Homework # 4

9.4*. Let m ≥ 1 be a fixed but arbitrary integer.

|EX1An| ≤ E|X |1An

= E|X |1|X|≤m1An+ E|X |1|X|>m1An

≤ mP (An) + E|X |1|X|>m

Thus,lim sup

n→∞|EX1An

| ≤ E|X |1|X|>m

It remains to show thatlim

m→∞E|X |1|X|>m = 0

First, notice thatlim

m→∞|X |1|X|>m = 0 a.s.

By the bound |X |1|X|>m ≤ |X | and by dominated convergence theorem,

limm→∞

E|X |1|X|>m = 0.

9.5* Clearly, Q(Ω) = 1. According to Definition 2.3, p.8, all we need is to verify thatfor every countable, pair wise sequence An in A

Q(

∞⋃

n=1

An

)

=

∞∑

n=1

Q(An)

Indeed, for each m ≥ 1, write Bm =⋃m

n=1An and B∞ =

⋃∞n=1

An. Notice that thedisjointness leads to

1Bm=

m∑

n=1

1An

Consequently,

EX1Bm=

m∑

n=1

EX1An=

m∑

n=1

Q(An)

It remains to show thatlim

m→∞EX1Bm

= EX1B∞

Indeed, it follows from the fact X1Bm↑ X1B∞

and monotonic convergence (or dominatedconvergence).

9.8. (a). We follow the definition that

EQX−1 = sup

EQY ; Y is simple and Y ≤ X−1

1

Let

Y =n

∑

k=1

ak1Ak

where ak are constant and Ak ∈ A are pairwise disjoint. We assume Y ≤ X . Consequently,ak ≤ X(ω)−1, or akX ≤ 1 for ω ∈ Ak. Notice that

EQY =

n∑

k=1

akQ(Ak) =

n∑

k=1

akEX1Ak≤

n∑

k=1

E1Ak=

n∑

k=1

P (Ak) = P(

n⋃

k=1

Ak

)

≤ 1

Therefore EQX−1 ≤ 1 < ∞.

(b) For each A ∈ A, taking Y = X−11A in 9.7* we have R(A) = EQX−11A =EX−11AX = P (A).

9.11. By definition V ar(X) = E[

X − µ]2

. By Expectation rule (Corollary 9.1) withh(x) = (x − µ)2 the right hand side is equal to

∫ ∞

−∞

(x − µ)2f(x)dx

9.16.

EXk =

∫ ∞

−∞

xkf(x)dx =1

Γ(α)

∫ ∞

0

xk+α−1e−xdx =Γ(k + α)

Γ(α)

By the relation Γ(θ + 1) = θΓ(θ) (θ > 0),

EX =Γ(1 + α)

Γ(α)= α, EX2 =

Γ(2 + α)

Γ(α)= (α + 1)α

Thus,V ar(X) = EX2 − (EX)2 = (α + 1)α − α2 = α

9.18.

Pµ − dσ < X < µ + dσ = P|X − µ| < dσ = 1 − P|X − µ| ≥ dσBy Chebyshev inequality,

P|X − µ| ≥ dσ ≤ σ2

(dσ)2=

1

d2

Therefore,

Pµ − dσ < X < µ + dσ ≥ 1 − 1

d2

9.19.

PX > x =1√2π

∫ ∞

x

e−u2/2du ≤ 1√2π

∫ ∞

x

u

xe−u2/2du

=1

x√

2π

∫ ∞

x

ue−u2/2du =1

x√

2πe−x2/2

2