7-1

Chapter 7 Special Continuous Distributions 7.1 Uniform Random Variable

Def: A continuous RV X is said to have a uniform distribution if the pdf of X is

We symbolize the distribution as X~U(a, b).

Example:

If X~U(a, b), evaluate (a) MX(t) (b) X (c)X2

others , 0

),( ,1

)( baxabxf X

7-2

Sol:

12

)(][][][

3

1

3

1

)0(''][(c)2

1

2

1

)0('][(b))(

1

][)((a)

222

2232

2

2

abXEXEXVar

baba

a

b

ab

xdx

abx

MXE

ba

a

b

ab

xdx

abx

MXEabt

eedx

abe

eEtM

b

a

X

b

a

X

b

a

atbttx

tXX

7-3

Ex:r.v. X 之分配為 U(-3,3) 。試問以下

方程式 有實根之機率？ Sol: 有實根，則

0)2x(xt4t4 2

2

1

6

1

6

2

13

1

0)2)(1(

0)2(16

0)2(44)4(2

2

有實根之機率＝

3x2 or x

2x or x

xx

xx

xx

7-4

Ex: Let X be a uniformly distributed continuous random variable with E [X] =1 and E [X2] =2 , what is the probability density function of X .

Sol: 令 r.v. X 之 p.d.f. 為

23

aabbdx

ab

xXE

12

abdx

ab

xXE

others , 0

bxa , ab

1)x(f

22b

a

22

b

a

7-5

others , 0

31x31, 32

1)x(f

31b,31a

故

聯立可得

7-6

7.2 Normal (Gaussian) Random Variable1.Def: A continuous RV X is said to have a normal

distribution with parameters and (or and 2), where - < < and > 0, if the pdf of X is

We symbolize the distribution as X~N(, 2).

xexf

x

X ,2

1)(

2

2

2

)(

7-7

Note: (1)

(2) fX(x+) = fX(-x+) (3) P[X ] = P[X ] =

X

fX(x)

2

1

7-8

(4) Remark:(a) If RV X~B(n, p) and = np, then X~P() when n and p 0 ( = np is constant).(b) If RV X~B(n, p), then X~N(, 2) when n (p 0), where = np and 2 = np(1- p) (c) If RV X~P(), then X~N(, 2) when .

7-9

Example:

If X~N(, 2), evaluate (a) MX(t) (b) X (c)X2

Sol:

):(

22

1

2

1

2

1

][)( (a)

2

2222

2

2222

2

2

2

1

2

1

2

)(

2

1

2

)(

adxeNote

ee

dxee

dxee

eEtM

ax

tttt

txtt

xtx

tXX

7-10

2. Proposition:

If X~N(0, 2), then

Proof:

2

222

1

)0(''][ (c))0('][

2

1]ln[

)](ln[)( (b)22

kXVarkXE

tte

tMtktt

X

!2

)!2(][

22

n

nXE

n

nn

7-11

!2

)!2(][

!2)!2(

][,2 let (2), (1) Compare

!

][]

!

)([][)((2)

!2!

)2

1(

)((1)

22

22

00

2

0

2

0

22

2

1 22

n

nXE

nn

XE nk

tk

XE

k

tXEeEtM

tnn

tetM

n

nn

n

nn

k

kk

k

ktX

X

n

nn

n

n

n

t

X

7-12

Ex: Let X have the normal density

Find the mean and variance of

X2

Sol:

),0(N 2

etM

NX

XEXE

XEXEXVar

XEXE

XEXEXVar

XE

t

X

,0~

)2(

)1(

0

22

2

1

2

224

22222

2222

222

7-13

4

224

2242

4

0t4

X4

2

)(3

XEXE XVar

3

tMXE

7-14

Example:

Let X(t) be a signal defined by X(t) = Acos(wt + ), where A and are independent random variables, and w is a positive constant. Suppose that A is a zero-mean gaussian random variable with unity variance and is uniformly distributed on [0, 2]. Calculate the mean and variance of X(t).

Sol: RV A~N(0, 1), RV ~U(0, 2)

(1) E[X(t)] = E[Acos(wt + )]

= E[A]E[cos(wt + )]

= 00 = 0

7-15

(2)

2

1

))]2sin()42(sin(4

1[

2

1

2

)22cos(1

2

1

2

1)(cos1

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

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

2

0

2

0

2

22

222

22

wtwt

dwt

dwt

wtEAEwtAEtXEtXEtXEtXVar

7-16

3. Thm: RV X~N(, 2)(1) If RV Z = aX + b Z~N(a + b, a22).

(2) If RV Z = Z~N(0, 1).

