Homework # 3

7.1. By additivity∞∑

n=1

p(An) = P(

∞⋃

n=1

An

)

≤ 1

This leads to limn→∞

P (An) = 0.

7.11. First, the integral

P (A) =

∫

A

f(x)dx A ∈ B(R)

defines an probability measure on (R,B(R)) with the continuous distribution function

F (x) =

∫ x

−∞

f(u)du x ∈ R

Second, for any integer n ≥ 1, x0 ⊂ (x0 − n−1, x0]. Subsequently,

0 ≤ P (x0) ≤ P(

(x0 − n−1, x0])

= F (x0) − F (x0 − n−1) n = 1, 2, · · ·

By continuity of F at x0, the right hand side tends to 0 as n → ∞. So we have P (x0) = 0.

7.14. Null set in a complete probability space (Ω,A, P ) means a set A ⊂ A withP (A) = 0. In this case, we have P (An) = 0 for n = 1, 2, · · ·. By sub-additivity

P (B) ≤∞∑

n=1

P (An) = 0

So B is null set.

Without completness of (Ω,A, P ), there are Cn ∈ A such that An ⊂ Cn and P (Cn) =

0 for n = 1, 2, · · ·. Let C =

∞⋃

n=1

Cn. Then C ∈ A and B ⊂ C. By sub-additivity

P (C) ≤

∞∑

n=1

P (Cn) = 0

So B is null set.

7.15. Since Ω is countable, X has at most countable many values, say, x1, · · · , xn, · · ·.

0 = E|X | =∑

n

|xn|PX = xn

1

This leads to PX = xn = 0 for xn 6= 0. In other words,

PX 6= 0 =∑

xn 6=0

PX = xn = 0

In general, we can not conclude X 6= 0 = φ. For example, randomly pick up a pointin (−∞,∞) and assume the picked number ξ obeys the standard normal distribution.Write the sample space Ω = ω0, ω1, where

ω0 = ξ is not an integer and ω1 = ξ is an integer,

Define X(ω0) = 0 and X(ω1) = 1. Then E|X | = 0 and X = 1 = ξ is an integer 6= φ.

9.2* Let x ∈ R be fixed but arbiotrary. By assumption X ≤ x ∈ F and X ≤ x ∈G. Therefore,

F (x) = PX ≤ x = P(

X ≤ x ∩ X ≤ x)

= PX ≤ x · PX ≤ x = F 2(x)

Thus, for any x, F (x) = 0 or 1. By the property of distribution function, F (·) is non-decreasing and right continuous on R and

limx→−∞

F (x) = 0 and limx→∞

F (x) = 1

Therefore, there is a constant c such that F (x) = 0 for x < c and F (x) = 1 for x > c.

We now claim that PX = c = 1. Indeed,

X = c =∞⋂

n=1

c −1

n< X ≤ c +

1

n

By continuity theorem,

PX = c = P(

∞⋂

n=1

c −1

n< X ≤ c +

1

n

)

= limn→∞

P

c −1

n< X ≤ c +

1

n

= limn→∞

F(

c +1

n

)

− F(

c −1

n

)

= limn→∞

(1 − 0) = 1

2