Probability theory

The department of math of central south university

Probability and Statistics Course group

1 、 Multi-dimensional random variable & distribution f

unction

2 、 Multi-dimensional marginal distribution of random

variables

3 、 The mutual independence of random variables

§3.3 Multi-dimensional random variable

When a random phenomenon is considered , many

random variables are often needed to be studied ,such as

launching a shell, the need to study the impact point by

several coordinates; study the market supply model, the

needs of the supply of goods, consumer income and

market Prices and other factors must be taken into

account

1 、 Multi-dimensional random variable & distributi

on function

definition 3.3 Suppose randm variable series ζ1 (

ω), ζ2 (ω),…, ζn (ω) are definited in the same prob

ability space (Ω, F, P ) ， Then ζ (ω)= (ζ1 (ω), ζ2 (

ω), …, ζn (ω)) is called a n-dimensional random ve

ctor or a n-dimensional random variable.

is called the distribution of n-dimensional rand

om variable

ζ(ω)= (ζ1(ω),ζ2(ω), …, ζn(ω))

The following function

))(,,)(,)((

),,,(

2211

21

nn

n

xxxP

xxxF

For a n-dimensonal random vector, each of its com

ponents is a one-dimensional random variables, and c

an be separately studied . In addition, most important

is that each pair of components are interrelated.

We will pay more attention to the two-dimensional

random variables. In fact The results about two-dim

ensional random variables can be applied to multi

-dimensional random variables.

Definition The basic space of random experiment

E isΩ, ξandηare random variables definited onΩ,th

en (ξ ， η) is called a two-dimensional variable

1.1 、 Two-dimensional random variable &

distribution function

a Two-dimensional random variable can be ragard

ed as a random point (ξ ， η) on x-o-y plane space with

value （ x ， y ）。 according whether the number of

points （ x ， y) that (ξ ， η) get is finite or not ,the tw

o-dimensional random variables are divided into two m

ajor discrete and continuous random variables

y

xo

),( yx

is called the distribution function of two-dimensi

onal random variable (ξ ， η) ， or the joint dist

ribution function of ξandη.

},{),( yxPyxF

Definition Suppose (ξ ， η)is a two-dimensional ra

ndom variable,for any real numbers x ， y ， binary

variable function

Please pay attention to these rules:

1° {ξ≤x , η ≤ y } express the product event of {ξ ≤x }an

d {η ≤ y } ．

2° The function value F(x ， y) is the probability that

(ξ ， η) get values on the following region :

－∞＜ ξ≤ x , －∞＜ η ≤ y ． y

xo

),( yx

The distribution function F(x, y) of two dimensional random variable (ξ,η)has the following properties :

1° 0≤F(x,y),and for any number x,y ,F(x,y) satisfies

,0),(,0),( xFyF

1),(,0),( FF

1.1.1 The distribution function properties of

two-dimensional random variable

2°F(x ， y) is a nondecreasing function for variavles x and y ．

),(),(),(),(},{ 111221222121 yxFyxFyxFyxFyyxxP y

xo

2y

1y

1x 2x

3°F(x ， y) is left continuous for x and y

4° The probability that (ξ ， η) satisfy

x1 ＜ ξ≤x2 ， y1 ＜ η≤y2 is

1.2 、 Two-dimensional continuous random variable

Definition Suppose (ξ ， η)is a two-dimensional random variable with distribution function F(x,y) ， if there exists nonnegative function f(x,y) for any x ， y ,and F(x,y)satisfies the following integral equation

x yyxyxfyxF dd),(),(

x ydudvvuf ),(

then (ξ ， η) is called a two-dimensional continuous random v

ariable ， f(x,y) is called the joint probability density function

of (ξ ， η) ． f(x,y) has the following features

1dd),( yxyxfFeature 2

0),( yxfFeature 1

Feature 3 f(x,y) meets the following expression

at continuous points

yx

yxFyxf

),(

),(2

G

yxyxfGYXP dd),(}),{(

Feature 4 Let G be a regional of x-o-y plane

the probability that points (X, Y) fell within G is

Example 6 It is given the probability density function

for a two-dimensional random variable (ξ ， η)

.,0,0,0,),(

)32(

othersyxkeyxf

yx

（ 1 ） What valute is k?

（ 2 ） What expression is the distribution function F(x,y)?

（ 3 ） What is the probability that ξis large than η?

Solution （ 1 ） We have

1dd),( yxyxf

（ 1 ） What valute is k?

（ 2 ） What expression is the distribution function F(x,y)?

（ 3 ） What is the probability that ξis large than η?

