Chapter 4-2 Continuous Random Variables 主講人 : 虞台文

Chapter 4-2Continuous Random Variables

主講人 :虞台文

Content Functions of Single Continuous Random Variable Jointly Distributed Random Variables Independence of Random Variables Distribution of Sums Distributions of Multiplications and Quotients Conditional Densities Multivariate Distributions Multidimensional Changes of Variables

Functions of Single Continuous Random Variable

The Problem

( )Y g X( )Xf x 已知

( )Yf y =?

Example 11

2Y X( )Xf x 已知

( )Yf y =?

Example 112Y X

( )Xf x 已知

( )Yf y =?2Y X

( )Xf x 已知

( )Yf y =?

( ) ?I Y [0, )

( ) ( )YF y P Y y 2( )P X y P y X y

X XF y F y

( ) ( )dY Ydyf y F y d

X Xdy F y F y

d yd yX Xdy dyf y f y

12 X Xy

f y f y 0y

Example 112Y X

( )Xf x 已知

( )Yf y =?2Y X

( )Xf x 已知

( )Yf y =?

( ) , 0Y X Xyf y f y f y y

Example 12 1

2( ) , 0Y X Xy

f y f y f y y

2 , ( ? )YY X f y ~ (0,1).X N2 / 21

2( ) , x

Xf x e x

/ 2 / 21 1 12 2 2

( ) y yY y

f y e e

21/ 2 1/ 21

2 , 0yy e

2~Y 1 1,

Example 12 1

2( ) , 0Y X Xy

f y f y f y y

2 , ( ? )YY X f y ~ (0,1).X N

2~Y 1 1,

請熟記 !

標準常態之平方為一個自由度的卡方

Example 13

21. , ( ) ?YY X f y

~ ( 2,4).X U

2. | |, ( ? )ZZ X f z

( ) , 0Y X Xyf y f y f y y

( ) ?I Y [0,16)

( )Y X Xyf y f y f y

0as 0 y <16

0as 0 y 4

Example 13

21. , ( ) ?YY X f y

~ ( 2,4).X U

2. | |, ( ? )ZZ X f z

( ) ?I Z [0,4)

( ) ( )ZF z P Z z (| | )P X z ( )P z X z ( ) ( )X XF z F z

( ) ( )dZ Zdzf z F z ( ) ( )d

X Xdz F z F z ( ) ( )X Xf z f z

0as 0 z <4

0as 0 z 21

0 2( )

Example 14

~ (0,1).X U

( ) ?I Y (0, )

( ) ( )YF y P Y y 1( ln(1 ) )P X y

1 ln(1 ) with 0, ( ) ?YY X f y

ln(1 )P X y

(1 )yP X e ( 1)yP X e ( 1 )yP X e

(1 )yXF e 1 ye

( ) , 0yYf y e y ~ ( )Y Exp

Example 14 x

~ ( )Y Exp

How to generate exponentially

distributed random numbers

using a computer?

How to generate exponentially

distributed random numbers

using a computer?

~ (0,1).X U 1 ln(1 ) with 0, ( ) ?YY X f y

Example 14 x

~ ( )Y Exp

~ (0,1).X U 1 ln(1 ) with 0, ( ) ?YY X f y

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1

y= 1ln(1-x ) = 2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1

y= 1ln(1-x ) = 2 1 ln(1 )Y X

ln(1 )Y X

1Ye X 1 YX e ( )YF Y

Theorem 1

Let g be a differentiable monotone function on an interval I, and let g(I) denote its range. Let X be a continuous r. v. with pdf fX such that fX(x) = 0 for x I. Then, Y

= g(X) has pdf fY given by

1 1( ) ( ) ( ) , ( )Y X

df y f g y g y y g I

Theorem 1

Let g be a differentiable monotone function on an interval I, and let g(I) denote its range. Let X be a continuous r. v. with pdf fX such that fX(x) = 0 for x I. Then, Y

