03probability nd statistics -Multiple Random Variables-II.ppt

its all about probability and statistics of multiple random variable

1. Chapter 3: Multiple Random Variables
Addis Ababa Science & Technology University
Department of Electrical & Electronics Engineering
Probability and Random Process (EEEg-2114)

2. Multiple Random Variables
Outline
Introduction
The Joint Cumulative Distribution Function
The Joint Probability Density and Mass Functions
Marginal Statistics
Independence
Conditional Distributions
Correlation and Covariance
Functions of Two Random Variables
Semester-II, 2013/14 2

3. 3
One Function of Two Random Variables
Given two random variables Xand Yand a function g(x,y),
we form a new random variable Zas
Given the joint pdf how does one obtain
the pdf of Z ? Problems of this type are of interest from a
practical standpoint. For example, a receiver output signal
usually consists of the desired signal buried in noise, and
the above formulation in that case reduces to Z = X + Y.).,(YXgZ ),,(yxf
XY ),(zf
Z
Semester-II, 2013/14

4. 4
It is important to know the statistics of the incoming signal
for proper receiver design. In this context, we shall analyze
problems of the following type:
The cdf of Zis given by:),(YXgZ YX )/(tan
1
YX
 YX XY YX/ ),max(YX ),min(YX 22
YX   



yx
XY
Z
dxdyyxf
zYXgPzZPzF
,
,),(
),()()( 
Semester-II, 2013/14

5. 5
Example -1: Let Z = X + Y. Find
Solution:
Since the required region in the xyplane where is
the shaded area shown belowto the left of the line
Integrating over the horizontal strip along the x-axis first
(inner integral) followed by sliding that strip along the y-axis
from to (outer integral) we cover the entire shaded
area. 









,),()(
y
yz
x
XYZ dxdyyxfzYXPzF zyx .zyx   yzx x y ).(zf
Z
Semester-II, 2013/14

6. 6
We can find by differentiating directly. In this
context, it is useful to recall the differentiation rule by
Leibnitz. Suppose
Then,
Using the above two equations, we get
Alternatively, the above integration can be carried out first
along the y-axis followed by the x-axis as below.)(zF
Z )(zf
Z 

)(
)(
.),()(
zb
za
dxzxhzH    



)(
)(
.
),(
),(
)(
),(
)()( zb
za
dx
z
zxh
zzah
dz
zda
zzbh
dz
zdb
dz
zdH



( , )
( ) ( , ) ( , ) 0
( , ) .
z y z y
XY
Z XY XY
XY
f x y
f z f x y dx dy f z y y dy
zz
f z y y dy
   
   


 
    
 
 

   

(i)
Semester-II, 2013/14

7. 7
In the second case
and differentiating the above
equation gives









,),()(
x
xz
y
XYZ dxdyyxfzF 
 






















.),(
),(
)(
)(
x
XY
x
xz
y
XY
Z
Z
dxxzxf
dxdyyxf
zdz
zdF
zf
(ii)
If Xand Yare independent, then
and inserting equation (iii) into equations (i) and (ii), we get)()(),( yfxfyxf
YXXY  .)()()()()(









x
YX
y
YXZ dxxzfxfdyyfyzfzf
(iii)
(iv)xzy x y
Semester-II, 2013/14

8. 8
The above integral is the standard convolution of the
functions and expressed two different ways. We
thus reach the following conclusion: If two random variables
are independent, then the density of their sum equals the
convolution of their density functions.
As a special case, suppose that for and
for then we can use the following figure to determine
the pdf of Z.)(zf
X )(zf
Y 0)(xf
X 0x 0)(yf
Y ,0y yzx x y )0,(z ),0(z
Semester-II, 2013/14

9. 9
In the above case,
or
On the other hand, by considering vertical strips first in
above figure, we get
or
if Xand Yare independent random variables.




z
y
yz
x
XYZ dxdyyxfzF

0

0
),()( 














 




.0,0
,0,),(
),()(

0

0

0
z
zdyyyzf
dydxyxf
z
zf
z
XY
z
y
yz
x
XYZ 






 



,0,0
,0,)()(
),()(

0

0
z
zdxxzfxf
dxxzxfzf
z
y
YX
z
x
XYZ 




z
x
xz
y
XYZ dydxyxfzF

0

0
),()(
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10. 10
Example-2: Suppose Xand Yare independent exponential
r.vs with common parameter , and let Z = X + Y.
Determine
Solution: We have
and we can make use of (13) to obtain the pdf of Z= X+ Y.
As the next example shows, care should be taken in using
the convolution formula for r.vs with finite range.
Example-3: Xand Yare independent uniform r.vs in the
common interval (0,1). Determine where Z= X+ Y.
Solution: Clearly, here, and as following
figure shows there are two cases of zfor which the shaded
areas are quite different in shape and they should be
considered separately.),()( ),()( yUeyfxUexf
y
Y
x
X



 20 zYXZ ),(zf
Z ).( )(
2

0
2

0
)(2
zUezdxedxeezf
z
z
z
z
xzx
Z



  ).(zf
Z
Semester-II, 2013/14

11. 11x y yzx 10 )( za x y yzx 21 )( zb
For
For notice that it is easy to deal with the unshaded
region. In that case,10z ,21z .10 ,
2
)( 1 )(
2

0

0

0
 



z
z
dyyzdxdyzF
z
y
z
y
yz
x
Z 
.21 ,
2
)2(
1)1(1
1 11)(
2
1
1z
1
1
1








 
z
z
dyyz
dxdyzZPzF
y
zy yzx
Z
Semester-II, 2013/14

12. 12
Finally, differentiating the cdf, we obtain
By direct convolution of and we obtain the
same result as above. In fact, for [Figure (a)]
and for [Figure (b)]
Figure (c) shows which agrees with the convolution
of two rectangular waveforms as well.





