Chapter 3Chapter 3 Random vectors Random vectors

and their numerical and their numerical

characteristicscharacteristics

3.1 Random vectors1.n-dimension variables

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

Let (X, Y) be bivariable and (x, y)R2, define

F(x,y)=P{Xx, Yy}

the bivariate joint distribution of (X, Y)

Bivariate distribution

00 , yx

00 ,,, yyxxyx

Geometric interpretation ： the val

ue of F( x, y) assume the probability

that the random points belong to eara

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 distribution of (X,Y) is (x,y), find the probability that (X,Y) stands in erea

G .

Answer

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

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

22313221

13323312

yxyxyxFyxF

yxyxyxFyxFGYXP

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-hand 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 of some bivariable.

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 that X

and Y have a discrete joint distribution .

The joint distribution is defined to be a function such t

hat 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 form 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

X

Please find the joint distribution 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

x y

dudvvufyxF ,),(),(

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) 若 f (x, y) 在 (x, y)R2 处连续，则有

( , ) 1;f x y dxdy

－ －

G

dxdyyxfGYXP .),(}),{(

(4) For any region G R2,

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 yx1

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}),{(

3. Bivariate uniform distribution Bivariate (X, Y) is said to follow uniform distribution if the density function of is specified by

21 ( , )

( , ) the measure of 0 ,

x y D Rf x y D

，

el se

D

G

S

SGYXP }},{(

By the definition, one can easily to find that if (X, Y) is uniformly distributed, then

Suppose that (X,Y) is uniformly distributed on area D, which is indicated by the graph on the left hand, try to determine:(1)the density function of (X,Y) ；(2)P{Y<2X} ；(3)F(0.5,0.5)

1DS

others

Dyxyxf

0

),(1),()1(

Answer

4

11

2

1

2

1GS

4

1

2

11

2

13 S

4

11

41

}2{)2( XYP

4

1)5.0,5.0()3( F

}2{ XYG

H

}5.0,5.0{ YXH

where ， 1 、 2 are constants and 1>0,2>0 、 |

|<1 are also constant , then, it is said that (X, Y)

follows the two-dimensional normal distribution

with parameters 1, 2, 1, 2, and denoted it by

),,,,(~),( 2

22121 NYX

(2)Two dimensional normal distribution

Suppose that the density function of (X, Y) is specified by

,e12

1)y,x(f

])y()y)(x(

2)x(

[)1(2

1

221

22

22

21

2121

21

2

The concept of joint distribution can be easily generalized to n-dimensional random vectors.

nnn bxabxaxxD ,...:,... 111

D nnn dxdxxxfDXXP ...),...,x(...... 1211

Definition. Suppose that (X1,X2,...Xn) is a n-dimensional random vector ， if for any n-dimensional cube

there exists a nonnegative f(x1,x2,...xn) such that

It is said that (X1,X2,...Xn) follows continuous n-dimensional distribution with density function f(x1,x2,...xn)

Definition. Suppose that (X1,X2,...Xn) assume finite or countable points on Rn . We call that X1,X2,...X

n) is a discrete distributed random vector and P{X1=x1,X2=x2,...Xn=xn}, (x1,x2,...xn) R∈ n

is the distribution law of (X1,X2,...Xn)

Multidimensional random varibables

Discrete d.f.

Distribution function

Continuous d.f.

Probability for area Standardized

P{(X,Y)G}

Determine: （ 1） P{X0},(2)P{X1},(3)P{Y y0}

others

yxeyxf

y

0

0),(

EX: Suppose that （ X ， Y ） has density funciton

x

y

DAnswer: P{X0}=0

11

0

1}1{

edyedxXPx

y

00

0}{

0

0

00

00

y

ydyedxyYP

y

x

yy

FY(y) ＝ F (+, y) ＝ ＝ P{Yy}

)y,x(Flimy

)y,x(Flimx

3.2 Marginal distribution and independence

FX(x) ＝ F (x, +) ＝ ＝ P{Xx}

Marginal distribution function for bivariate

Define

the marginal distribution of (X, Y) with respect to X and Y respectively

Example 1. Suppose that the joint distribution of (X,Y) is specified by

1 0

( , ) 1 0

0

x y

y y

e xe x y

F x y e ye y x

else

Determine FX(x) and FY(y) 。Answer:

FX(x)=F(x,)=

00

01

x

xe x

FY(y)=F(,y)=

00

01

y

yyee yy

Marginal distribution for discrete distribution

Suppose that

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

2, … Define P{X ＝ xi} ＝ pi. ＝ ， i ＝ 1, 2, … P{Y ＝ yj} ＝ p.j ＝ ， j ＝ 1, 2, …

1j

ijp

1i

ijp

the marginal distribution of (X, Y) with respect to X and Y respectively.

