = 2, n−α+2−1
−α+2is replaced by log n.)
∑∞n=1E(Xn/n1{|Xn|/n≤1}) < ∞ only if
α = β. Condition (ii) holds only when α = β.
For condition (i), now that α = β,∑n
P (|Xn|/n > 1) =∑n
P (|X1| > n) = 2c∑n
∫ ∞n
x−βdx = 2c∑n
n−β+1/(β−1) <∞,
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since β > 2.
For condition (iii),∑n
var(Xn/n1{|Xn|≤n}) =∑n
(2c/n2)[
∫ n
1
x2x−βdx+
∫ 1
0
x2dx]
=∑n
(2c/n2)(n−β+3 − 1
−β + 3+ 1/3
)<∞,
since β > 2. (Here n−β+3−1−β+3
is replaced by log(n) if β = 3.
Overall, the condition on α and β is α = β to ensure∑
nXn/n <∞, a.e..
7. What condition on α and β would ensure Sn/√n→ N(0, σ2) in Exercise 6? Express
σ2 in terms of α and β.
Solution. To require the random variables Xn to be mean 0 with finite varianceimplies α = β and β > 3. By the CLT for iid r.v.s, we have Sn/
√n → N(0, σ2)
where σ2 is the variance of Xn, which is
2c[
∫ 1
0
x2dx+
∫ ∞1
x2x−βdx] = 2c[1/3 + 1/(β − 3)].
Since 1 =∫∞−∞ f(x)dx = c
∫∞−∞ g(x)dx = 2c[1 +
∫∞1x−βdx] = 2c[1 + 1/(β − 1)], it
follows that
σ2 =1/3 + 1/(β − 3)
1 + 1/(β − 1).
8. 0 < α < 1/2. Suppose X1, ...Xn, ... are independent such that P (Xn = nα+1) =1/(6n2α) = P (Xn = −nα+1) and P (Xn = 0) = 1 − 1/(3n2α). Show that Sn/bn →N(0, 1) when b2n = var(Sn). (Hint: verify the Lindeberg condition.)
Solution. Xn are mean 0 with variance σ2n = E(X2
n) = 2n2α+2/(6n2α) = n2/3.Hence, b2n = var(Sn) =
∑nj=1 j
2/3, which is at the order of n3. (To be exact,
b2n = (1/18)n(n + 1)(2n + 1).) For any fixed ε > 0, nα+1 ≤ εbn for all large n. Asa result, for large n
∑nj=1E(X2
j 1{Xj>bnε}) = 0. Therefore, the Lindeberg conditionholds, and CLT holds thereby.
9. For X1, ..., Xn, ... independent with mean 0 and finite variances, show that the Lin-deberg condition implies max{var(Xj) : 1 ≤ j ≤ n}/var(Sn)→ 0.
Solution. Let a2n = var(Sn).
max{var(Xj) : 1 ≤ j ≤ n} = max{E(X2j 1{|Xj |>εan}) + E(X2
j 1{|Xj |≤εan}); 1 ≤ j ≤ n}
≤ max{E(X2j 1{|Xj |>εan}); 1 ≤ j ≤ n}+ ε2a2
n ≤n∑j=1
E(X2j 1{|Xj |>εan}) + εa2
n.
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When divided by a2n the first term tends to 0 by the Lindgeberg condition and the
second term is bounded by ε. The desired result follows since ε is arbitrary.
10. Show that the central limit theorem implies WLLN.
Solution. CLT implies Sn/√n → N(0, σ2) in distribution. For any x > 0,
P (Sn/√n > x) → 1 − Φ(x) where Φ is the cdf of N(0, 1). For any fixed ε > 0,
ε√n > x for large n. It follows that, for large n, P (Sn/n > ε) = P (Sn/
√n >
ε√n) > P (Sn/
√n > x) → 1 − Φ(x). As x > 0 is arbitrary, P (Sn/n > ε) → 0 as
n→∞. Analogously, P (Sn/n < −ε)→ 0. Therefore, Sn/n→ 0 in probability andthe WLLN holds.
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