0 0

)32( dddd),( yxkeyxyxf yx and

（ 2 ） When x ＞ 0 ， y ＞0

x y yxx yyxeyxyxfyxF

0 0

)32( dd6dd),(),(

.,0,0,0),1)(1(),(

32

othersyxeeyxF

yx

6dd

0

3

0

2 kyexek yx

So k =6 ．

As to other points (x ， y) ， for f (x,y) =0 ， then F(x,y)=0 ． The distribution function can be given as follows:

)1)(1( 32 yx ee

（ 3 ） Grapy the regional G={(x,y)|x ＞ y },and

we have

}),{(}{ GPP

G G

yxyxfyxyxf1

dd),(dd),(

5

3d6d

0

)32(

0 yex

x yx

y

xo

1G

(1) ． Uniform distribution

Let D be a bound regional in x-o-yplane with area S ， (ξ ， η) is a two-dimensional continuous random variab

le with density function

others

DyxSyxf

,0

,),(,1

),(

1.3 、 Several common two-dimensional continuous random variable

then (ξ,η) is called to subject to uniform distribution

（ 1 ） What is the probability density function of (ξ,η) ?

Example 7 A two –dimensional random variable (ξ,

η) subjects to uniform in region

}10,0|),{( xxyyxD

4

30,

4

3

2

1

（ 2 ） What is the probability that (ξ,η) gets value in region

Solution （ 1 ） Draw the graph of regional D

and acounting the area,then the probability dens

ity function of (ξ,η) is given as follows

y

xoD

1

1

.,0,),(,2

),(Others

Dyxyxf

（ 2 ） Marking the following ragionals

4

30,

4

3

2

1),( yxyxG

4

3

2

1,0),(1 xxyyxG

4

3

y

xo1G

2G

2

1

4

3 1

4

3

2

1,

4

3),(2 xyxyxG

we have

}),({4

30,

4

3

2

1GPYXP

16

5dd0dd2dd),(

21

GGG

yxyxyxyxf

4

3

y

xo1G

2G

2

1

4

3 1

Then (ξ ， η) is subjected to a two-dimensional normal

distribution with parameters

(2) ． Normal distribution

Suppose (ξ ， η) is a two-dimensional random variable wih probability density function

22

22

21

2121

12

)())((2

)(

)1(2

1

221 12

1),(

yyxx

eyxf

yx ,

Here are constants ， and ,,,, 2121

,11,0,0 21

,,,, 2121

),,,,( 22

2121 Nand denoted as (ξ,η) ~

Example 8 suppose (ξ ， η) is a two-dimension random variable with density function

yxeyxfyx

,,2

1),(

)(2

1

2

222

Solution ．

G

yxyxfGP dd),(}),{(

rre

yxe

r

yx

G

dθd2

1

dd2

1

0

22

02

)(2

1

2

2

2

222

2

1

1

e

}|),{( 222 yxyxG}),{( GP Please calculate

Have a break !

§ 3.3 the distribution of multi-dimensional

random variables (continued)

2 、 the marginal distribution of two-dimensional random variables

Given the distribution function F(x ， y) of (ξ ，η) ， then the marginal distribution function

of random variable ξ is as follows:

),(},{}{)( xFxPxPxF

),(lim yxFy

(1)Discrete random variables

),2,1,(},{ jipyxP ijji

},{}{ ii xPxPThen

It is known the joint distribution law of random variable (ξ ， η) in the following

),(lim),()( yxFyFyFx

the marginal distribution function of random

variable η can also be expressed as follows:

Let’s racall the marginal distribution of discrete raandom variables.

11

},{})(,{j

jij

ji yxPyxP

11

},{j

iijj

ji ppyxP

),2,1(}{1

ippxP

jijii

That is ,the marginal distribution law of random variable ξ can be expressed as

Similarly ,the marginal distribution law of random variable ηis as follows

),2,1(}{1

jppyP

iijjj

Example 9 The joint distribution law of (ξ ，η) is as follows.

  η ξ 1 2

1 1 / 6 0 2 0 1 / 6 3 1 / 6 0 4 0 1 / 6 5 1 / 6 0 6 0 1 / 6 Please calculate the marginal distribution law of random variable of ξ 、 η,respectively.