Proof:

(1)

(2)

X

),(~ RV][

)()(

22

)(2

1)()(

2

1)( 22222

abaNZeee

atMetMtatbaatat

bt

Xbt

Z

)1,0(~),(~ RV

,1

Let

22 NabaNZ

ba

7-17

4. Standardizing

When X~N(, 2). The standardized variable

is . Let Z = ,

Then Z is a standard normal RV.

(E[Z] = 0, Var[Z] = 1.)

X

X

7-18

5. Function (x):

Def: dtexx

t

2

2

2

1)(

Xx

(x)

Standard normal

density function

Note:(a) (-x)=1- (x)

(b) When RV X~N(0, 1), P[X x] = (x)

-x

7-19

Note:(1) If RV Z~N(0, 1),

(2) If RV X~N(, 2),

)()( )()(][

abaFbFbZaP ZZ

)()(

][

][

ab

bXaP

bXaP

7-20

(3) Table of Function:

7-21

Example: A fair coin is drawn 1000 times. Let RV X be the

number that the head appears. Evaluate P[460 X 535].

Sol: Method 1: RV X~B(1000, 0.5)

535

460

10001000

10001000

)2

1(]535460[

)2

1()(

xx

xX

C

C

XP

xf

7-22

Method 2: RV X~B(1000, 0.5)

9807.00057.09864.0

)53.2()21.2()5

104()

10

107(

]250

500535

250

500

250

500460[

]535460[)250,500(~

2502

1

2

11000

5002

11000 where

),(~

2

2

XP

XPNX

NX

7-23

Ex: 某輪胎公司宣稱該公司所製的輪胎有90% 機率在 25000 哩和 35000 哩之間損壞 ,假設其為常態分佈 ,試求和。

Sol:

645.1X

645.1r

9.0rZP

1,0N~ZX

Z

300002

2500035000

由查表得

令

7-24

5.3039645.1

5000

3500025000

645.1645.1

X

X

又

7-25

Ex: 某次選舉有甲乙參加。若已知有 55%民眾支持乙候選人 ,試問在抽樣調查 100人中。至少半數支持甲候選人之機率。

Sol:

1.005ZP

975.4

4550ZP50XP

1,0N~X

Z

4.975p-1np , 45

X(r.v. 45.0p,100n

則

令

不適用卜瓦松分配）使用常態分配來模擬（

支持甲之人數）可得

7-26

0.16

由查表可知）(84.01

1.005-1

1.005ZP-1

7-27

Ex:Suppose that the weight of a person

selected at random from some

population is normally distributed with

parameter and . Suppose also that

and2

1]160[ XP

4

1]140[ XP

7-28

(1)Find and , and

(2)find

(3)Of all the people in population weighting

at least 200 pounds, what percentage

will weigh over 220 pounds?

]200[ XP

7-29

Sol:

4

1201

2020

160140160]140[

.

160

160 2

1)1(

,~..

ZP

XP XP

160xZ vr

XPXP

NXvr

令

7-30

%69.80869.0

9131.01

1.36-1

1.36ZP

41.29

160200

41.29

160200)2(

41.2968.0

2068.0

20

75.04

320

function error

XPXP

z

查表得

為其中

7-31

%82.23

0869.0

0207.0200220

0207.0

9793.01

04.21

04.2

41.29

160220

41.29

160220

200

220200220)3(

XXP

ZP

XPXP

XP

XPXXP

7-32

Ex: (a) A random variable X has mean =15 and variance 2=9, and an unknown probability distribution .Find the value of c using the Chebyshev’s theorem such that (b) If the random variable X in (a) has the binomial distribution, compute the value of ( c is the same as in (a), and assume

, where Z is a random variable having the standard normal distribution.)

75.0)|15(| cXP

)|15(| cXP

)()( aaZP

7-32-A

Thm: Chebyshev’s Inequality

Let X be a RV with mean and variance 2.

Then

proof:

So the theorem results.

2

2

] |-XP[|

dxxfxXE X

)()(])[( 222

dxxfxdxxfx XX

)()()()( 22

)22)(

22)((

xxx

xxx

dxxfdxxf XX

)()( 22

]|[|2 XP

7-32-B

Note:

(1)

(2)

x

fX(x)

- +

2

2

] |-XP[|:figure

2

1] |-XP[|

kk

2

11] |-XP[|

kk

7-33

Sol:( 接 p7-32) (a)

625.0

9

925.075.0115

75.0]15[

15115

:theorem sChebyshev'

9XVar

15..