= g(X) has pdf fY given by

11( ) , (( ) ( ) )XY

df g y y g I

dyf y g y 11( ) , ( ) ) )( (Y X

dyff y yg y gg y I 11( ) , ( ) ) )( (Y X

dyff y yg y gg y I 1 1 , ( ) ( ) ( )) (Y X

dyff y g y g Ig yy 1 1( ) ( ) ( ) , ( )XY

dyff y g y g y y g I

Y = g(X)

Theorem 1

g oror 11( ) , (( ) ( ) )XY

df g y y g I

dyf y g y 11( ) , ( ) ) )( (Y X

dyff y yg y gg y I 11( ) , ( ) ) )( (Y X

dyff y yg y gg y I 1 1 , ( ) ( ) ( )) (Y X

dyff y g y g Ig yy 1 1( ) ( ) ( ) , ( )XY

dyff y g y g y y g I Y = g (X) and

Case 1:Pf) g ( ) ( )YF y P Y y ( ( ) )P g X y

1( )P X g y 1( )XF g y

1 1( ) ( ) ( )Y X

df y f g y g y

positive

Y = g(X)

Theorem 1

g oror 11( ) , (( ) ( ) )XY

df g y y g I

dyf y g y 11( ) , ( ) ) )( (Y X

dyff y yg y gg y I 11( ) , ( ) ) )( (Y X

dyff y yg y gg y I 1 1 , ( ) ( ) ( )) (Y X

dyff y g y g Ig yy 1 1( ) ( ) ( ) , ( )XY

Case 2:Pf) ( ) ( )YF y P Y y ( ( ) )P g X y

1( )P X g y 11 ( )XF g y

1 1( ) ( ) ( )Y X

df y f g y g y

negative

Example 15Redo Example 14 using Theorem 1.

~ (0,1).X U 1 ln(1 ) with 0, ( ) ?YY X f y

?X Y 1( ) 1 YX g Y e 11. ( ) 1 yg y e

12. ( ) yddy g y e

( ) ?I Y (0, )

13. ( ) yddy g y e

14. ( ) 1Xf g y

g oror 11( ) , (( ) ( ) )XY

df g y y g I

dyf y g y 11( ) , ( ) ) )( (Y X

dyff y yg y gg y I 11( ) , ( ) ) )( (Y X

dyff y yg y gg y I 1 1 , ( ) ( ) ( )) (Y X

dyff y g y g Ig yy 1 1( ) ( ) ( ) , ( )XY

0, ( )y I Y

(0,1), ( )y I Y

( ) , 0yYf y e y

Example 16

?X Y 1 2( )X g Y Y 1 21. ( )g y y

12. ( ) 2ddy g y y

( ) ?I Y (0, )

13. ( ) 2ddy g y y

214. ( ) yXf g y e

g oror 11( ) , (( ) ( ) )XY

df g y y g I

dyf y g y 11( ) , ( ) ) )( (Y X

dyff y yg y gg y I 11( ) , ( ) ) )( (Y X

dyff y yg y gg y I 1 1 , ( ) ( ) ( )) (Y X

dyff y g y g Ig yy 1 1( ) ( ) ( ) , ( )XY

0, ( )y I Y

( ) 2 , 0yYf y ye y

1. , ( ) ?YY X f y ~ ( ).X Exp 22. 1 ( ?)ZZ X f z

( ) , 0xXf x e x

Example 16

?X Z 1( ) 1X g Z Z 11. ( ) 1g z z

1 12 1

2. ( )ddz z

( ) ?I Z ( ,1)

1 12 1

3. ( )ddy z

1 14. ( ) zXf g z e

g oror 11( ) , (( ) ( ) )XY

df g y y g I

dyf y g y 11( ) , ( ) ) )( (Y X

dyff y yg y gg y I 11( ) , ( ) ) )( (Y X

dyff y yg y gg y I 1 1 , ( ) ( ) ( )) (Y X

dyff y g y g Ig yy 1 1( ) ( ) ( ) , ( )XY

0, ( )z I Z

2 1( ) , 1z

Z zf z e z

1. , ( ) ?YY X f y ~ ( ).X Exp 22. 1 ( ?)ZZ X f z

( ) , 0xXf x e x

Example 17

( ) , 0xXf x e x

1/ , 0Y X

Example 17

( ) , 0xXf x e x

1/ , 0Y X

?X Y 1( )X g Y Y 11. ( )g y y

1 12. ( )ddy g y y

( ) ?I Y (0, )