6

10

6

1}1{ 12111 pppP Solution

6

1

6

10}2{ 22212 pppP

6

1

6

10}6{ 62616 pppP

ξ 1 2 3 4 5 6

P 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6

… …

So we can easily outline the distribution law of ξ in the following tableau

61514131211111 pppppppP }{

62524232221222 pppppppP }{

The marginal distribution of η is :

Y 1 2

P 1 / 2 1 / 2

2

10

6

10

6

10

6

1

2

1

6

10

6

10

6

10

(2) 、 continuous random variable

Random variable (ξ ， η) has jointed density f

unction f (x, y) ， then the marginal distribution

funtion of ξ can be expressed as

xxyyxfxFxF dd),(),()(

the marginal probability density function of ξ is

)(d),()(

xyyxfxf

With the same ， the marginal probability density of η is

)(d),()(

yxyxfyf

Example 10 Suppose (ξ ， η) subject to uniform distribution on regi

on D surrounded by the curve y = x2 and y = x .What are the margina

l density functions of random variablesξ 、 η.

y

xo

1

2xy

x

D

Solution the area of region D is

6

1d)(

1

0

2 xxxS

so the jointed density function of (ξ ， η) is

othersDyx

yxf,0

),(,6),(

When 0 ＜ x ＜ 1 ， )(6d6d),()( 2

2xxyyyxfxf

x

x

0d),()( yyxfxf

y

xo

1

2xy

x

D

When x≤0 or x ≥1 ，

Therefore

.,0,10),(6)(

2

Othersxxxxf

Similarly

.,0,10),(6)(

othersyyyyf

1 ． Suppose (ξ ， η) is subjected to uniform

distribution on a region surrounded by linears x

=0 ， y=0 ， x+y=1.Find the marginal distributi

on of random variablesξ 、 η ．

.,0,10),1(2

)(.,0,10),1(2

)(othersyy

yfothersxx

xf

Exercise

2 ． If (ξ ， η)~ ),,,,( 22

2121 N

),( yxf

),( yx

Find the marginal distributions of ξ 、 η.

22

22

21

2121

21

2221

212

1

12

1

)y()y)(x()x(

)(exp

The marginal density functions of ξ 、 η are outlined as follows ,respectivily. ， ．

yeyf

xexf

x

x

,2

1)(

,2

1)(

22

22

21

21

2

)(

2

2

)(

1

),(~ 211 N

),(~ 222 N

That is ,

Definition (ξ ， η)is a two –dimensional ra

ndom variable ， if the joint distribution of (ξ ，η) equal the product of marginal distribution o

f ξ and η , then ξ and η are independent of eac

h other.

3, The mutual independence of random variables

If is the jointed distribution of (ξ ， η) , Fξ

( x) 、 Fη(y) are the marginal distribution fun

ction of ξ 、 η ,respectivily,then the necessar

y and sufficient conditions of thatξ and η are

mutually independent is

)()(),( yFxFyxF

Especially,For a two-dimensional discrete random varia

ble (ξ ， η) ， then the necessary and sufficient condit

ions of thatξ and η are mutually independent is

,2,1

,2,1}{}{},{j

iyPxPyxP jiji .

Moreover,for a two-dimensional continuous rand

om variable (ξ ， η) ， then the necessary and

sufficient conditions of thatξ and η are mutually

independent is

y

xyfxfyxf )()(),(

Example 11 A two-dimensional random variable (ξ ， η) has probability density function as follows

Judge whether ξ,ηare mutual independent or not

.,0,0,0,),(

)(

Othersyxxeyxf

yx

,0,0

,0,d),()(

x

xxeyyxfxf

x

.0,0

,0,d),()(

y

yexyxfyf

y

)()(),( yfxfyxf Based on this point ,we know that the ξ,ηare mutual independent.

Solution For any x ， y,

Then

Solution }20{2

1020,

2

10

PPP

2

1

0

22

0

2

1

0

2

0)1(

2

1dd1d)(d)( eyexyyfxxf y

.

,,0,10,1)(

othersxxf

0,0,0,)(

yyeyf

y

Example 12 Suppose ξ and ηare mutual independent,

20,

2

10 PCalculate the probability of

Proof ：

22

22

21

2121

21

2

)())((2

)(

)1(2

1

221 12

1),(

yyxx

eyxf

1°sufficient condition ： It is known that ， then

0

22

22

21

21

2

)(

2

)(

212

1),(

yx

eyxf

Example 13 Suppose (ξ, η) ～ ， ),,,,( 22

2121 N

0

Proof the necessary and sufficient conditio

n of that ξand η are mutual independent i

s

and ，

so ，

Which means ξand ηare independent ．

22

22

21

21

2

)(

2

2

)(

1 2

1)(,

2

1)(

yx

eyfexf

)()(),( yfxfyxf

21 , yx

212

212

1

12

1

0Therefore ．

Especially,let , We can get the following equation ,

2°Necessary condition ： It is known that ξand ηare independent ,then for any number x ， y,the following equation is established

)()(),( yfxfyxf

22

22

21

2121

21

2

)())((2

)(

)1(2

1

221 12

1

yyxx

e

22

22

21

21

2

)(

2

2

)(

1 2

1

2

1

yx

ee

Have a break !