math 6710 homework 5: solutions fall, 2016 grading: …jerison/math6710/hw5-sol.pdfmath 6710...

Math 6710 Homework 5: Solutions Fall, 2016

Grading: 1, 2, 3, 4, 5 (each 8 pts).

Problem 1: a) P (|Xn − X| > ε) = P (|an − a| > ε) = 1{|an − a| > ε}. Xn →p X if the

left side goes to zero as n → ∞ for each fixed ε, and an → a if the right side goes to zero as

n → ∞ (i.e. is eventually all zeros) for each fixed ε. b) For the first statement, fix ε > 0. By

the triangle inequality and a union bound, P (|Zn + Z ′n| > ε) ≤ P (|Zn| > ε/2) + P (|Z ′n| > ε/2)

and both those terms tend to zero as n → ∞. For the second statement, take Zn = Xn − Xand Z ′n = Yn − Y . Then Zn →p 0 and Z ′n →p 0, so Zn + Z ′n →p 0 which is equivalent to the

desired statement Xn + Yn →p X + Y .

Problem 2: Fix ε, δ > 0 and choose k = k(δ) sufficiently large that P (|X| > k) < δ and

P (|Y | > k) < δ. By additivity, P (|XnYn − XY | > ε) ≤ P (|XnYn − XY | > ε, |X| ≤ k, |Y | ≤k)+P (|X| > k)+P (|Y | > k). Next, for |X| ≤ k and |Y | ≤ k, note that by the triangle inequal-

ity |XnYn−XY | = |(Xn−X)(Yn−Y )+X(Yn−Y )+Y (Xn−X)| ≤ |Xn−X||Yn−Y |+k|Xn−X|+k|Yn−Y |. If both |Xn−X|, |Yn−Y | ≤ ε/(3k) then this sum is at most ε2/(9k2)+ε/3+ε/3 which is

at most ε as long as ε/(3k2) ≤ 1 (which we may assume by increasing k if necessary). Therefore

P (|XnYn −XY | > ε, |X| ≤ k, |Y | ≤ k) ≤ P (|Xn −X| > ε/(3k)) + P (|Yn − Y | > ε/(3k)) which

tends to zero as n → ∞ for fixed ε, k. We conclude that limn→∞ P (|XnYn − XY | > ε) ≤ 2δ.

Since δ was arbitrary this limit must be zero, so XnYn →p XY .

Problem 3: For any 0 < ε < 1, 0 ≤ d(X, Y ) = E(min{1, |X − Y |}) = E(1|X−Y |>1 + |X −Y |1|X−Y |≤1) = P (|X−Y | > 1)+E(|X−Y |1|X−Y |≤1) = P (|X−Y | > 1)+E(|X−Y |1ε<|X−Y |≤1)+

E(|X − Y |1|X−Y |≤ε). Next, suppose Xn →p X. For any 0 < ε < 1, if n is sufficiently large

then P (|X −Xn| > ε) ≤ ε. Thus, d(X,Xn) = P (|X −Xn| > 1) + E(|X −Xn|1ε<|X−Xn|≤1) +

E(|X −Xn|1|X−Xn|≤ε) ≤ P (|X −Xn| > 1) + 1 · P (ε < |X −Xn| ≤ 1) + ε · P (|X −Xn| ≤ ε) ≤P (|X −Xn| > ε) + ε · 1 ≤ 2ε. It follows that d(X,Xn)→ 0. To prove the reverse, suppose that

d(X,Xn)→ 0 and fix ε > 0. For every ε1 > 0 there exists n0 s.t. n ≥ n0 implies d(X,Xn) < εε1.

So, εε1 > d(X,Xn) = P (|X −Xn| > 1) + E(|X −Xn|1ε<|X−Xn|≤1) + E(|X −Xn|1|X−Xn|≤ε) ≥P (|X−Xn| > 1)+εP (|X−Xn| > ε) ≥ ε(P (|X−Xn| > 1)+P (|X−Xn| > ε)) = εP (|X−Xn| > ε),

that is P (|X−Xn| > ε) < ε1 for n ≥ n0. Now since ε1 was arbitrary we get P (|Xn−X| > ε)→ 0

as n→∞, as desired.

Problem 4: Since the Xi are uncorrelated, E[(Sn/n− νn)2] = Var(Sn/n) = (1/n2)[Var(X1) +

· · · + Var(Xn)] = (1/n)[Var(X1)/n + · · · + Var(Xn)/n] ≤ (1/n)[Var(X1)/1 + Var(X2)/2 +

· · · + Var(Xn)/n]. If an = Var(Xn)/n then an → 0 and so the averages (1/n)(a1 + · · · + an)

also tend to 0. (Proof: For any ε > 0 there is n0 such that |an| < ε for all n > n0.

Write (1/n)(a1 + · · · + an) = (1/n)(a1 + · · · + an0) + (1/n)(an0+1 + · · · + an). The first

term tends to 0 as n → ∞ and the second term is less than ε in absolute value, hence

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Math 6710 Homework 5: Solutions Fall, 2016

lim supn→∞(1/n)|a1 + · · · + an| ≤ ε. Since ε was arbitrary the limit must be zero.) It fol-

lows that E[(Sn/n− νn)2]→ 0, hence Sn/n− νn → 0 in L2 and thus also in probability.

Problem 5: a) Because it is integer valued, then X =∑∞

n=1 1X≥n, so E[X] = E[∑∞

n=1 1X≥n] =∑∞n=1 P (X ≥ n).

b) E[X2] =∑∞

n=1 n2P (X = n) =

∑∞n=1

∑n2

m=1 P (X = n) =∑∞

m=1

∑n:n2≥m P (X = n) =∑∞

m=1 P (X ≥√m) =

∑∞k=1

∑k2

m=(k−1)2+1 P (X ≥ k) =∑∞

k=1[k2 − (k − 1)2]P (X ≥ k) =∑∞

k=1(2k − 1)P (X ≥ k). Alternatively, using the Lemma 2.2.8 of the textbook and the fact

that X is integer-valued we have E[X2] =∫∞0

2xP (X > x)dx =∑∞

n=1

∫ nn−1 2xP (X > x)dx =∑∞

n=1 P (X ≥ n)[n2 − (n− 1)2], which is the same as above.

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