1 13. ( )ddy g y y

14. ( ) yXf g y e

0, ( )y I Y

1( ) , 0yYf y y e y

Example 17

( ) , 0xXf x e x

1/ , 0Y X

1( ) , 0yYf y y e y

( )0 0

x e dx yF y

Example 17

( ) , 0xXf x e x

1/ , 0Y X

1( ) , 0yYf y y e y

1 0( )

e yF y

( )( )

f th t

Example 17

( ) , 0xXf x e x

1/ , 0Y X

1( ) , 0Yh t t t

0 1 2 3 4 5

h (t )

0 1 2 3 4 5

h (t )

0.750.50

0.250.10

< 1 : DFR = 1 : CFR > 1 : IFR

Example 18

Let X be a continuous random variable. Define Y to be the cdf of X, i.e., Y = FX(X). Find fY(y).

( ) ?I Y [0,1]

( ) ( )YF y P Y y ( ( ) )XP F X y 1( )XP X F y 1( )X XF F y y

( ) ( )Y Yf y F y 1 0 1y

( )XY F X ~ (0,1)Y U

Random Number Generation

The method to generate a random number X such that it possesses a particular distribution by a computer:

1. Let Y = FX(X).

2. Find

3. Generate a random variable by a computer in interval (0, 1). Let y be such a random number.

4. Computing , we obtain the desired random number x.

( )XY F X ~ (0,1)Y U

1( ).XX F Y

1( )XX F Y

Example 19

How to generate a random variable X by a

computer such that X ~ Exp()?

How to generate a random variable X by a

computer such that X ~ Exp()?

Let Y = FX(X) = 1 eX. So, Y ~ U(0, 1).

Assume U(0, 1) can be generated by a computer.

By letting X = 1ln(1Y), we then have X ~ Exp().

( )XY F X ~ (0,1)Y U

Jointly DistributedRandom Variables

Definition Joint Distribution Functions

The joint (cumulative) distribution function (jcdf) of random variables X and Y is defined by:

FX,Y(x, y) = P(X x, Y y), < x < , < y < .

(x, y)

Properties of a jcdf

1. 0 ( , ) 1, ,F x y x y

1 1 2 2 1 2 1 22. ( , ) ( , ), ,F x y F x y x x y y

(x1, y1)

(x2, y2)

Properties of a jcdf

1. 0 ( , ) 1, ,F x y x y

1 1 2 2 1 2 1 22. ( , ) ( , ), ,F x y F x y x x y y

, ,3. lim ( , ) 0, lim ( , ) 1

x y x yF x y F x y

4. ( , )P a X b c Y d

d(b, d)

(b, c)

(a, d)

(a, c)

( , )F b d ( , )F a d ( , )F b c ( , )F a c

Definition Marginal Distribution Functions

Given the jpdf F(x, y) of random variables X, Y. The marginal distribution functions of X and Y are defined respectively by

( ) ( , )XF x P X x Y ( , )P X x Y

lim ( , )y

( ) lim ( , )Y xF y F yx

Definition Joint Probability Density Functions

A joint probability density function (jpdf) of

continuous random variable X, Y is a nonnegative

function fX,Y(x, y) such that

, ,( ( ), ) ,y x

X XY Yf uF x y dudvv

Properties of a Jpdf

2. ( , ) 1f x y dxdy

3. ( , ) ( , )d b

c aP a X b c Y d f x y dxdy

( , )1. ( , ) X Y

F x yf x y

4. ( ) ( , ) ( ) ( , )x y

X YF x f u y dydu F y f x v dxdv

5. ( ) ( , ) ( ) ( , )X Yf x f x y dy f y f x y dx

f X(u) f Y(v)

Properties of a Jpdf

2. ( , ) 1f x y dxdy

3. ( , ) ( , )d b

c aP a X b c Y d f x y dxdy

( , )1. ( , ) X Y

F x yf x y

4. ( ) ( , ) ( ) ( , )x y

X YF x f u y dydu F y f x v dxdv

5. ( ) ( , ) ( ) ( , )X Yf x f x y dy f y f x y dx

f X(u) f Y(v)

