4. Distribution Functions and Discrete Random Variableshtwu/courses/SP2017PB/CHAP4...

1

2010/4/16 1

4. Distribution Functions and Discrete Random

Variables

p2.

4.1 Random variables

• Def A real-valued function X: S R is called a random variable of the experiment if, for each interval I R,is an event.

• In probability, the set is often abbreviated as or simply as

⊆ })(:{ IsXs ∈

})(:{ IsXs ∈}{ IX ∈

IX ∈

2

p3.

Random variables: Example

• Ex 4.1 Suppose that 3 cards are drawn from an ordinary deck of 52 cards, 1-by-1, at random and with replacement.

Let X be the number of spades drawn; then X is a random variable.

p4.

Random variables: Example (Cont.)

If an outcome of spades is denoted by s, and other outcomes are represented by t, then X is a real-valued function defined on the sample space

S={(s,s,s), (t,s,s), (s,t,s), (s,s,t),(s,t,t), (t,s,t), (t,t,s), (t,t,t)},

by X(s,s,s)=3, X(s,t,s)=2, X(s,s,t)=2, X(s,t,t)=1, and so on.

3

p5.


P(X=0)=P({(t,t,t)})=27/64P(X=1)=P({(s,t,t),(t,s,t),(t,t,s)})=27/64P(X=2)=P({(s,s,t),(s,t,s),(t,s,s)})=9/64P(X=3)=P({(s,s,s)})=1/64

If the cards are drawn without replacement,

P(X=i)=C(13,i)C(39,3-i)/C(52,3)

for i=0,1,2,3.

p6.

Random variables: Example

• Ex 4.3 In the U.S., the number of twin births is approximately 1 in 90.

Let X be the number of births in a certain hospital until the first twins are born. X is a random variable.

4

p7.


Denote twin births by T and single births by N. The X is a real-valued function defined on the sample space

The set of all possible values of X is {1, 2, 3, …}

,...},,,{ NNNTNNTNTTS =

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛===

−

− 901

9089)...()(

1

1

i

iTNNNNPiXP 43421

p8.

Random variables: Example• Ex Three balls are selected randomly without

replacement from a box containing 20 balls number 1 through 20. What is the probability that the largest number of the drawn balls is as large as or larger than 17?

Sol: Let X denote the largest drawn number

Sol. (a)

Sol. (b)

∑=

==≥⇒=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛ −

==20

17

)()17(20,....,4,3,

3202

1

)(k

kXPXPk

k

kXP

,

3203

16

1)16(1)17(

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

−=≤−=≥ XPXP

5

p9.

4.2 Distribution functions

• Def If X is a random variable, then the function F defined on by Fx(t)=F(t) is called the (cumulative) distribution function of X, or CDF of X, where

),( +∞−∞

)()()( tXPtFtFX ≤==

p10.

Distribution functions

Properties of the distribution functions:

1. F is nondecreasing; that is, if t<u, then F(t)<=F(u).

2.

3.

4. F is right continuous; that is, for every t in R, F(t+)=F(t)

1)(lim =∞→

tFt

0)(lim =−∞→

tFt

6

p11.


F(a) – F(a-)X = a

F(b-) –F(a-)a <= X < b1 – F(a-)X >= a

F(b) – F(a-)a <= X <= bF(a-)X < a

F(b-) – F(a)a < X < b1 – F(a)X > a

F(b) – F(a)a < X <= bF(a)X <= a

Probability of the event in terms of F

Event concerning X

Probability of the event in terms of F

Event concerning X

p12.

Distribution functions: Example

• Ex 4.7 The distribution function of a random variable X is given by

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

≥<≤+<≤<≤

<

=

31322/112/212/1104/

00

)(

xxxxxx

x

xF

7

p13.

Figure of Ex 4.7

p14.

Distribution functions: Example (Cont.)

Compute the following quantities:

(a)P(X<2) = 1/2(b)P(X=2) = 2/3-1/2(c)P(1<=X<3) = P(X<3)-P(X<1)(d)P(X>3/2) = 1- F(3/2)(e)P(X=5/2) = F(5/2)-P(X<5/2)(f)P(2<X<=7) = F(7)-F(2)=1-2/3

8

p15.


