Math 3338, Homework 3

1. Let 〈S, E , P 〉 be a probability space. Recall that a function X :S → R is called a random

variable if for each x ∈ R the set {s ∈ S : X(s) ≤ x} is a member of the σ–algebra E .Deduce the following from the properties of such.

(a) For any x ∈ R the set {s ∈ S : X(s) > x} is a member of E .(b) For any x ∈ R the set {s ∈ S : X(s) < x} is a member of E .(c) lim

n→∞P ({s ∈ S : X(s) ≤ −n}) = 0.

(d) limn→∞

P ({s ∈ S : X(s) > n}) = 0.

(e) limn→∞

P ({s ∈ S : x < X(s) ≤ x+ 1/n}) = 0 for any x ∈ R.

Also, prove the following if true, give an example if not.

(f) Is P ({s ∈ S : X(s) ≤ x}) = P ({s ∈ S : X(s) < x}) always true?(g) Is lim

n→∞P ({s ∈ S : x ≤ X(s) ≤ x+ 1/n}) = 0 always true?

Hint: For (a) recall that when E ∈ E so is Ec . For (b) consider the countable union⋃∞

n=1{s ∈ S : X(s) ≤ x−1/n} . Try to use exercise 10 from your first homework assignment

to show (c)–(e). Both (f) and (g) are not true in general.

2. Recall that the distribution function for a random variable X , denoted by FX , is

defined for every x ∈ R by FX(x) = P (X ≤ x) where P (X ≤ x) is shorthand notation

for P ({s ∈ S : X(s) ≤ x}) . Conclude the following for any two real numbers a < b .

(a) P (a < X ≤ b) = FX(b)− FX(a).

(b) P (a ≤ X ≤ b) = P (X = a) + FX(b)− FX(a).

(c) P (a < X < b) = FX(b)− FX(a)− P (X = b).

3. This is an interesting problem. Part (a) is straight forward. Use results from exercise 1

above to do (b) and (c). For any distribution function, FX , prove the following.

(a) FX(x) is a nondecreasing function of x.

(b) limx→−∞

FX(x) = 0, limx→+∞

FX(x) = 1.

(c) FX(x) = limy↓x

FX(y) for every x ∈ R.

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I’ll give a quick outline of Lebesgue measure. If you’re interested, see [1] for an excellent

set of detailed notes concerning this topic.

The measure of a bounded interval I , denoted by m(I) , with endpoints a < b is given by

b−a . That is, m(I) is just the length of I . The measure of the empty set, m(∅) , is definedto be zero. One can prove that any bounded and nonempty open subset of real numbers,

say O , is composed of a unique countable (or possibly finite) union of disjoint nonempty

open intervals, O =⋃

n(an, bn) . The measure of O is given by m(O) =∑

n(bn − an)

and is a well defined real number. Now, suppose E denotes a subset of S where S is a

nonempty bounded interval. The outer measure of E is denoted by mo(E) and is given

by the well defined number

mo(E) ≡ inf{m(O) : E ⊆ O ∩ S, O is open relative to S } ≤ m(S).

(While not technically correct, you may think of inf as meaning: the minimum of.) The

inner measure of E is denoted by mi(E) and is given by

mi(E) ≡ m(S)−mo(Ec),

where Ec = {s ∈ S : s 6∈ E} . For any subset E ⊆ S we always have 0 ≤ mi(E) as well as

mi(E) ≤ mo(E) . In the case when mi(E) = mo(E) we say E is Lebesgue measurable, and

we denote its measure m(E) as this common value. It is very difficult to even imagine a

bounded set of real numbers which is not Lebesgue measurable, yet in a weird sense such

sets do exist; for those interested see [2] for example.

• Don’t let this concept of measurability confuse you. In essentially all applications we

will consider, events E will be composed of a finite (or at worst a countably infinite) union

of disjoint intervals. Such sets are obviously measurable.

Some important consequences of Lebesgue’s theory include:

(L.1) The family E of all Lebesgue measurable subsets of a bounded interval is a σ -algebra.

