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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.
Random variables: Example (Cont.)
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.
Random variables: Example (Cont.)
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.
Distribution functions
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.
Distribution functions
• 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.
Expectations of discrete random variables
• 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.
Expectations of discrete random variables
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.
Expectations of discrete random variables
• 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.
Expectations of discrete random variables
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.
Expectations of discrete random variables
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.
Expectations of discrete random variables
• 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.
Expectations of discrete random variables
• 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.
Expectations of discrete random variables
• 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.
Variances and moments of discrete random variables
• 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.
Variances and moments of discrete random variables
• Theorem 4.4 Var(X)=0 if and only ifX is constant with probability 1
• Theorem 4.5 Var(aX+b)=a2Var(X)
p38.
Variances and moments of discrete random variables
• 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.
Variances and moments of discrete random variables
• 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.
Variances and moments of discrete random variables
• 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