• Ex 4.9 Suppose that a bus arrives at a station every day between 10:00 A.M. and 10:30 A.M., at random. Let X be the arrival time; find the distribution function of X and sketch its graph.

p16.

Solution of Ex 4.9Ans:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

≥<≤−

<=

5.1015.1010)10(2

00)(

ttt

ttF

9

p17.

4.3 Discrete random variables• Def Whenever the set of possible values

that a random variable X can assume is at most countable, X is called discrete.

• Examples of set measure

finite set {0, 1, 2} countable infinite set {1, 2, 3, 4, … }uncountable set {x: x >= 0}

Note: finite must be countable

p18.

Discrete random variables• Def The probability mass function (p.m.f)

p of a discrete random variable X whose set of possible values is {x1, x2, x3, …} is a function from R to R that satisfies the following properties.

(a) p(x)=0 if x {x1, x2, x3, …}(b) p(xi)=P(X=xi) and hence p(xi)>=0

(c) 1)(1

=∑∞

=iixp

∉

10

p19.

Discrete random variables: Example• Ex 4.11 Roll a fair die twice. Let X be the

maximum of the two number obtained. Determine the probability mass function and the cumulative distribution function of X.

Sol: The sample space is{ (1,1), (1,2), ………, (1,6)

(2,1), (2,2), ………, (2,6)…………………………….(6,1), (6,2), ………, (6,6)}, Each with p=1/36

p20.

Figure of Ex 4.11

11

p21.

Discrete random variables

• Ex 4.12 Can a function of the form

be a probability mass function?

• Sol: Yes, if we let

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧ ==

.0,...3,2,1)3/2()(

elsewherexcxp

x

21

)3/2(

11)(

1

==⇒=

∑∑ ∞

=i

ix

cxpQ

p22.

4.4 Expectations of discrete random variables

• Def The expected value of a discrete random variable X with the set of possible values A and probability mass function p(x) is defined by

We say that E(X) exists if this sum converges absolutely.

• E(X) is also called the mean or the expectationof X and is also denoted by EX, or .

∑∈

=Ax

xxpXE )()(

Xμ μ

12

p23.

Expectations of discrete random variables

• Ex 4.14 Flip a fair coin twice and let X be the number of heads obtained. What is the expected value of X?

Sol: p(0)=P(X=0)=1/4p(1)=P(X=1)=1/2p(2)=P(X=2)=1/4

E(X)= 0*p(0)+1*p(1)+2*p(2)=1

p24.

ExpectationsEx In the lottery game, players pick 6 integers between 1 and 10. The cost of one bet is NT$ 50. A player wins NT$ 10000 for the grand prize of 6 matches, wins NT$ 200 for the 2nd prize of 5 matches, wins NT$ 100 for the 3rd prize of 4 matches. What is the expected amount of money that a player can win in a game?

13

p25.

Solution of example

)50(*)50()50(*50)150(*150)9950(*9950)(

)50()150()9950(1)50(,

610

24

46

)5050100(

610

14

56

)15050200(,

6101)99505010000(

−−+++=

−−−=−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

==−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

==−=

⎟⎟⎠

⎞⎜⎜⎝

⎛==−=

ppppXE

pppXPXP

XPXP

Sol:

p26.


• Ex 4.18 (St. Petersburg Paradox) In a game, the player flips a fair coin successively until he gets a heads. If this occurs on the kth flip, the player wins 2k

dollars.

• Question: To play this game, how much should a person, who is willing to play a fair game, pay?

14

p27.


Sol: Let X be the amount of money the player wins. Then X is a random variable with the set of possible values {2, 4, 8, …} and P(X=2k)=1/2k, k=1, 2, 3, …Therefore,

This shows that this game remains unfair even if a person pays the largest possible amount to play it.

∞=+++=== ∑∑∞

=

∞

=

L1111)21(2)(

11 k

k

k

kXE

p28.