(L.2) If E1, E2, E3, . . . are disjoint members of E then m(⋃

n En) =∑

n m(En) .

When the collection of sets E1, E2, E3, . . . from E is not necessarily disjoint we may only

conclude the inequality m(⋃

nEn) ≤∑

n m(En) provided∑

nm(En) < ∞ .

4. Suppose E1, E2, E3, . . . each has measure zero, that is m(En) = 0 for each n ∈ N .

(Recall that just because m(E) = 0, in general this does not imply that E is empty.)

(a) Show that⋃∞

n=1En also must have measure zero. (b) The measure of a set containing

[1] http://web.media.mit.edu/∼lifton/snippets/measure theory.pdf

[2] http://en.wikipedia.org/wiki/Vitali set

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one point has measure zero. The set of rational numbers, denoted by Q , is a countable

union of numbers of the form r = p/q with q ∈ N and p ∈ Z . Show that m(Q∩ [0, 1]) = 0.

(c) The set irrational numbers is denoted by Qc . Show that m(Qc ∩ [0, 1]) = 1.

5. Consider the sample space S given by the interval of real numbers [−1, 1] . For each of

the following functions, X : [−1, 1] → R , calculate explicit formulae for the set

{s ∈ [−1, 1] : X(s) ≤ x} as x ∈ R varies. (Write as a union of intervals in case form.)

(a) X(s) = s (c) X(s) =

{

0 if s < 0s if s ≥ 0

(e) X(s) ={

1/s if s 6= 00 if s = 0

(b) X(s) = s2 (d) X(s) =

{

0 if s < 01 if s ≥ 0

(f) X(s) ={

log(|s|) if s 6= 022 if s = 0

Therefore, each of these is a measurable function on S .

6. Consider the probability space 〈S, E , P 〉 , where the sample space S is given by the

interval [−1, 1] , the associated family of events E is given by the collection of all measurable

subsets of S and the probability measure P is defined by P (E) = 1

2m(E) for every E

from E . Determine the distribution function FX(x) for each random variable given in the

previous exercise.

7. Consider the probability space 〈S, E , P 〉 , where the sample space S is given by the

interval [0, 1] , the associated family of events E is given by the collection of all measurable

subsets of S and the probability measure P is defined by

P (E) =

{

3

4m(E) + 1

4if 1

2∈ E

3

4m(E) if 1

26∈ E.

Determine P (E) when E is the given subset of S .

(a) E = [0, 1/4] (b) E = [1/4, 3/4] (c) E = [1/2, 1] (d) E = [1/2, 1/2]

8. For the probability space defined in the previous exercise, determine the distribution

function for the given random variable X .

(a) X(s) = s (b) X(s) = 1− s

9. This exercise is optional. Suppose f is a nonnegative function on the interval [0, 1] .

Define Λf (y) = m({x ∈ [0, 1] : f(x) > y}) . Conclude that∫∞

0Λf (y) dy =

∫ 1

0f(x) dx .

Hint: View computing the area under the graph of f by a ”calculus 3”∫ ∫

∗ dxdy type

double integral.

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Answers. (Please let me know if you find errors or typos here.)

2.(a) {s ∈ S : X(s) ≤ b} = {s ∈ S : X(s) ≤ a} ∪ {s ∈ S : a < X(s) ≤ b} and these

are disjoint. Therefore FX(b) ≡ P (X ≤ b) = P (X ≤ a) + P (a < X ≤ b) which is

equal to FX(a) + P (a < X ≤ b) . (b) {a ≤ X(s) ≤ b} = {X = a} ∪ {a < X(s) ≤ b} .(c) {a < X(s) ≤ b} = {a < X(s) < b} ∪ {X(s) = b} .3. Outline of proofs. (a) FX(x) ≡ P (Ex) where Ex = {s ∈ S : X(s) ≤ x} . Show that

if x1 ≤ x2 then Ex1⊆ Ex2

. Therefore FX(x1) = P (Ex1) ≤ P (Ex2

) = FX(x2) . (b) Let

ǫ > 0. From 1(d) find N such that P (X > N) < ǫ . Therefore 0 ≤ 1 − P (X ≤ x) =

P (X > x) ≤ P (X > N) < ǫ , which from (a), is true for all x ≥ N . Use 1(c) to show the

limit x → −∞ . (c) For y > x see that 0 ≤ P (X ≤ y)− P (X ≤ x) = P (x < X ≤ y) . Let

