Download 3.4 The distribution of random variable function

Probability
theory
The department of math of central
south university
Probability and Statistics Course group
§3.4 The distribution of random variable function
1.The sum of random variables
2.The ratio of random variables
3.The extreme value of random variables
Of discrete random variables, we have discussed the
distribution of random variables function problems.
In fact ,the same condition can also be applicated to
the continuous variables .
For example,if random variable ξis a normal
random variable with parameters
( , 2 )
,in
order to obtain the probability of ξeasily ,it is
convenient by getting another random variable
 a


 


Here ηis a function of random variable
ξ,and also a random variable 。We now
turn to solve the distribution problems of
random variable function.
Firstly ,a theorem must be introduced.
Theorem 3.1 Suppose ξis acontinuous random variable
with density function p (x) .Function f(x) is Strictly
monotonous,the inverse function h(y) is continuous and
differentable,then function   f ( ) is also a random
variable and the density function is
 p[h( y ) | h ' ( y ) |],   y  
 ( y)  
0, others

Here
  min{ f (), f ()}
  max{ f (), f ()}
Proof is omitted
Examples，Suppose ξ ~ N ( ,2) , η = a ξ +b,
then
1 1

f ( y)  f  ( y  b)  
|a| a

1
2  | a |
η ~ N ( a +b, a22 )
especially ，if ξ ~ N (  , 2) ,
then
X 

~ N (0,1)


e
( y b  a ) 2
2 a 2 2
  y  
The above theorem is restricted to strong conditions that
f(x)must be strictly monotonous,and the inverse function
h(y) be continuous and differentable,we also know these
are difficulty to be satisfied ,So another method is hoped to
be producted to get the distribution of random variable
function based on the above analysis ,which will be
introduced as follows
f (x)
Here is a question thatξ is a random variable
with density function p(x) (or distribution
function F(x) )
Find the density function aor distribution of
η=g(ξ)
The starting point to solve the problem has ：
（1）The distribution function
（2）The density function
（1）The distribution function
The distribution function Fη(y) must be firstly
obtained from the definition of distribution
function,that is
F  y   P  y  Pg    y

 p ( x)dx
g ( x ) y
Seconly, the density function Fη(y) of η=g(ξ) is
obtained in the following through the relation
between the density function and the
distribution function
p  y   F  y .
Example 14
Suppose ξis a random variable with
density function
2 x, 0  x  1,
p ( x)  
其它.
 0,
Please the density function of η =ξ -4
Solution
F ( y )  P{  y}
 P{  4  y}  P{  y  4}
F ( y)  
y4

p ( x)dx.
f ( y )  p ( y  4)  ( y  4)
then
2( y  4) 1, 0  y  4  1,

0,
其它.

That is
2 y  8,  4  y  3,
f ( y )  
其它.
 0,
Example 15 Given the density function f ξ(x) of
random variable ξ,let
  a  b
Here,a and b are constant ， a  0, What is the
density function fη( y ) of η?
Solution F ( y )  P(  y )  P(a  b  y )
When a > 0 ，
1

  F  1 ( y  b) 
F ( x)  P   ( y  b)  
a

a


1 1

f ( y )  f  ( y  b) 
a a

When a < 0 ，
1

1

  1  F  ( y  b) 
F ( y )  P   ( y  b) 
a


a


1 1

f ( y )   f  ( y  b) 
a a

Then
1 1

f ( y ) 
f   ( y  b) 
|a| a

1.The distribution of Z= ξ+η
Suppose (ξ,η)is a two -dimension rndom variable
random variables with joint density functin P(x,y),
What is the distribution of function Z=ξ+η?
FZ ( z )  P{ Z  z } 
 p( x, y) d x d y
y
x y z

z y


  [
x  u y

z


 [
p(u  y, y) d u] d y
z



x y z
p( x, y) d x] d y
  [ p(u  y, y) d y] d u.
O
x
The density function can be drawn from above
analysis

pz ( z )   p( z  y, y)dy

For ξand is symmetrical,
When ξis independent of η
expressed as follows

pZ ( z )    p( x , z  x ) d x .
pZ (zcan
) be

pZ ( z )    p ( z  y ) p ( y ) d y ,
or

pZ ( z )   p ( x) p ( z  x) d x.

Example 16 What is the distribution of
Z=ξ+η,here ξandηare given two standared
normal random variables?
Solution
1
For p ( x) 
e
2
p ( y ) 
then
1
e
2

y2

2
x2

2
,  x  ,
,    y  ,
pZ ( z )   p ( x) p ( z  x) d x.