2

22

2

2

2

2

cc

cccXP

cXP

cXPcXP

XP

XEXvr

由

7-34

ondistributi Normal : ,,

small very not is 4.0

large very

0.4p2

75n

91

15

,...,2,1,0 , 1

,~..)(

2NpnB

p

n

pnpXVar

npXE

nxppCxf

pnBXvrb

n

xnxnx

7-35

122

212

22

22 615

ondistributi Normal Standard : 1,0~3

15..

23

152615

ZPXP

NZ

XZvr

XPXP

　 令

7-36

7.3 Exponential Random Variable7.4 Gamma Distribution 1. Gamma Function:

Def: For > 0, the gamma function () is defined by

Note:

(1) For any > 1, () = ( - 1)(- 1). (2) For any possible integer n, (n) = (n - 1)!

(3)

（請同學自證！）

0

1)( dxex x

.)2

1(

7-37

2. Gamma distribution: Def: A continuous RV X is said to have a gamm

a distribution if the pdf of x is

We symbolize the distribution as X~(,) or X~Gamma(,).Note: The standard gamma distribution has = 1, so the pdf of a standard gamma RV is

.0,0 where

0,)(

1)( 1

xexxf

x

X

0,)(

)(1

xex

xfx

X

7-38

3. Negative Exponential Distribution: Def: A continuous RV X is said to have a negati

ve exponential distribution if the pdf of x is

We symbolize the distribution as X~NE().

Note:

(1) (1, ) = NE().

0 where

0,)(

xexf x

X

1

7-39

(2) Let the failure rate of a system be per unit time. (a) If the system can not used again after it was failed, then the pdf of it’s life is ~NE().

(b) If the system has fault tolerable protection that allows -1 failures, then the pdf of it’s life is ~Gamma(,), where = 1/.

Example:

If X~Gamma(,), evaluate (a) MX(t) (b) X (c)X2

7-40

Sol: (a)

)1(

1

)1)((

1

)/1(])/1[()(

1)(

)/1(,

)/1()

1(Let

)(

1

)(][)(

)(

0

1

0 1

1

)1

(1

0

1

tdueu

t

t

due

t

utM

t

dudx

t

uxxtu

dxex

dxex

eeEtM

u

uX

o

xt

x

txtXX

7-41

(b) Let k(t) = ln[MX(t)] = -ln(1 - t) E[X] = k’(0) = (c) Var[X] = k’’(0) = 2

Note:If RV X~NE(), then

2

1][(c)

1][(b)

1

1)((a)

XVar

XE

ttM X

7-42

Example: Suppose that the number of miles that a car can

run before its battery wears out is exponentially distributes with an average value of 10,000 miles. If a person desires to take a 5000-mile trip, what is the probability that he or she will be able to complete the trip without having to replace the car battery?

Sol: Let RV X be the number of miles that a car can

run before the battery wears out. X~NE(), E[X] = 10000 = 1/10000

7-43

607.0)(]5000[

0,10000

1)(

2

1

5000

10000

edxxfXP

xexf

X

x

X

7-44

4. Thm: If the RV X~NE(), then RV X is memoryless. That is:

P[X > s + t | X > s] = P[X > t], s and t.

Proof:

][

][

][]|[

][

)(

tXPee

e

sXP

tsXPsXtsXP

et

edxetXP

ts

ts

tx

t

x

7-45

5. Thm: Assume X(t) means the number of events occur in [t1, t2] and X(t)~P(t),where t = t2 - t1.(1) If RV T means the time interval of length bet

ween two events occurred. Then T~NE().(2) If RV T means the time interval of length bet

ween + 1 events occurred. Then T~Gamma(, ), where = 1/.

Proof: (1)

0,)(

)(

1]0)([1 ][1][)(

4)-ch4 Process,(Poisson !

)()(

tedt

tdFtf

etXPtTPtTPtF

x

texf

tTT

tT

xt

X

7-46

(2)

),(Gamma~

0,)1)((

)()(

!

)(1

]1)([1 ][1][)(

)(~

1

1

0

T

tet

dt

tdFtf

x

tetXP

tTPtTPtFNET

tT

T

x

xt

T

1 where !

)(1 )(

then, and ),,(Gamma If:1

0

x

xt

T x

tetF

NX~Note

7-47

Proof:

!

