Probabilit y theory The department of math of central south university Probability and Statistics Course group
Jan 03, 2016
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 !