ǫ > 0 and from 1(e) find N such that P (x < X ≤ x + 1/N) < ǫ Therefore, again using

part (a), for all y in x < y < x+ 1/N we have 0 ≤ FX(y)− FX(x) < ǫ ,

4. Outline. (a) By (L.1) E =⋃∞

n=1En is measurable. Moreover 0 ≤ m(E) ≤ ∑∞

n=1m(En)

and, since m(En) = 0 for each n , the sum is equal to zero. (b) m([a, a]) = a − a = 0.

(c) Q ∩ [0, 1] =⋃∞

n=1{rn} where {rn : n ∈ N} are the rationals in [0, 1] . (d) [0, 1] =

(Q∪Qc)∩[0, 1] = (Q∩[0, 1])∪(Qc∩[0, 1]) and these are disjoint. Therefore 1 = m([0, 1]) =

m(Q ∩ [0, 1]) +m(Qc ∩ [0, 1]) = m(Qc ∩ [0, 1]) .

5. Let Ex = {s ∈ [−1, 1] : X(s) ≤ x} . (a) If x < −1 then Ex = ∅ . If −1 ≤ x < 1 then

Ex = [−1, x] . If x ≥ 1 then Ex = [−1, 1] . (b) If x < 0 then Ex = ∅ . If 0 ≤ x < 1 then

Ex = [−√x,

√x] . If x ≥ 1 then Ex = [−1, 1] . (c) If x < 0 then Ex = ∅ . If 0 ≤ x < 1

then Ex = [−1, x] . If x ≥ 1 then Ex = [−1, 1] . (d) If x < 0 then Ex = ∅ . If 0 ≤ x < 1

then Ex = [−1, 0] . If x ≥ 1 then Ex = [−1, 1] . (e) If x < −1 then Ex = [1/x, 0).

If −1 ≤ x < 0 then Ex = [−1, 0). If 0 ≤ x < 1 then Ex = [−1, 0] . If x ≥ 1 then

Ex = [−1, 0] ∪ [1/x, 1] . (f) If x ≤ 0 then Ex = [−ex, 0) ∪ (0, ex] . If 0 < x < 22 then

Ex = [−1, 0) ∪ (0, 1] . If x ≥ 22 then Ex = [−1, 1] .

6. Let FX(x) = P (Ex) = 1

2m(Ex) . (a) If x < −1 then FX(x) = 0. If −1 ≤ x < 1

then FX(x) = (x + 1)/2. If x ≥ 1 then FX(x) = 1. (b) If x < 0 then FX(x) = 0. If

0 ≤ x < 1 then FX(x) =√x . If x ≥ 1 then FX(x) = 1. (c) If x < 0 then FX(x) = 0.

If 0 ≤ x < 1 then FX(x) = (x + 1)/2. If x ≥ 1 then FX(x) = 1. (d) If x < 0 then

FX(x) = 0. If 0 ≤ x < 1 then FX(x) = 1/2. If x ≥ 1 then FX(x) = 1. (e) If x < −1 then

FX(x) = −1/(2x) . If −1 ≤ x < 1 then FX(x) = 1/2. If x ≥ 1 then FX(x) = 1− 1/(2x) .

(f) If x < 0 then FX(x) = ex . If x ≥ 0 then FX(x) = 1.

7. (a) 3/16, (b) 5/8, (c) 5/8, (d) 1/4.

8. (a) If x < 0 then FX(x) = 0. If 0 ≤ x < 1/2 then FX(x) = 3x/4. If 1/2 ≤ x < 1 then

FX(x) = 1/4 + 3x/4. If x ≥ 1 then FX(x) = 1. (b) Same as (a).

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