Marginal density function

the joint distribution of (X,Y) with respec to X and Y.

dyyxfxf X ),()(

dxyxfyfY ),()(

Suppose that (X, Y) ～ f (x, y), (x, y)R2, define

One canc easily to find that the marginal dist

ribution of N(1, 2, 12, 2

2, ) is N(1, 12) and

and N(2, 22).

Example 3. Suppose that the joint density

function of (X,Y)is specified by

others

xyxcyxf

0),(

2

Determine (1) the value of c;

(2)the marginal distribution of (X,Y) with respect to X

2

1

0

(1) 1x

x

dx cdy 6 c

dyyxfxf X ),()()2(100 xorx

10)(66 2

2

xxxdyx

x

Independence of random vectors

Definition It is said that X is independent of Y for any real nu

mber a<b, c<d,p{a<Xb,c<Yd} =p{a<Xb}p{c<Yd},i.e.

event{a<Xb}is independent of {c<Yd}.

Theorem A sufficient and necessary condition for random variables X and Y to be independent is

F(x,y)=FX(x)FY(y)

Remark

(1) If (X,Y) is continuously distributed, then a suffici

ent condition for X and Y to be independent is f(x,y)=

fX(x)fY(y)

(2) If (X,Y) is discrete distributed with law Pi,j=

P{X=xi,Y=yj},i,j=1,2,...then a sufficient and continousl

y for X and Y to be independent is Pi,j=Pi.Pj

EX ： try to determine that whether the (X,Y) in

Example1 ， Exmaple2 ， Example3 are independent or not?

Example 4. Suppose that the d.f. of (X,Y) is give by the following chart and that X is independent of Y, try to determine a and b.

x 1 20 0.15 0.15

1 a b

0.15 0.15 1a b 0.7a b 0.15 ( 0.15) 0.3a

0.35, 0.35a b

§ 3.3 CONDITIONAL DISTRIBUTION

Definition 3.7 Suppose is an discrete two-dimensional distribution, for given , if, then the conditional distribution law for given can be represented by which is defined as (3.14)

and (1)

(2)

( , )X Yj

( , )( ) ,

( )i j ij

i jj j

P X x Y y pP X x Y y

P Y y p

| 0;i jp

| 1.i ji

p

| 0i jp

jY y |i jp

Definition 3.9 Suppose that for any , holds, if

exist, then the limit is called the conditional distribution of for given and denoted it by or .

0x ( ) 0P x X x x

0 0

( , )lim ( | ) lim

( )x x

P Y y x X x xP Y y x X x x

P x X x x

Y

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

Theorem 3.6 Suppose that has continuous density function and then (3.15)

is a continuous d.f. and its density function is which is

called the conditional density function of given the condition

and denoted it by

( , )X Y( , )p x y ( ) 0,Xp x

|

( , )( | )

( )

y

Y XX

p x v dvF y x

p x

( , ),

( )X

p x y

p x

Y

X x

| ( | ).Y Xp y x

3.4 Functions of random vectorsFunctions of discrete random vectors

Suppose that

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

then Z ＝ g(X, Y) ～ P{Z ＝ zk} ＝ ＝ pk ,

k ＝ 1, 2, …

kji zyxgkiijp

),(:,

(X,Y) (x1,y1) (x1,y2) … (xi,yj) …

pij p11 p12 pij

Z=g(X,Y) g(x1,y1) g(x1,y2

)g(xi,y

j)

or

EX Suppose that X and Y are independent a

nd both are uniformly distributed on 0-1 with law X 0 1

P q p Try to determine the distribution law of (1)W ＝ X ＋ Y ; (2) V ＝ max(X, Y) ；(3) U ＝ min(X, Y);(4)The joint distribution law of w and V .

(X,Y) (0,0) (0,1) (1,0) (1,1)

pij

W ＝ X ＋YV ＝max(X, Y)

U ＝ min(X, Y)

2q pq pq 2p

0 1 1 2

0 1 1 1

0 0 0 1

VW

0 1

0 1 2

2q 0 0

0 pq22p

Example 3.17 Suppose and are independent of

then each other, then

1 2~ ( ), ~ ( )X P Y P

1 2~ ( ).Z X Y P

Example 3.18 Suppose and are independent of , then

~ ( , ), ~ ( , )X b n p Y b m p

~ ( , ).Z X Y b m n p

Let be the joint density function of , then the density function of can be given by the following way.

and

( , )p x y ( , )X Y

Z X Y

( ) ( ) ( ) ( , )

( , ) [ ( , ) ]

z

D

z x

x y z

F z P Z z P X Y z p x y dxdy

p x y dxy p x y dy dx

( , ) ( , )z z

u y x dx p x u x du p x u x dx du

令

( ) ( , )Zp z p x z x dx

Suppose is the density function of and

has continuous partial derivatives, the inverse transformation

exists, the Jacobian determinant J is defined as follows:

Define then the joint density function of can

be determined as

( , )p x y ( , )X Y1

2

( , ),

( , ).

