Chapter 3 Multivariate Random Variables

1.n-dimensional variables

n random variables X1， X2， ...,Xn compose a n-dimensional vector (X1,X2,...,Xn), and the vector is named n-dimensional variables or random vector.

2. Joint distribution of random vector

Define function

F(x1,x2,…xn)= P(X1≤x1,X2 ≤ x2,...,Xn ≤ xn)

the joint distribution function of random vector (X1,X

2,...,Xn).

3.1 Two-Dimensional Random Variables

Definition 3.1-P53

Let (X, Y) be 2-dimensional random variables. Define F(x,y)=P{Xx, Yy}

the bivariate cdf of (X, Y) .

The Joint cdf for Two random variables

00 ,,, yyxxyx

Geometric interpretation ： t

he value of F( x, y) assume th

e probability that the random p

oints belong to area in dark

For (x1, y1), (x2, y2)R2, (x1< x2 ， y1<y2 ), then

P{x1<X x2 ， y1<Yy2 }

＝ F(x2, y2) － F(x1, y2) － F (x2, y1) ＋ F (x1, y1).

(x1, y1)

(x2, y2)

(x2, y1)

(x1, y2)

1x 2x 3x

1y

2y

3ySuppose that the joint cdf of (X,Y) is F(x,y), find the probability that (X,

Y) stands in area G .

Answer

2 1 3 3 2 3 3 1

1 2 2 3 1 3 2 2

{( , ) } [ ( , ) ( , ) ( , ) ( , )]

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

P X Y G F x y F x y F x y F x y

F x y F x y F x y F x y

L

Joint distribution F(x, y) has the following characteristics:

0),(lim),(

yxFFyx

1),(lim),(

yxFFyx

0),(lim),(

yxFyFx

0),(lim),(

yxFxFy

(1) For all (x, y) R2 , 0 F(x, y) 1,

(2) Monotonically increment

for any fixed y R, x1<x2 yields

F(x1, y) F(x2 , y) ；

for any fixed x R, y1<y2 yields

F(x, y1) F(x , y2).

);y,x(F)y,x(Flim)y,0x(F 0xx

00

).y,x(F)y,x(Flim)0y,x(F 0yy

00

(3) right continuous for xR, yR,

(4) for all (x1, y1), (x2, y2)R2, (x1< x2 ， y1<y2 ),

F(x2, y2) － F(x1, y2) － F (x2, y1) ＋ F (x1, y1)0.

Conversely, any real-valued function satisfied the aforementioned 4 characteristics must be a joint distribution function of 2-dimensional variables.

Example 1. Let the joint distribution of (X,Y) is

)]3

()][2

([),(y

arctgCx

arctgBAyxF

1) Find the value of A ， B ， C 。2) Find P{0<X<2,0<Y<3}

Answer 1]2

][2

[),(

CBAF

0)]3

(][2

[),( y

arctgCBAyF

0]2

)][2

([),(

Cx

arctgBAxF

2

1

2

ACB

16

1)0,2()3,0()3,2()0,0(}30,20{ FFFFYXP

Discrete joint distribution

If both x and y are discrete random variable,

then,(X, Y) take values in (xi, yj), (i, j ＝ 1, 2, … ), it is said t

hat X and Y have a discrete joint distribution .

Definition 3.2-P54

The joint distribution is defined to be a function such th

at for any points (xi, yj),

P{X ＝ xi, Y ＝ yj,} ＝ pij ， (i, j ＝ 1, 2, … ).

That is

(X, Y) ～ P{X ＝ xi, Y ＝ yj,} ＝ pij ， (i, j ＝ 1, 2, … ) ，

X Y y

1 y

2 … y

j …

p11 p12 ... P1j ...

p21 p22 ... P2j ...

pi1 pi2 ... Pij ...

......

...

...

...

...

... ...

Characteristics of joint distribution :(1) pij 0 , i, j ＝ 1, 2, … ； (2) 1

1 1

＝ i j

ijp

x1

x2

xi

The joint distribution can also be specified in the following table

Example 2. Suppose that there are two red balls and three white balls in a bag, catch one ball from the bag twice without put back, and define X and Y as follows:

ball whiteisput second the0

ball red isput second the1

ball whiteisput first the0

ball red isput first the1

Y

XPlease find the joint pmf of (X,Y)

XY

1 0

1 0

10

110

3

10

3

10

3

25

22}1,1{

P

PYXP

25

32}0,1{

PYXP

25

23}1,0{

PYXP

25

23}0,0{

P

PYXP

Continuous joint distributions and density functions

1. It is said that two random variables (X, Y) have a continuous joint distribution if there exists a nonnegative function f (x, y) such that for all (x, y)R2 ， the distribution function satisfies

and denote it with

(X, Y) ～ f (x, y) ， (x, y)R2

2. characteristics of f(x, y)

(1) f (x, y)0, (x, y)R2;

(2)

);,(),(2

yxfyx

yxF

(3)

( , ) 1;f x y dxdy

－ －

(4) For any region G R2,

G

dxdyyxfGYXP .),(}),{(

G

dxdyyxfGYXP .),(}),{(

Let

others

yxyxfYX

0

10,101),(~),(

Find P{X>Y}

2

11}{

0

1

0

x

dydxYXP

1

1

x

y

Find (1)the value of A ； (2) the value of F(1,1) ； (3) the probability of (X, Y) stand in

region D ： x0, y0, 2X+3y6

casesother for ,0

0,0,),(~),(

)32( yxAeyxfYX

yx

Answer (1) Since

6 A

1

0

1

0

32)32( )1)(1(6)1,1()2( eedxdyeF yx

1

1

－ －

-(2x+3y)

0 0

f(x, y)dxdy = Ae dxdy = 1

(3)

3

0

322

0

)32(6 dyedx

x

yx

671 e

dxdyeDYXPD

yx )32(6}),{(