chapter 4-1 continuous random variables 主講人 : 虞台文

Chapter 4-1Continuous Random Variables

主講人 :虞台文

Content Random Variables and Distribution Functions Probability Density Functions of Continuous Rando

m Variables The Exponential Distributions The Reliability and Failure Rate The Erlang Distributions The Gamma Distributions The Gaussian or Normal Distributions The Uniform Distributions

Random Variables and Distribution Functions


The Temperature in Taipei

今天中午台北市氣溫為25C之機率為何 ?

今天中午台北市氣溫小於或等於 25C之機率為何 ?

Renewed Definition of Random Variables

A random variable X on a probability space (, A,

P) is a function

X : Rthat assigns a real number X() to each sample point , such that for every real number x, the

set {|X() x} is an event, i.e., a member

of A.

The (Cumulative) Distribution Functions

The (cumulative) distribution function FX of a rando

m variable X is defined to be the function

FX(x) = P(X x), − < x < .

Example 1

Example 1

3 31 18 8 8 8

0 1 2 3

( )X

x

p x

18

48

78

0 0

0 1

( ) 1 2

2 3

1 3

X

x

x

F x x

x

x

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

Example 1

R

y

( ) ( )YF y P Y y 2

2

y

R

2

2

y

R 0 y R

Example 1

R

y

( ) ( )YF y P Y y

2

2

0 0

0

1 1

y

yy R

Ry

Example 1

( ) ( )YF y P Y y

2

2

0 0

0

1 1

y

yy R

Ry

0

0.5

1

y

F Y (y )

RR /20

0.5

1

y

F Y (y )

RR /2

Example 1

R

RY

R/2

( ) ( )ZF z P Z z

2

2

2

2

( )3

59 3 2

8 29 2 3

( ) 23

0 0

1 0

1

1

R z RR

R R

R R

R z RR

z

z

z

z

z R

R z

Example 1

( ) ( )ZF z P Z z

2

2

2

2

( )3

59 3 2

8 29 2 3

( ) 23

0 0

1 0

1

1

R z RR

R R

R R

R z RR

z

z

z

z

z R

R z

0

0.5

1

z

F Z (y )

RR /2R /3 2R /3

0

0.5

1

z

F Z (y )

RR /2R /3 2R /3

Example 1

0

0.5

1

z

F Z (y )

RR /2R /3 2R /3

0

0.5

1

z

F Z (y )

RR /2R /3 2R /30

0.5

1

z

F Z (y )

RR /2R /3 2R /3

0

0.5

1

z

F Z (y )

RR /2R /3 2R /30

0.5

1

y

F Y (y )

RR /2

0

0.5

1

y

F Y (y )

RR /2-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

( )XF x ( )YF y ( )ZF z

Properties of Distribution Functions

1. 0 F(x) 1 for all x;2. F is monotonically nondecreasing;3. F() = 0 and F() =1;4. F(x+) = F(x) for all x.

0

0.5

1

z

F Z (y )

RR /2R /3 2R /3

0

0.5

1

z

F Z (y )

RR /2R /3 2R /30

0.5

1

z

F Z (y )

RR /2R /3 2R /3

0

0.5

1

z

F Z (y )

RR /2R /3 2R /30

0.5

1

y

F Y (y )

RR /2

0

0.5

1

y

F Y (y )

RR /2-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

Definition Continuous Random Variables

0

0.5

1

z

F Z (y )

RR /2R /3 2R /3

0

0.5

1

z

F Z (y )

RR /2R /3 2R /30

0.5

1

z

F Z (y )

RR /2R /3 2R /3

0

0.5

1

z

F Z (y )

RR /2R /3 2R /30

0.5

1

y

F Y (y )

RR /2

0

0.5

1

y

F Y (y )

RR /2-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

A random variable X is called a continuous random variable if

( ) ( ) ( ) 0P X x F x F x

Example 2

0

0.5

1

z

F Z (y )

RR /2R /3 2R /3

0

0.5

1

z

F Z (y )

RR /2R /3 2R /30

0.5

1

z

F Z (y )

RR /2R /3 2R /3

0

0.5

1

z

F Z (y )

RR /2R /3 2R /30

0.5

1

y

F Y (y )

RR /2

0

0.5

1

y

F Y (y )

RR /2-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

-3 -2 -1 0 1 2 3 4 5 6 7

x

F X (x )1

Probability Density Functions of Continuous

Random Variables


Probability Density Functions of Continuous Random Variables

A probability density function (pdf) fX(x) of a continuous random variable X is a nonnegative function f such that

( ) ( )x

X XF x f u du

Properties of Pdf's

1. ( ) 0;f x

2. ( ) 1;f x dx

3. ( ) ( ) ( ) ( )P a X b P a X b P a X b P a X b

( ) ( )P X b P X a ( ) ( )F b F a

( ) ( )x

X XF x f u du

( ) ( )b af x dx f x dx

( )

b

af x dx

( )4. ( ) ( ).

dF xf x F x

dx

Remark: f(x) can be larger than 1.