Marginal Probability Density Functions(see next page)

Marginal Probability Density Functions

( ) ( , )

f x f x y dy

f y f x y dx

Example 20

Example 20( , ) 1f x y dxdy

0 01 ( )k x y dxdy 11 21

20 0k x yx dy

k y dy 121 1

2 2 0k y y k

Example 20

0( ) ( )Xf x x y dy

( ) ( , )

f x f x y dy

f y f x y dx

0( ) ( )Yf y x y dx

2 0xy y

2 0x xy

Example 2012

( ) , 0 1

( ) , 0 1X

f x x x

f y y y

( )XF x 120

( ) 0 1

u du x

( )YF y 120

( ) 0 1

v dv y

Example 2012

( ) , 0 1

( ) , 0 1X

f x x x

f y y y

( )XF x 2

( )YF y2

Example 20

, ( , ) , 0 , 1X Yf x y x y x y

, ) ?( ,X YF x y

, ( , )X Yf x y

Example 20

, ( , ) , 0 , 1X Yf x y x y x y

, ) ?( ,X YF x y

, ( , ) ( , )X YF x y P X x Y y , ( , ) ( , )X YF x y P X x Y y

00 0 0

Example 20

, ( , ) , 0 , 1X Yf x y x y x y

, ) ?( ,X YF x y

00 0 0

1(x, y)

, 0 0( , ) ( )

X YF x y u v dudv

Example 20

, ( , ) , 0 , 1X Yf x y x y x y

, ) ?( ,X YF x y

00 0 0

1(x, y)

(x, y)

, 0 0( , ) ( )

X YF x y u v dudv 1

, 0 0( , ) ( )

X YF x y x v dxdv

Example 20

, ( , ) , 0 , 1X Yf x y x y x y

, ) ?( ,X YF x y

00 0 0

1(x, y)

(x, y)

, 0 0( , ) ( )

X YF x y u v dudv 1

, 0 0( , ) ( )

X YF x y x v dxdv

(x, y)1

, 0 0( , ) ( )

X YF x y u y dydu

Example 20

, ( , ) , 0 , 1X Yf x y x y x y

, ) ?( ,X YF x y

Example 21

, ( , ) , 0xX Yf x y e y x

( ) , ( )? ?X Yf x f y

( 5 1 ?| 0)P Y X

Example 21

, ( , ) , 0xX Yf x y e y x

( ) , ( )? ?X Yf x f y

( 5 1 ?| 0)P Y X

Y ( ) ( , )Xf x f x y dy

x xe dy, 0xxe x

Example 21

, ( , ) , 0xX Yf x y e y x

( ) , ( )? ?X Yf x f y

( 5 1 ?| 0)P Y X

Y ( ) , 0xXf x xe x

fY(y)y

( ) ( , )Yf y f x y dx

, 0ye y

Example 21

, ( , ) , 0xX Yf x y e y x

( ) , ( )? ?X Yf x f y

( 5 1 ?| 0)P Y X

Y ( ) , 0xXf x xe x

( ) , 0yYf y e y

( 10, 5)( 5 | 10)

P X YP Y X

e dydx

0.00647

Independence of Random Variables

Definition Independence of Random Variables

Two random variables X and Y are said

to be independent, denoted as , if X Y

, ( , ) ( ) ( ),

,X Y X YF x y F x F y

Theorem 2

X Y , ( , ) ( ) ( ),

,X Y X Yf x y f x f y

, ( , ) ( ) ( ),

,X Y X Yf x y f x f y

Example 22

2 2( ) / 2( , )

x xy yf x y ce

Example 22

2 2( ) / 2( , )

x xy yf x y ce

2 2/ 2 3 / 4 / 2( , )

x y yf x y ce

2 2/ 2 3 / 4 / 2( )