u g x y

v g x y

( , ),

( , )

x x u v

y y u v

1

1( , ) ( , )

0.( , ) ( , )

u ux yx yx y u vu uJ

x y v vu v x y

v v x y

1

2

( , )

( , ),

U g X Y

V g X Y

( , )U V

( , ) ( ( , ), ( , )) .p u v p x u v y u v J

Example 3.23 Suppose that is independent of with marginal density and respectively, then the density function of is specified by

Remark Under the conditions of Example 3.23, one can easily find the density function of is

( )Xp xX Y

( )Yp y

U XY

1( ) ( ) ( ) .U X Yp u p u v p v dv

v

U X Y

( ) ( ) ( ) .U X Yp u p uv p v v dv

1.Definition Suppose that the variance of r.v. X and Y exist,

define the expectation E{[XE(X)][YE(Y)]} is the covariance

of X and Y, Cov(X, Y)=E(XY-E(X)E(Y).

If Cov(X,Y)=0 ， It is said that X and Y are uncorrelated

What is the different between the concept of “X and Y are in

dependent”and the one of “X and Y are uncorrelated”?

Numerical characteristics of random vectorsNumerical characteristics of random vectors

Example 2 Suppose that (X, Y) is uniformly

distributed on D={(X, Y) ： x2+y21} .Prove that X

and Y are uncorrelated but not independent.

Proof

others

yxyxf0

11

),(22

11

1

1

01

)(

2

2

1

1

1

1

dyxdxXEx

x

0)(

2

2

1

1

1

1

dyxy

dxXYEx

x

0)()()(),( YEXEXYEYXCov

Thus X and Y are uncorrelated. Since

others

xxdyxf

x

xX

0

11121

)(

2

2

1

1

2

others

yydyyf

y

yY

0

11121

)(

2

2

1

1

2

)()(),( yfxfyxf YX

Thus, X is not independent of Y.

2. Properties of covariance:

(1) Cov(X, Y)=Cov(Y, X);

(2) Cov(X,X)=D(X);Cov(X,c)=0

(3) Cov(aX, bY)=abCov(X, Y), where a, b are constants

Proof Cov(aX, bY)=E(aXbY)-E(aX)E(bY)

=abE(XY)-aE(X)bE(Y)

=ab[E(XY)-E(X)E(Y)]

=abCov(X,Y)

(4) Cov(X+Y ， Z)=Cov(X, Z)+Cov(Y, Z);

Proof Cov(X+Y ， Z)= E[(X+Y)Z]-E(X+Y)E(Z)

=E(XZ)+E(YZ)-E(X)E(Z)-E(Y)E(Z)

=Cov(X,Z)+Cov(Y,Z)

(5) D(X+Y)=D(X)+D(Y)+2Cov(X, Y).

Proof

)()(2)(2)()()( YEXEXYEYDXDYXD

),(2 YXCovRemark D(X-Y)=D[X+(-Y)]

=D(X)+D(Y)-2Cov(X,Y)

Definition of covariance

Properties of Covariance

Independent and c uncorrelated

Coefficient

Coefficients

Definition Suppose that r.v. X ， Y has finite variance, dentoed by DX>0,DY>0 ， respectively, then,

DYDX

)Y,Xcov(XY

is name the coefficient of r.v. X and Y .

Obviously, EX*=0 ， DX*=1 and

).(),cov( **** YXEYXXY

DX

XEXX

)(* Introduce which is the standardized of X

Properties of coefficients (1) |

XY|1 ；

(2) |XY

|=1There exists constants a, b such that P {Y=

aX+b}=1 ； (3) X and Y are uncorrelated XY;

1. Suppose that (X,Y) are uniformly distributed on D:0<x<1,0<y<x, try to determine the coefficient of X and Y.

D

1

x=y

others

Dyxyxf

0

),(2),(Answer

3

22)(

0

1

0

x

dyxdxXE

3

12)(

0

1

0

x

ydydxYE

4

12)(

0

1

0

x

ydyxdxXYE

1( , ) ( ) ( ) ( )

36COV X Y E XY E X E Y

18

1

9

42)(

0

1

0

2 x

dydxxXD

18

1

9

12)(

0

21

0

x

dyydxYD

)()(

),(

YDXD

YXCOVXY

2

1D

1

2

2

1) ~ (0,1), , determine

2 ~ ( 1,1), , determine

XY

XY

X U Y X

X U Y X

）

What does Example 2 indicate ？

Answer 1)

45

4)(,

12

1)(,

4

1)(,

3

1)(,

2

1)( YDXDXYEYEXE

968.0

454

121121

XY 2) 0)(,0)( XYEXE

0XY

2 21 2 1 2Example 3 Suppose ( , ) ~ ( , , , , ), .XYX Y N then

Thus, if （ X， Y） follow two-dimensional distribution, then “X and Y are independent” is equvalent to “X and Y are uncorrelated” 。