Example 3

Example 3

(1 ) 0 1( )

0

kx x xf x

otherwise

1 ( )f x dx

1

0(1 )kx x dx

12 3

02 3

x xk

6

k

6k

6k

Example 3

6 (1 ) 0 1( )

0

x x xf x

otherwise

6k

0

0 0

( ) 6 (1 ) 0 1

1 1

x

x

F x u u du x

x

2 3

0 0

3 2 0 1

1 1

x

x x x

x

0

0.5

1

1.5

2

-1 0 1 2

x

f (x )

0

0.2

0.4

0.6

0.8

1

1.2

-1 0 1 2

F (x )

x

0

0.5

1

1.5

2

-1 0 1 2

x

f (x )

0

0.2

0.4

0.6

0.8

1

1.2

-1 0 1 2

F (x )

x

Example 3

6 (1 ) 0 1( )

0

x x xf x

otherwise

6k

2 3

0 0

( ) 3 2 0 1

1 1

x

F x x x x

x

0

0.5

1

1.5

2

-1 0 1 2

x

f (x )

0

0.2

0.4

0.6

0.8

1

1.2

-1 0 1 2

F (x )

x

0

0.5

1

1.5

2

-1 0 1 2

x

f (x )

0

0.2

0.4

0.6

0.8

1

1.2

-1 0 1 2

F (x )

x

Example 3

6 (1 ) 0 1( )

0

x x xf x

otherwise

6k

2 3

0 0

( ) 3 2 0 1

1 1

x

F x x x x

x

2 31 1 13 3 3( ) 3( ) 2( )P X

70.25926

27

1/3

0.25926

The Exponential Distributions



The following r.v.’s are often modelled as exponential:

1. Interarrival time between two successive job arrivals.

2. Service time at a server in a queuing network.

3. Life time of a component.


A r.v. X is said to possess an exponential distribution and to be exponentially distributed, denoted byX ~ Exp(), if it possesses the density

( ) , 0xf x e x


( ) , 0xf x e x

~ ( )X Exp

: arriving rate: failure rate

pdf

cdf1 0

( )0 0

xe xF x

x


( ) , 0xf x e x

~ ( )X Exp


pdf

cdf1 0

( )0 0

xe xF x

x

0

0.5

1

1.5

2

2.5

3

3.5

-2 0 2 4 6

x

f (x )

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-2 0 2 4 6

F (x )

x

0

0.5

1

1.5

2

2.5

3

3.5

-2 0 2 4 6

x

f (x )

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-2 0 2 4 6

F (x )

x

~ ( )2X Exp

Memoryless or Markov Property

~ ( )X Exp

( | ) ( )P X a b X a P X b 0

0

a

b


~ ( )X Exp

( | ) ( )P X a b X a P X b 0

0

a

b

( and )( | )

( )

P X a b X aP X a b X a

P X a

( )

( )

P X a b

P X a

( ) , 0xf x e x

~ ( )X Exp

pdf

cdf1 0

( )0 0

xe xF x

x

( ) , 0xf x e x

~ ( )X Exp

pdf

cdf1 0

( )0 0

xe xF x

x

1 ( )

1 ( )

P X a b

P X a

1 ( )

1 ( )

F a b

F a

( )a b

a

e

e

be

( )P X b


~ ( )X Exp

( | ) ( )P X a b X a P X b 0

0

a

b

Exercise:

連續型隨機變數中，唯有指數分佈具備無記憶性。

The Relation Between Poisson and Exponential Distributions : arriving rate

: failure rate

Nt

Let r.v. Nt denote #jobs arriving to a computer system in the interval (0, t].

t0

?~tN ( )P t

The Relation Between Poisson and Exponential Distributions

Let X denote the time of the next arrival.


The next arrival

Nt


t0

X

?~tN ( )P t

?~X or ( ? ?( ) )f t F t

?~tN ( )P t




The next arrival

Nt


t0

X

?~X or ( ? ?( ) )f t F t

能求出 P(X > t)嗎 ?能求出 P(X > t)嗎 ?