Yf y c e dx

22 / 2 / 23 /8 x yyce e dx

23 /82 yc e 2

22 4/3

2 4 / 3c

22 4/33

Example 22

2 2( ) / 2( , )

x xy yf x y ce

2 2/ 2 3 / 4 / 2( , )

x y yf x y ce

22 4/33

( ) , 2 2

Yf y e y

22 4/33

( ) , 2 2

Xf x e x

( ) ( )X Xf x f y

Example 23

1 2( )1 2( , ) , 0 ,x yf x y e x y

?( , )F x y

Example 23

1 2( )1 2( , ) , 0 ,x yf x y e x y

?( , )F x y

1 2( )1 20

( ) x yXf x e dy

1 21 20

x ye e dy

1 2( )1 20

( ) x yYf y e dx

2 12 10

y xe e dx

Example 23

1 2( )1 2( , ) , 0 ,x yf x y e x y

?( , )F x y

11( ) , 0x

Xf x e x

22( ) , 0x

Yf y e y

( , ) ( ) ( )X Yf x y f x f y

1~ ( )X Exp

2~ ( )Y Exp ( , ) ( ) ( )X YF x y F x F y

1 21 1

x ye e

Distribution of Sums

The Problem

Given f(x, y) or F(x, y), and Z = (X, Y),

FZ(z) = ?

fZ(z) = ?

The Problem

FZ(z) = ?

fZ(z) = ?

( ) ( )

( , )z

F z P Z z

f x y dxdy

( ) ( , ) : ( , )zA z x y x y z

The Distribution of Sums

FZ(z) = ?

fZ(z) = ?

( ) ( )

( , )z

F z P Z z

f x y dxdy

( ) ( , ) :z yA z x y x z

( ) ( )

( , )z

F z P Z z

f x y dxdy

Z = X + Y

( ,( ))z x

Z f x y dyF z dx

Z = X + Y

( ,( ))z x

Z f x y dyF z dx

yy dxyx df

y t x ( , )t x z x

t xd dt tf x xx

( , )z

dxdt x tf x

( , )z

f x t x dxdt

Zf t dt

Z = X + Y

( ,( ))z x

Z f x y dyF z dx

yy dxyx df

y t x ( , )t x z x

t xd dt tf x xx

( , )z

dxdt x tf x

( , )z

f x t x dxdt

Zf t dt

( ) ( , )Zf z f x z x dx

( , )f z y y dy

Z = X + Y

( ) ( , )Zf z f x z x dx

I(X), I(Y) 00

( ) ( , )z

Zf z f x z x dx

( ) ( ) ( )z

Z X Yf z f x f z x dx and

Example 24

Let X ~ Exp(), and Y ~ Exp() be independent. Let Z = X + Y. Find fZ(z).

2 ( )( , ) , 0 ,x yf x y e x y X YI(X), I(Y) 0

0( ) ( , )

Zf z f x z x dx

z zzf z e dx 2 , 0zze z

Fact: ~ (2, )X Y

Example 25

Let X ~ U(0, 1) and Y ~ U(0, 1) be two independent variables. Find fX+Y.

X ~ U(0, 1), Y ~ U(0, 1) X Y

fX+Y = ?

Example 25X ~ U(0, 1), Y ~ U(0, 1) X Y

fX+Y = ?

( , ) 1, 0 , 1f x y x y I(X), I(Y) 0

.Z X Y Define ( ) ?I Z (0,2)

[0 1][0 1]I x z x 積分區間分析

0( ) ( , )

Zf z f x z x dx

[0 1][ 1 ]x z x z [0 1][ 1 ]I x z x z

Example 25X ~ U(0, 1), Y ~ U(0, 1) X Y

fX+Y = ?

[0 1][ 1 ]I x z x z 積分區間分析

Case 1: Case 2:

( , ) 1, 0 , 1f x y x y I(X), I(Y) 0

.Z X Y Define ( ) ?I Z (0,2)0

( ) ( , )z

Zf z f x z x dx 0< z 1 1 < z < 2

[0 ]I x z [ 1 1]I z x

Example 25X ~ U(0, 1), Y ~ U(0, 1) X Y

fX+Y = ?

