Probability theory
-
Upload
tatyana-watkins -
Category
Documents
-
view
21 -
download
0
description
Transcript of Probability theory
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 !