.21,2
,10)(
)(
zz
zz
dz
zdF
zf
Z
Z )(xf
X ),(yf
Y 10z 21z . 1 )()()(

0
zdxdxxfxzfzf
z
YXZ   .2 1 )(
1
1
zdxzf
z
Z 
 )(zf
Z
Semester-II, 2013/14

13. 13)(xf
Y x 1 )(xzf
X x z )()( xfxzf
YX x z 1z 10 )( za )(xf
Y x 1 )(xzf
X x )()( xfxzf
YX x 1 1z z 1z 21 )( zb
(c))(zf
Z z 2 0 1
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14. 14
Example-4: Let Determine its pdf
Solution:
and hence
If Xand Yare independent, then the above formula reduces
to
which represents the convolution of with.YXZ   









),( )(
y
yz
x
XYZ dxdyyxfzYXPzF

()
( ) ( , ) ( , ) .
zy
Z
Z XY XY
yx
dF z
f z f x y dx dy f y z y dy
dz z
  
  

   


  

( ) ( ) ( ) ( ) ( ),
Z X Y X Y
f z f z y f y dy f z f y


     )(zf
X ).(zf
Y y x zyx zyx y ).(zf
Z
Semester-II, 2013/14

15. 15
As a special case, suppose
In this case, Zcan be negative as well as positive, and that
gives rise to two situations that should be analyzed
separately, since the region of integration for and
are quite different. For from Figure (a)
and for from Figure (b)
After differentiation, this gives






0

0
),( )(
y
yz
x
XYZ dxdyyxfzF 








0
),( )(
zy
yz
x
XYZ dxdyyxfzF .0 ,0)( and ,0 ,0)(  yyfxxf
YX 0z 0z ,0z ,0z 












.0,),(
,0,),(
)(



0
zdyyyzf
zdyyyzf
zf
z
XY
XY
Z
(b)y x yzx z y x yzx z z
(a)
Semester-II, 2013/14

16. 16
Example-5: Given Z = X / Y, obtain its density function.
Solution: We have
The inequality can be rewritten as if
and if
Figure (a) shows the area corresponding to the first term,
and Figure (b) shows that corresponding to the second term
in the above equation.  . /)( zYXPzF
Z  zYX/ YzX ,0Y YzX .0Y y x yzx
(a)y x yzx
(b)   
  . 0,0,
0,/0,/ /


YYzXPYYzXP
YzYXPYzYXPzYXP
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17. 17
Integrating over these two regions, we get
Differentiation with respect to zgives
Note that if Xand Yare nonnegative random variables, then
the area of integration reduces to that shown below.),( ),( )(
0




0






 

y yzx
XY
y
yz
x
XYZ dxdyyxfdxdyyxfzF . ,),(||
),()(),()(


0


0








zdyyyzfy
dyyyzfydyyyzyfzf
XY
XYXYZ y x yzx
Semester-II, 2013/14

18. Exercise
The joint pdf of two random variables Xand Yis given by:
Find the pdf of Zif :
otherwise , 0
10 ,10 ,
),(


 

yxyx
yxf
XY
18YXZd
Y
X
Zb
XYZcYXZa


. .
. . Semester-II, 2013/14

19. Assignment-III
1.Suppose that two continuous random variables Xand Yhave
joint pdf given by:
)20( . )3( .
)4( . )2 ,43( .
ies.probabilit following theEvaluate .
. and of pdf lconditiona theFind .
t?independen and Are .
. and of pdf and cdf marginal theFind .
. and of cdfjoint theFind .
.constant theFind .
constant. a is re whe
otherwise , 0
50 ,62 , )2(
),(




 

YPivXPii
YXPiiiYXPi
e
YXe
YXd
YXc
YXb
ka
k
yxyxk
yxf
XY
Semester-II, 2013/14 19

20. Assignment-III Cont’d……
2.Let the joint pmf of two discrete random variables Xand Yis
given by: t?independen and Are .
. and of pmf marginal theFind .
. of value theFind .
constant. a is re whe
otherwise , 0
2 ,1 3; 2, ,1 , )(
),(
YXc
YXb
ka
k
yxyxk
yxP
iji
jiXY


 

Semester-II, 2013/14 20

21. Assignment-III Cont’d…..
3.Suppose that two continuous random variables Xand Yare
independent and uniform in the interval (0, 4).
Find the pdf of Zif :
.
.
.
.
.
X
Y
Ze
XYZd
YXZc
XYZb
YXZa





21Semester-II, 2013/14