Case 1: Case 2:

( , ) 1, 0 , 1f x y x y I(X), I(Y) 0

.Z X Y Define ( ) ?I Z (0,2)0

( ) ( , )z

Zf z f x z x dx 0< z 1 1 < z < 2

[0 ]I x z [ 1 1]I z x

Zf z dx z1

1( )Z z

f z dx

Example 25X ~ U(0, 1), Y ~ U(0, 1) X Y

fX+Y = ?

Case 1: Case 2:

( , ) 1, 0 , 1f x y x y I(X), I(Y) 0

.Z X Y Define ( ) ?I Z (0,2)0

( ) ( , )z

Zf z f x z x dx 0< z 1 1 < z < 2

[0 ]I x z [ 1 1]I z x

Zf z dx z1

1( )Z z

f z dx

0 1( )

2 1 2Z

z zf z

Example 26

f(x, y)

X ~ U(0, 1), Y ~ U(0, 1) X Y

(2 2) ?P X Y

0. ?5P X Y

( , ) 1, 0 , 1f x y x y

Example 26

f(x, y)

X ~ U(0, 1), Y ~ U(0, 1) X Y

(2 2) ?P X Y

0. ?5P X Y

( , ) 1, 0 , 1f x y x y

0.5 2 2(2 2)

xP X Y dydx

2x + y = 2

Example 26

f(x, y)

X ~ U(0, 1), Y ~ U(0, 1) X Y

(2 2) ?P X Y

0. ?5P X Y

( , ) 1, 0 , 1f x y x y

y 0.5P X Y 0.5 0.5P X Y

Distributions of Multiplications and Quotients

Distributions of Multiplications and and Quotients

Z XY /Z Y X( , )f x y 已知

) ?(Zf z ( , )f x y 已知

) ?(Zf z

|( ) ,

x xz f x dx

( ) ,| |Zf z f x dxx zx

|( ) ,

xxz f x dx

( ) ,| |Zf z f x dxzxx

|( ) ,

xxz f x dx

( ) ,| |Zf z f x dxzxx

I(X), I(Y) 0 I(X), I(Y) 0

xz f dx

0( ) ,Zf z f xx zx dx

X Yand X Yand

1( ) )Z X Yf

xz x f x

( ) ( ) ( )Z X Y zxf dxz f x f x

Example 27

Let X ~(1, ) and Y ~(2, ) be independent random variables. Find the pdf of Y/X.

X Y1 2~ ( ), ~ ( )X Y

( ?/ , )ZZ Y X f z

Example 27

1 2 1 21 1 ( )

1( , ) , , 0

( ) ( )x yf x y x y e x y

( ) ,Zf z f xx zx dx

1 2 1 (1 )

( ) ( )( ) ( )

x zZf z x zx e dx

11 (1 )

01 2( ) ( )

x zzx e dx

1 2 1 2

( ) ( ) (1 )

Chapter 2Exercise

X Y1 2~ ( ), ~ ( )X Y

( ?/ , )ZZ Y X f z

Example 27

1 2 1 21 1 ( )

1( , ) , , 0

( ) ( )x yf x y x y e x y

( ) ,Zf z f xx zx dx

1 2 1 (1 )

( ) ( )( ) ( )

x zZf z x zx e dx

11 (1 )

01 2( ) ( )

x zzx e dx

1 2 1 2

( ) ( ) (1 )

X Y1 2~ ( ), ~ ( )X Y

( ?/ , )ZZ Y X f z

Conditional Densities

Let X and Y be continuous random variables having j

pdf f. The conditional density fY|X is defined by

( , )( ) 0

( )( | )

f x yf x

f xf y x

otherwise

Facts | ( | )

( , )( ) 0

f x yf x

otherwise

|1. ( , ) (( | ) ).Y X Xf yf fxx y x

|2. ( )( | ) .Y X Yf yX fxY y

3. ( )Yf y ( , )f x y dx

| | ( )( )Y X Xf x dxf y x

|4. ( | ) ( | )Y XF y x P Y y X x ( , )

P X x Y y

( , )lim

P x X x x Y y

P x X x x

( , )lim

f u v dudv

f u du

( , )lim

x f x v dv

f x v dv

f x | ( | )Y X

yf v x dv

Facts | ( | )