( ) ( 0)tP X t P N

?~tN ( )P t




The next arrival

Nt


t0

X

?~X or ( ? ?( ) )f t F t

能求出 P(X > t)嗎 ?能求出 P(X > t)嗎 ?

( ) ( 0)tP X t P N

( ) ( 0)tP X t P N 0( )

0!

tt e

te

( ) 1 tF t e

( ) tf t e

0t

0t

0t

( ) , 0xf x e x

~ ( )X Exp

pdf

cdf1 0

( )0 0

xe xF x

x

( ) , 0xf x e x

~ ( )X Exp

pdf

cdf1 0

( )0 0

xe xF x

x

?~tN ( )P t




Nt


t0

X

~ ( )X Exp

The next arrival

The Relation Between Poisson and Exponential Distributions : arriving rate

: failure ratet1 t2 t3 t4 t5

The interarrival times of a Poisson process are exponentially distributed.

Example 50

10 secs

= 0.1 job/sec = 0.1 job/sec

P(“No job”) = ?

Example 50

10 secs

= 0.1 job/sec = 0.1 job/sec

P(“No job”) = ?

Method 1:

Method 2:

Let N10 represent #jobs arriving in the 10 secs.

Let X represent the time of the next arriving job.

10( ) ( 0)P P N N o "b" o j

( ) ( 10)P P X " "No job

10 ~ (1)N P

~ (0.1)X Exp

0 11

0!

e

1e

1 ( 10)P X 101 (1 )e

1e

The Reliability and

Failure Rate


Definition Reliability

Let r.v. X be the lifetime or time to failure of a component. The probability that the component survives until some time t is called the reliability R(t) of the component, i.e.,

R(t) = P(X > t) = 1 F(t)

Remarks:

1. F(t) is, hence, called unreliability.

2. R’(t) = F’(t) = f(t) is called the failure density function.

The Instantaneous Failure Rate

剎那間，ㄧ切化作永恆。

( | )t X tP X tt


0 t

t

t+t

( | )P t X t t X t

生命將在時間 t後瞬間結束的機率


( | )P t X t t X t

生命將在時間 t後瞬間結束的機率

and( )

( )

P t X t t X t

P X t

( )

( )

P t X t t

P X t

( ) ( )

( )

F t t F t

R t

( ) ( )( | )

( )

F t t F tP t X t t X t

R t


( | )P t X t t X t

瞬間暴斃率 h(t)

( ) ( )( | )

( )

F t t F tP t X t t X t

R t

0

( | )limt

P t X t t X t

t

( )h t

0

( ) ( )lim

( )t

F t t F t

tR t

( )

( )

F t

R t

( )

( )

f t

R t


瞬間暴斃率 h(t)

( )( )

( )

f th t

R t

Example 6

Show that the failure rate of exponential distribution is characterized by a constant failure rate.

Show that the failure rate of exponential distribution is characterized by a constant failure rate.

( ) , 0xf x e x

~ ( )X Exp

pdf

cdf1 0

( )0 0

xe xF x

x

( ) , 0xf x e x

~ ( )X Exp

pdf

cdf1 0

( )0 0

xe xF x

x

( )( )

( )

f th t

R t

( )

1 ( )

f t

F t

t

t

e

e

0t

以指數分配來 model物件壽命之機率分配合理嗎 ?

More on Failure Rates

t

h(t)

CFR ( ) 0h t


t

h(t)

CFR ( ) 0h t

Useful Life

CFR ( ) 0h t

DFR

( ) 0h t IFR

( ) 0h t


t

h(t)

CFR ( ) 0h t

Useful Life

CFR ( ) 0h t

DFR

( ) 0h t IFR

( ) 0h t

Exponential

Distribution

Exponential

Distribution? ?

Relationships among F(t), f(t), R(t), h(t)

( )F t

( )f t

( )R t

( )h t

1 ( )F t

( )dF t

dt

( )

( )

f t

R t


( )F t

( )f t

( )R t

( )h t

( )f t 1 ( )F t

( )

( )

R t

f t


( )F t

( )f t

( )R t

( )h t

( )

( )

f t

R t

( )dF t

dt

1 ( )R t


( )F t

( )f t

( )R t

( )h t

?

?

?