)(1)(

!

)(1

)!1(

)!1(])!1(...))(1()[(

)()(

1)(

Let

)1)(()()(

1

0

1

0

21

0

1

0

1

0

k

kt

T

k

kx

x

x

X

xt

x

XX

k

tetF

k

xe

xxe

dexF

t

dtet

dttfxF

7-48

Note: If RV Xi~NE(), i = 1, 2, …, , then Z~Gamma

(, 1/ ) if Z = X1 + X2 + … + X.6. Def: A continuous RV X is said to have a chi-squ

ared distribution with parameter v if the pdf of x is the gamma density with = v/2 and = 2. The pdf of a chi-squared RV is thus

The parameter v is called the number of degrees

of freedom (df) of X. We symbolize the distribution as X~2(v).

0,2)

2(

1)( 2

1)2

(

2

xexv

xfxv

vX

7-49

Ex: 已知卡車通過地磅站符合卜瓦松試驗 ,且 每小時平均有六部卡車通過 ,試問 : (1) 以指數分配來模擬卡車之通過間隔時間 ? (2) 已知 10:15 有一輛卡車通過 ,試問在 10:45 以

後才有一輛車通過之機率 ? (3) 自某一輛卡車通過後開始計算 ,並取其後 第十輛卡車之通過時間為隨機變數 X。求 E[X],Var[X], 及 P[X2] 。 Sol:

tetf tTvr

λ

66:..

6)1(

小時　隔時間表示兩部卡車通過之間

小時卡車

7-50

)(18

5

36

103

5

6

10

!9

6

10

6

6

11,10~ )3(

62

1

)2(

22

2

69106910

3

2

16

小時

（小時）

屬於伽傌分配

由於指數分配無記憶性

XVar

XE

exex

xf

XX

edteTp

xx

t

7-51

9

0

12

7896

610

67

63

7

62

869

2

0

6910

!

121

0

2]1

!1

6...

!7

)6(

!8

)6(

!9

)6([

0

2)]

6

1)(!9()

6

1)(!9(

)6

1)(89(

)6

1)(9()

6

1)([(

!9

!9

62

k

k

x

xx

x

xx10

x

ek

xxxxe

eex

... ex

exex6

dxexXP

7-52

Ex: 已知單位時間顧客前往櫃臺結帳之人數符合 卜瓦松程序 ,且平均每二分鐘一人 ,試問 :

(a) 十分鐘內結帳人數之方差為 ? (b) 第三位結帳的客人和第四位 ,其間隔時間大於三分鐘之機率 ?

(c) 在二十分鐘內有多於 20 人結帳之機率 ? (d) 第三位和第五位結帳客人之時間間隔方差為何 ?

Sol:

)(5!

5

)(10),/(5.0

10..)(

25

分

分分人分鐘內結帳人數表示令

tXVarx

exf

t

Xvra

x

7-53

!

10xf

)(20),/(5.0

..)(2

13

5.0)(

..)(

10

3

2

3

3

2

5.0

x

xe

t

Xvrc

edtedttfTP

etf

)/0.5(

e~T

Tvrb

t

t

t

分分人數表示二十分鐘內結帳人令

分人其中

間間隔表示每位客人結帳的時令

7-54

)(822

42)2(

21

,2,~

)(!20

10...

!2

101011

!

101

!

1020

222

2

2

2

2210

20

0

10

21

10

分

其中

時間間隔表示每兩位客人結帳的令

βαTVar

te

Γ

tetf

λβ α βαGammaT

r.v.Td

e

x

xe

x

xeXP

tt

xx

7-55

Ex:Seven light bulbs are turned on at t=0. The lifetime of any particular bulb is independent of the lifetimes of all other bulbs and is decreased by the probability density function.

Determine the mean and variance of random variable Y, the time until the third failure .

otherwise , 0

0 if , )(

tetf

t

7-56

Sol:

)5()6()(

21,0

..