• Ex 4.19 The tanks of a country’s army are numbered 1 to N. In a war this country loses n random tanks to the enemy, who discovers that the captured tanks are numbered. If X1, X2, …, Xn are the numbers of the captured tanks, what is E(max Xi)? How can the enemy use E(max Xi) to find an estimate of N, the total number of this country’s tanks?

15

p29.


Sol: Let Y=max Xi; then for k=n, n+1, n+2, …, N,

∑∑∑===

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

===

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

==

N

nk

N

nk

N

nk nk

nNn

nN

nk

kkYkPYE

nN

nk

kYP

11

)()(

11

)(

p30.


If enemy captures 12 tanks and the maximum of the numbers of the tanks captured is 117, then N is around (13/12)117-1 = 126

1)(1

1)1(1

1

)(

11

that noted isIt

−+

=

++

=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛++

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛++

=⎟⎟⎠

⎞⎜⎜⎝

⎛

∑

∑

=

=

YEn

nN

nNn

nN

nN

n

nk

nNnYE

nN

nk

N

nk

N

nk

16

p31.


• Theorem 4.1 If X is a constant random variable, that is, if P(X=c)=1 for a constant c, then E(X)=c

• Theorem 4.2 Let g be a real-valued function. Then g(X) is a random variable with

∑∈

=Ax

xpxgXgE )()()]([

p32.


• Coro Let g1 , g2 , …, gn be real-valued functions, and let a1 , a2 , …, an be real numbers. Then

)]([)]([)]([)]()()([

2211

2211XgEaXgEaXgEa

XgaXgaXgaE

nn

nn+++=

+++L

L

17

p33.


• Theorem If X is a random variable. Then for any constants a and b,

E(aX+b)=aE(X)+b

Therefore, E(X) obeys the linear rule

p34.

4.5 Variances and moments of discrete random variables

• Def Variance of X

Standard deviation of X

)()(

])[(]))([()(2

222

xpx

XEXEXEXVar

x

X

∑ −=

−=−==

μ

μσ

)(XVarX =σ

18

p35.

Variances and moments of discrete random variables

• Theorem 4.3 Var(X) = E(X2)– (E(X))2

Pf: Var(X) = E[(X-E(X))2]= E[X2 – 2XE(X) + (E(X))2]= E(X2) – 2E(X)E(X) +(E(X))2

= E(X2) – (E(X))2

• Application: (E(X))2 <= E(X2)

p36.


• Ex 4.27 What is the variance of the random variable X, the outcome of rolling a fair die?

Sol: E(X)=(1+2+3+4+5+6)/6=7/2E(X2)=(1+4+9+16+25+36)/6=91/6Var(X)=91/6-(7/2)2=35/12

19

p37.


• Theorem 4.4 Var(X)=0 if and only ifX is constant with probability 1

• Theorem 4.5 Var(aX+b)=a2Var(X)

p38.


• Ex 4.25 E(X)=2 and E[X(X-4)]=5. Var(–4X+12)=?

Sol: E[X2-4X]=E(X2) –4E(X)=5so E(X2) =5+4x2=13Hence Var(X)=E(X2) –(E(X))2

=13-22=9By Theorem 4.5 Var(–4X+12)=16x9=144

20

p39.


• Def Let w be a given point. X is more concentrated about w than is Y. If for all t > 0

P(|Y-w|<=t) <= P(|X-w|<=t)

• Theorem 4.6 Suppose that E(X)=E(Y)=a. If X is more concentrated about a than is Y, then Var(X)<=Var(Y)

p40.


• Def Let c be a constant, n>=0 be an integer, and r>0 be any real number.

The nth moment of XThe rth absolute moment of XThe 1st moment of X about cThe nth moment of X about cThe nth central moment of X

E(Xn)E(|X|r)E(X-c)E[(X-c)n]E[(X-E(X))n]

DefinitionE[g(X)]

21

p41.

4.6 Standardized random variables

• Def Let X be a random variable with mean and standard deviation . The random variable is called the standardized X. We have

σμ /)(* −= XX

μσ

1)(1)1()(

0)(1)1()(

2*

*

==−=

=−=−=

XVarXVarXVar

XEXEXE

σσμ

σ

σμ

σσμ

σ

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