( , )( ) 0

f x yf x

otherwise

|1. ( , ) (( | ) ).Y X Xf yf fxx y x

|2. ( )( | ) .Y X Yf yX fxY y

3. ( )Yf y ( , )f x y dx

| | ( )( )Y X Xf x dxf y x

|4. ( | ) ( | )Y XF y x P Y y X x | ( | )Y X

yf v x dv

|5. ( | ) ( | )Y X

aP a Y b X x dvf y x

Example 28

2( , ) 10 , 0 1.f x y xy x y

( 0.5 | .25 ?0 )P Y X

Example 28

2( , ) 10 , 0 1.f x y xy x y

( 0.5 | .25 ?0 )P Y X

1 2( ) 10X xf x xy dy 2

0( ) 10

Yf y xy dx1

xy 2 2

310(1 ),

3x x 45 ,y0 1x 0 1y

| 3103

10( | )

(1 )Y X

xyf y x

0 1.x y

Example 28

2( , ) 10 , 0 1.f x y xy x y

( 0.5 | .25 ?0 )P Y X

| 3103

10( | )

(1 )Y X

xyf y x

0 1.x y

|0.5( 0.5 | 0.25) ( | 0.25)Y XP Y X f y dy

1 0.25

Example 29

~ ( , ), | ~ ( )Y Exp

Example 29

, |( , ) ( | ) ( )Y Yf y f y f

~ ( , ), | ~ ( )Y Exp

| ( | ) , 0yYf y e y

( ) , 0( )

,( ) ( , )Y Yf y f y d

( ) ( )y

, 0( )

0( )ye d

Example 29

, |( , ) ( | ) ( )Y Yf y f y f

~ ( , ), | ~ ( )Y Exp

1( ) , 0

( )Yf y yy

( , )( | )

f yf y

1 ( )( )

Example 29

~ ( , ), | ~ ( )Y Exp

( , )( | )

f yf y

1 ( )( )

| ~ ( 1, )Y y

Multivariate Distributions

Definitions

Properties of Multivariate Distributions

( , , )1. ( , , )

F x xf x x

1 12. ( , , ) 1n nf x x dx dx

11 1 1 1 13. ( , , ) ( , , )

n n n n na aP a X b a X b f x x dx dx

1 1 1 14. ( ) ( , , , , )iX i i n i i nf x f x x x dx dx dx dx

Definition Independence

Random variables X1, …, Xn are called independent if

11 1 1( , , ) ( ) ( ), , ,nn X X n nF x x F x F x x x

1 nX X

11 1 1( , , ) ( ) ( ), , ,nn X X n nf x x f x f x x x

Example 30

( ) , 0i

xX i if x e x

1( )1 1( , , ) , 0 , ,nx xn

n nf x x e x x

Example 311 nX X1 nX X ~ ( )iX Exp

min , , ( )

max , ,

?( )n Y

Y X X f y

Z X X f z

min , , ( )

max , ,

?( )n Y

Y X X f y

Z X X f z

( ) ( )YF y P Y y

( )( ) Y

dF yf y

1min , , nP X X y

11 min , , nP X X y

11 , , nP X y X y

11 nP X y P X y

( ) ( )

1 , 0iX i

F x P X x

( ) ( )

1 , 0iX i

F x P X x

1 , 0n ye y

, 0n yn e y

~ ( )Y Exp n

min , , ( )

max , ,

?( )n Y

Y X X f y

Z X X f z

( ) ( )ZF z P Z z

( )( ) Z

dF zf z

1max , , nP X X z

1 , , nP X z X z

1 nP X z P X z ( ) ( )

1 , 0iX i

F x P X x

( ) ( )

1 , 0iX i

F x P X x

11 , 0

nz zn e e z

Important Theorem of Sums

To be proved in the next chapter.