1
pZ ( z )  
e
 2
1

e
2
z
t  x
2
1
e
2
z2

4
z2

4
x2
( z  x)2


2
2
e

e

e
z 2

 x  
2

t 2
dt
This means that Z ~ N (0,2).
dx
dx
1
2 
e
z2

4
.
Explantation
Generally,when random variables ξand η
are independent ,and
 ~ N(1 , ), ~ N(2 , )
2
2
2
1
then,
Z ~ N(1  2 ,    ).
2
1
2
2
The linear combination of finite independent
normal random variable is still subject to the
normal distribution.
Example 17
Supposeξandηare independent
random variables，andξ~U(0,1),η~U(0, 1).Let Z
= ξ+η, please identify the density function of
random variable Z .
Solution
1, 0  y  1,
1, 0  x  1,


f  ( x)  
f ( y)  
0, others,
0, others.
By using convolution formula,we have

1

0
f z ( z )   f ( z  y ) f ( y )dy   f ( z  y )dy
t  z  y
z
z 1
f (t )dt.
(1) When z  0 ，then z  1  0 and
(2) When 0  z  1 ，then
(3) When
z 1  0
f Z ( z)  0
z
and
f Z ( z )   1 dt  z
0
1
 2 z
1  z  2 ， then 0  z  1  1 and f Z ( z )   z 11 dt ．
(4) When z  2 ，then z  1  1 and
f Z ( z ) ．0
Therefore

 z,
f Z ( z )  2  z ,

 0,
0  z  1,
1  z  2,
z  2 or z  0 .
2.The distribution of the ratio Z= ξ/η
Suppose (ξ,η) be a two -dimension rndom variable
random variables with joint density functin P(x,y),
What is the distribution of function Z=ξ/η?

FZ ( z)  P{Z  z}  P{  z}

x
z
y
Let u  x y ,
y
  p( x, y) d x d y   p( x, y ) d x d y
G2
G1



0
yz

O

p( x , y) d x d y   yz p( x , y) d x d y,
0
x
 p( x, y) d x d y
G1


0


0

z


yz

p( x, y) d x d y
z

z
0
yp( yu, y) d u d y   0 yp( yu, y) d y d u
by the same reason
 p( x, y ) d x d y     yp( yu, y ) d y d u,
G2
therfore
FZ ( z )  P{ Z  z }
  p( x, y )dxdy   p( x, y) d x d y
G1
G2
z

0

0

  [ yp ( yu, y) d y   yp ( yu, y) d y] d u.
Then the peobability density function can be
obtained as follows
p( z ) 


0
yp( yz, y ) d y 


0

yp( yz, y ) d y
y p( yz, y ) d y.
Especially,whenξis independent of η,
p( z )   y pX ( yz ) pY ( y) d y.
Example 18 Suppose ξand η are The lives
of of two different kinds of light bulbs
respectivily,when they are independent ,the
denstities function are outlined in the following
e  x , x  0,
f ( x)  
0, others,
2e  2 y , y  0,
g ( y)  
others
0,
What is the density function of Z=ξ/η
Solution
According to previous analysis

0
0

f Z ( z )   yf ( yz , y ) d y   yf ( yz , y ) d y ,
 2e  x e  2 y , x  0, y  0,
f ( x, y)  
其它.
0,
Therefore the density function isas follows .
When
z0

f Z ( z )   2 ye e
0
When
z0
and
 yz  2 y

d y   2 ye
 y ( 2 z )
0
f Z ( z )  0,
 2

2 , z  0,
f Z ( z )  (2  z )
0,
z  0.
2
d y  ( 2  z )2 ,
3.The distribution for extreme value of R.V.
Suppose ξand η are two independent random
variable,the distributions are F (x) and F ( y )
respectivly.
Let
M  max(  , )
N  min(  , )
Here M and N are called the extrem value
of ξand η .Please find the distributions of
M,N.
According to the definition of distribution
function,we can get
Fmax ( z )  P{ M  z }  P{  z,  z}
 P{  z}P{  z}  F ( z ) F ( z ).
Fmin ( z )  P{ N  z } 1  P{ N  z }
 1  P{  z,  z}
 1  P{  z}  P{  z}
 1  [1  P{  z}]  [1  P{  z}]
then
Fmax ( z )  F ( z ) F ( z ),
Fmin ( z )  1  [1  F ( z )][1  F ( z )].
Generally, suppose 1 ,  2 ,,  n be independent
random variable series with distribution functions
Fi ( xi ), (i  1,2, , n)
Let
M  max( 1 ,  2 ,,  n )
N  min( 1 ,  2 ,,  n )
then the distributions of M,N are outlined as follows
Fmax ( z )  F1 ( z ) F 2 ( z )  F n ( z ),
Fmin ( z )  1  [1  F1 ( z )][1  F 2 ( z )][1  F n ( z )].
Especially, suppose 1 ,  2 ,,  n be independent
random variable series with same distribution
functions F(x),then
Fmax ( z )  [ F ( z )]'' ,
Fmin ( z )  1  [1  F ( z )]'' .
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