Cumulative Hazard

0( ) ( )

tH t h x dx 0

( )

( )

t f xdx

R x 0

( )

( )

t R xdx

R x

0

1( )

( )

tdR x

R x 0

ln ( )t

R x

ln ( ) ln (0)R t R ln1 ln ( )R t ln ( )R t

( )( ) H tR t e 0( )

th x dx

e


( )F t

( )f t

( )R t

( )h t

0( )

th x dx

e 1 ( )R t

( )dF t

dt

Example 7

0( ) ( )

tH t h x dx 0 0

txdx

2

0

02

tx

20 , 02

tt

20( ) exp , 02

tR t t

The Erlang Distrib

utions


我的老照相機與閃光燈

它只能使用四次每使用一次後轉動九十度使用四次後壽終正寢

它只能使用四次每使用一次後轉動九十度使用四次後壽終正寢

time

The Erlang Distributions

The lifetime of my flash (X)

I(X)=? fX(t)=?[0, )


Consider a component subjected to an environment so that Nt, the number of peak stresses in the interval (0, t], is Poisson distributed with parameter t.

Suppose that the rth peak will cause a failure. Let X denote the lifetime of the component. Then,

Nt ~ P(t)

?( )P X t ( )tP N r1

0

( )

!

k tr

k

t e

k

1

0

( )( ) 1 , 0

!

k tr

k

t eF t t

k

cdf




Nt ~ P(t)

Exercise ofChapter 2

cdf

1

( ) , 0( 1)!

r r tt ef t t

r

pdf

1

0

( )( ) 1 , 0

!

k tr

k

t eF t t

k

The r-Stage Erlang Distributions



cdf

1

( ) , 0( 1)!

r r tt ef t t

r

pdf

1

0

( )( ) 1 , 0

!

k tr

k

t eF t t

k


cdf

1

( ) , 0( 1)!

r r tt ef t t

r

pdf

1

0

( )( ) 1 , 0

!

k tr

k

t eF t t

k

( , )X Erlang r

( ) (1, )Exp Erlang


1

, 0( 1)!

( )r r tt e

tr

f t

pdf

( , )X Erlang r

( ) (1, )Exp Erlang

1

, 0( 1)!

( )r r tt e

tr

f t

1

, 0( 1)!

r r tt et

r

Example 8

In a batch processing environment, the number of jobs arriving for service is 9 per hour. If the arrival process satisfies the requirement of a Poisson experiment. Find the probability that the elapse time between a given arrival and the fifth subsequent arrival is less than 10 minutes.

In a batch processing environment, the number of jobs arriving for service is 9 per hour. If the arrival process satisfies the requirement of a Poisson experiment. Find the probability that the elapse time between a given arrival and the fifth subsequent arrival is less than 10 minutes.

Let X represent the time of the 5th arrival.

= 9 jobs/hr.

16( ?hr)P X

~ (5,9)X Erlang

1

( ) , 0( 1)!

r r xx ef t x

r

( , )X Erlang r

5 4 91/ 6

0

9

4!

xx edx

4

9/ 6

0

(9 / 6)1

!

k

k

ek

0.0285

The Gamma

Distributions


Review

1

( ) , 0( 1)!

r r xx ef t x

r

pdf

( , )X Erlang r

r為一正整數欲將之推廣為正實數

r為一正整數欲將之推廣為正實數

Review

1

( ) , 0( 1)!

r r xx ef t x

r

pdf

( , )X Erlang r

( )

0

The Gamma Distributions

1

( ) , 0( )

xx ef x x

pdf

( , ), 0X

Review1

( ) , 0( )

xx ef t x

( , )X

1

0( ) xx e dx

1. (1) 1

2. ( ) ( 1) ( )

3. ( ) ( 1)!, N

14.

2

1 12

( 1)!5.

2 2 !n n

n n

Chi-SquareDistributions

2X

1

( ) , 0( )

xx ef t x

( , )X

1 1,

2 2

2vX 1

,2 2

v

The Gaussian or

Normal Distributions


The Gaussian or Normal Distributions

德國的 10馬克紙幣 , 以高斯 (Gauss, 1777-1855)為人像 , 人像左側有一常態分佈之 p.d.f.及其圖形。


2( ) ( ; , )f x n x pdf

2( , )X N 2

2

( )

21

2

x

e

x


2( ) ( ; , )f x n x

2( , )X N

2

2

( )

21

2

x

e

x

2( ) ( ; , )f x n x

2( , )X N

2

2

( )

21

2

x

e

x

: mean : standard deviation2: variance

Inflectionpoint

Inflectionpoint


2( ) ( ; , )f x n x

2( , )X N

2

2

( )

21

2

x

e

x

2( ) ( ; , )f x n x

2( , )X N

2

2

( )

21

2

x

e

x


15 varying

100 varying


2( ) ( ; , )f x n x

2( , )X N

2

2

( )

21

2

x

e

x

2( ) ( ; , )f x n x

2( , )X N

2

2

( )