:iX 有何意義

1 2 :nX X X 有何意義

:iX 有何意義

2 1 1,

熟記 !!! 靈活的將它們用於解題

Multidimensional Changes of Variables

Let X1, X2, …, Xn be continuous r.v.’s with jpdf 1 2, , , 1 2( , , , )

nX X X nf x x x

Y g X X

1 2, , , 1 2 ) ?( , , ,nY Y Y nf y y y

nX X X nf x x x

Y g X X

1 2, , , 1 2 ) ?( , , ,nY Y Y nf y y y

假設此函式為一對一

求反函式求反函式

nX X X nf x x x

1 2, , , 1 2 ) ?( , , ,nY Y Y nf y y y

Y g X X

一對一

求反函式求反函式1 1 1

x x xy y y

JacobinMatrix

1 2 1 2, , , 1 2 , , , 1 2( , , , ) det( ,) ,( ),n nY Y Y nn X X Xf y y y f hJ h h

Example 34

1 2, 1 2 1 2 1 2( , ) 4 , 0 , 1.X Xf x x x x x x

1 2, 1 2( , ) ?Y Yf y y

Example 34

1 2, 1 2 1 2 1 2( , ) 4 , 0 , 1.X Xf x x x x x x

1 2, 1 2( , ) ?Y Yf y y

求反函式求反函式1 1

x x xy y y

JacobinMatrix

1/ 2112

3/ 2 1/ 211 2 12

Example 34

1 2, 1 2 1 2 1 2( , ) 4 , 0 , 1.X Xf x x x x x x

1 2, 1 2( , ) ?Y Yf y y

JacobinMatrix

1/ 2112

3/ 2 1/ 211 2 12

1det( )

1 2 1 2, 1 2 , 1 2 11

1( , ) ,

2Y Y X Xf y y f y y yy

1 2 11

2y y y

Example 34

1 2, 1 2 1 2 1 2( , ) 4 , 0 , 1.X Xf x x x x x x

1 2, 1 2( , ) ?Y Yf y y

JacobinMatrix

1/ 2112

3/ 2 1/ 211 2 12

1det( )

1 2 1 2, 1 2 , 1 2 11

1( , ) ,

2Y Y X Xf y y f y y yy

1 2 11

2y y y

y 2 10 1y y

Example 35

1 23 21 2 1 2( , ) 6 , , 0.x xf x x e x x

1 2Y X X ) ?(Yf y

1 1 10( ) ( , )

Yf y f x y x dx 1 13 2( )10

6y x y xe dx

6y xye e dx 12

yxye e

26 1y ye e 2 36 , 0y ye e y

Example 35

1 23 21 2 1 2( , ) 6 , , 0.x xf x x e x x

1 2Y X X ) ?(Yf y

11 1( ) ( )Y Yf y f y

Define 1 1 2

1 2, 1 2 2( , )Y Yf y y dy

x x xy y y

JacobinMatrix

Example 35

1 23 21 2 1 2( , ) 6 , , 0.x xf x x e x x

1 2Y X X ) ?(Yf y

Define 1 1 2

1 2, 1 2 2( , )Y Yf y y dy

JacobinMatrix

det( ) 1J

1 2 1 2, 1 2 , 2 1 2( , ) 1 ( , )Y Y X Xf y y f y y y 2 1 23 2( )6 y y ye 1 22

2 16 , 0y ye y y

?11 1( ) ( )Y Yf y f y

Example 35

1 23 21 2 1 2( , ) 6 , , 0.x xf x x e x x

1 2Y X X ) ?(Yf y

Define 1 1 2

1 2, 1 2 2( , )Y Yf y y dy

JacobinMatrix

det( ) 1J

2 16 , 0y ye y y

y y ye dy

1 12 316 , 0y ye e y

11 1( ) ( )Y Yf y f y

Example 35

1 23 21 2 1 2( , ) 6 , , 0.x xf x x e x x

1 2Y X X ) ?(Yf y

Define 1 1 2

1 2, 1 2 2( , )Y Yf y y dy

JacobinMatrix

det( ) 1J

2 16 , 0y ye y y

y y ye dy

1 12 316 , 0y ye e y

11 1( ) ( )Y Yf y f y

2 3( ) 6 , 0y yYf y e e y

Example 36

此例非一對一，以上方法非直接可用，請參考講義。

Chapter 4-2 Continuous Random Variables 主講人 : 虞台文

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