yU yUy7

yyYP

Yvr

個燈泡壽命大於個燈泡壽命大於個燈泡壽命大於

個燈泡壞掉個或　個時間，只有經過了

止之時間表示直到第三個壞掉為

7-57

yFeyYP

yYPe

e

eCeC

ppCppCp

yYP

edteytPp

y

Yy

y

y

yy

y

yt

y2-y-5

y2-y-5

2y-y-y-y2-5

2y-572

y-671

7y-

2572

671

7

e1535e-211

1e1535e-21

e121e1e7e

e1e1e

)1()1(

][

][

的機率一個燈泡壽命大於

7-58

49

1

18

1

25

1

!2!4

!7

e1e!2!4

!7)1(

0 , e1e!2!4

!7

e105e210-105

e30e35e75e175-105

e30e35

e1535e-215

0

2y-y5-

2y-y5-

y2-y-5

y2-y-y2-y-5

y2-y-5

y2-y-5

dyyYE

y

e

e

e

e

dy

ydFyf

y

y

y

y

YY

7-59

22

222

0

2y-y5-22

343

1

54

1

125

1

!2!4

!7

e1e!2!4

!7)2(

YEYEYVar

dyyYE

7-60

7.5 Beta Distribution 1. Beta function:

Note: For any positive r and s, 2. Def: A RV X is said to have a Beta distribution

with parameters , (both positive) if the pdf of X is

We symbolize the distribution as X~Be(,).

1

0

11 )1(),( dxxxsrB sr

)(

)()(),(

sr

srsrB

10,)1()()(

)(

),(

)1()( 11

11

xxxB

xxxf X

7-61

Example: If X~Be(,), evaluate (a) X (b)X

2.Sol: (a)

(b)

)1(

)()1(

)()(

)(

),1()()(

)(

)1()()(

)(][

1

0

11

B

dxxxxXE

)1()(][ ][ ][

(a)) as same (The )1)((

)1(][

222

2

XEXEXVar

XE

7-62

7.6 Weibull Distribution1. Def: A RV X is said to have a Weibull distributi

on with parameters and if the pdf of X is

We symbolize the distribution as X~W(,). Note:

In general, there are two forms of pdf of Weibull distribution. One is shown above, and the other form is

(5.4.1) 0,)( 1 xexxf xX

(5.4.2) 0,)()(

1 xex

a

bxf

b

a

xb

bX

7-63

Note:

(1) Consider of (5.4.1), if we let = 1/ab and

= b, then it will become the form of

(5.4.2).

(2) W(, 1)~NE().

(3) If X~W(, ), then

0

)1(

!)( (a)

nn

nn

n

ttM

7-64

2. Failure Rate:

Def: If the pdf and cdf of a RV X is fX(x) and FX

(x), then the failure rate R(x) is defined by

2

2)]11([)21(][ (c)

XVar

)(1

)()(

xF

xfxR

X

X

1

)11(][ (b)

XE

7-65

Note:

If RV X means the life time and the pdf of X is fX

(x). Then the probability that X fails in the time interval (x, x + dx) is fX(x)dx.Hence the failure rate of X is

)(1

)(

][1

)(1lim

][

][1lim

]|[lim)(

0

0

0

xF

xf

xXP

xxf

x

xXP

xxXxP

x

x

xXxxXxPxR

X

XX

x

x

x

7-66

Note: (1) If RV X~NE(),

(2)

Proof:

]1[1)(1

)()(

x

x

X

X

e

e

xF

xfxR (constant)

0,)()( 0)(

xexRxfx

dR

X

)](1ln[0

)](1ln[

)(1

)()(

00

xFx

F

dF

fdR

XX

x

X

Xx

7-67

x

x

x

dRXX

dRX

dRX

exRxFdx

dxf

exF

exF

0

0

0

)(

)(

)(

)()()(

1)(

)(1

7-68

Ex: 某電氣用具的壽命的機率密度函數為 t0, 其中 t以年為單位 ,若公司欲使不多於 5% 的用具要求保固服務 (warranty services), 則其保固其應至多設為幾個月 ?

Sol:

ttetf 5.025.0)(

2,2),(~

2)2(

1

25.0)()1(

22

5.0

其中 gammaT

te

tetft

t

7-69

05.01

2

4

122

1

421

05.01

)2(

2

1

2

1

222

2

et

te

tR

et

te

tet

dtte

tTPtF

tF

tftR

ttt

t

t

7-70

（月）即

（年）

3

89

205.0

42

t

tt

t

7-71

7.7 Lognormal Distribution Def: A nonnegative RV X is said to have a logno

rmal distribution if the RV Y = ln(X) has a normal distribution. The resulting pdf of a lognormal RV when ln(X) is normally distributed with parameter and is

0,2

1)(

2

2

2

])[ln(

xe

xxf

x

X

xexf

NXComparex

X ,2

1)(

then,),(~ If :

2

2

2

)(

7-72

Note: If ln(X)~N(, 2), then

)(not )1(][

)(not ][22

2

22

2

eeXVar

eXE