21

2

x

e

x


2 / 2 2xe dx

Facts:

2 / 211

2xe dx

2 2( ) / 211

2xe dx


2( ) ( ; , )f x n x

2( , )X N

2

2

( )

21

2

x

e

x

2( ) ( ; , )f x n x

2( , )X N

2

2

( )

21

2

x

e

x


2 2( ) / 211

2xe dx

2( 9) /8 ?xe dx

2 /18

0?xe dx

2 2

3 2

2

Standard Normal Distribution

2( ) ( ; , )f x n x

2( , )X N

2

2

( )

21

2

x

e

x

2( ) ( ; , )f x n x

2( , )X N

2

2

( )

21

2

x

e

x

( ) ( ;0,1)Zf z n z

(0,1)Z N

2 / 21

2ze

z

Table of N(0, 1)

( ) ( ;0,1)Zf z n z

(0,1)Z N

2 / 21

2ze

z

z

( ) ( )ZF z z2 / 21

2

z xe dx

Table of N(0, 1)

z

( ) ( )ZF z z2 / 21

2

z xe dx

( 1.6 ?25) 0.0521

(1.62 ?5) 0.9479

Fact: ( ) 1 ( )z z

Probability Evaluation for N(, 2)

x 2

2

( )

21( )

2

tx

XF x e dt

?

2

2

( )

21( )

2

x

f x e

x

2

2

( )

21

2

tt x

te dt

2 2

2

( )

2 2

t y

t

y

t y


x 2

2

( )

21( )

2

tx

XF x e dt

?

2

2

( )

21( )

2

x

f x e

x

2

2

( )

21

2

tt x

te dt

t y

2 / 21

2

y x y

ye dy

2 / 21

2

xye dy

x

?


x

2( , )X N

( )X

xF x

Z-Score: 表距離中心若干個標準差

Fact: (0,1)X

N

Example 9

X ~ N(12.00, 0.202)

1. (11.92 1 27 ?2. )P X

2. ( 12.45 ?)P X

3. ( 11.70 ?)P X

Example 9X ~ N(12.00, 0.202)

1. (11.92 1 27 ?2. )P X

2. ( 12.45 ?)P X

3. ( 11.70 ?)P X

(11.92 12.27) ( 12.27) ( 11.92)P X P X P X

12.27 12 11.92 12

0.2 0.2

1.35 0.40 0.9115 0.3446 0.5669

0.5669

Example 9X ~ N(12.00, 0.202)

1. (11.92 1 27 ?2. )P X

2. ( 12.45 ?)P X

3. ( 11.70 ?)P X

( 12.45) 1 ( 12.45)P X P X

12.45 121

0.2

1 0.9878 0.0122

0.5669

1 2.25

0.0122

Example 9X ~ N(12.00, 0.202)

1. (11.92 1 27 ?2. )P X

2. ( 12.45 ?)P X

3. ( 11.70 ?)P X

11.70 12( 11.70)

0.2P X

0.0668

0.5669

1.50

0.0122

0.0668

Example 10

|X | < |X | < 2|X | < 3

2( , )X N

Example 10

|X | < |X | < 2|X | < 3

|X | < |X | < 2|X | < 3

2( , )X N

(| | )P X k ( )P k X k

XP k k

( ) ( )k k

1 ( ) ( )k k

1 2 ( )k

Example 10

|X | < |X | < 2|X | < 3

|X | < |X | < 2|X | < 3

2( , )X N

(| | )P X k 1 2 ( )k

(| | )P X 1 2 ( 1)

(| | 2 )P X 1 2 ( 2)

(| | 3 )P X 1 2 ( 3)

1 2 0.1587 0.6826

1 2 0.0228 0.9544

1 2 0.0013 0.9974

(| | )P X k

Example 10

(| | )P X k 1 2 ( )k

(| | )P X 1 2 ( 1)

(| | 2 )P X 1 2 ( 2)

(| | 3 )P X 1 2 ( 3)

1 2 0.1587 0.6826

1 2 0.0228 0.9544

1 2 0.0013 0.9974

The Uniform

Distributions


The Uniform Distributions

~ ( , ), X U a b a bpdf

cdf

1( ) , f x a x b

b a

0

( )

1

a x

x aF x a x b

b ab x

a bx

f(x)

1

b a

a bx

F(x)1

Summary

The Exponential Distributions The Erlang Distributions The Gamma Distributions The Gaussian or Normal Distributions The Uniform Distributions

chapter 4-1 continuous random variables 主講人 : 虞台文

Documents