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One Function of Two Random Variables X and Y : Two random variables g(x,y): a function We form a new random variable Z as Given the joint p.d.f how does one obtain the p.d.f of Z ? A receiver output signal usually consists of the desired signal buried in noise the above formulation in that case reduces to Z = X + Y. Practical Viewpoint
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One Function of Two Random Variables
One Function of Two Random Variables
X and Y : Two random variables
g(x,y): a function
We form a new random variable Z as
Given the joint p.d.f how does one obtain the p.d.f of Z ?
A receiver output signal usually consists of the desired signal buried in noise the above formulation in that case reduces to Z = X + Y.
).,( YXgZ
),,( yxf XY ),(zfZ
Practical Viewpoint
We have:
Where in the XY plane represents the region where . need not be simply connected
First find the region for every z, Then evaluate the integral there.
zDyx XY
zZ
dxdyyxf
DYXPzYXgPzZPzF
, ,),(
),(),()()(
X
Y
zD
zD
zD zyxg ),(
)(zFZ
zD
zD
Z = X + Y. Find
Integrating over all horizontal strips(like the one in figure) along the x-axis We can find by differentiating directly.
yzx
x
y
Example
,),()(
y
yz
x XYZ dxdyyxfzYXPzF
).(zfZ
)(zFZ)(zfZ
Suppose
Then
Using the above,
Recall - Leibnitz Differentiation Rule
)(
)( .),()(
zb
zadxzxhzH
)(
)( .),(),()(),()()( zb
zadx
zzxhzzah
dzzdazzbh
dzzdb
dzzdH
( , )( ) ( , ) ( , ) 0
( , ) .
z y z yXY
Z XY XY
XY
f x yf z f x y dx dy f z y y dyz z
f z y y dy
Example – Alternate Method of Integration
,),()(
x
xz
y XYZ dxdyyxfzF
.),(
),( )()(
x XY
x
xz
y XYZ
Z
dxxzxf
dxdyyxfzdz
zdFzf
)()(),( yfxfyxf YXXY
.)()()()()(
x YXy YXZ dxxzfxfdyyfyzfzf
xzy
x
y
If X and Y are independent, then
This integral is the standard convolution of the functions and expressed in two different ways.
)(zf X )(zfY
If two r.vs are independent, then the density of their sum equals the convolution of their density functions. As a special case, suppose that
- for
- for
then using the figure we can determine the new limits for
Example – Conclusion
yzx
x
y
)0,(z
),0( z
0)( xf X 0x0)( yfY ,0y
.zD
In that case
or
On the other hand, by considering vertical strips first, we get
or
if X and Y are independent random variables.
z
y
yz
x XYZ dxdyyxfzF
0
0 ),()(
.0,0,0,),( ),()(
0
0
0 zzdyyyzfdydxyxf
zzf
z
XYz
y
yz
x XYZ
,0,0
,0,)()(),()(
0
0 zzdxxzfxfdxxzxfzf
z
y YXz
x XYZ
z
x
xz
y XYZ dydxyxfzF
0
0 ),()(
Example – Conclusion
X and Y are independent exponential r.vs with common parameter ,
let Z = X + Y. Determine
Example
),()( ),()( yUeyfxUexf yY
xX
).( )( 2
0
2
0
)(2 zUezdxedxeezf zzzz xzxZ
).(zfZ
Solution
This example shows that care should be taken in using the convolution formula for r.vs with finite range.
X and Y are independent uniform r.vs in the common interval (0,1).
let Z = X + Y. Determine
Clearly,
There are two cases of z for which the shaded areas are quite different in shape and they should be considered separately.
Example
Solution
).(zfZ
20 zYXZ
x
y
yzx
10 )( zax
y
yzx
21 )( zb
Example – Continued
.10 ,2
)(
1 )(
2
0
0
0
zz
dyyz
dxdyzF
z
y
z
y
yz
xZ
.21 ,2
)2(1
)1(1
1 1
1)(
2
1
1z
1
1
1
zz
dyyz
dxdy
zZPzF
y
zy yzx
Z
So, we obtain
By direct convolution of and we obtain the same result. In fact, for
.21,2,10)()(
zzzz
dzzdFzf Z
Z
)(xf X ),( yfY
10 z
Example – Continued
)(xfY
x1
)( xzf X
xz
)()( xfxzf YX
xz1z
10 z
. 1 )()()(
0 zdxdxxfxzfzf
z
YXZ
)(xfY
x1
)( xzf X
x
)()( xfxzf YX
x11z z 1z
21 z
)(zfZ
z20 1
and for
Example – Continued
21 z .2 1 )(1
1 zdxzf
zZ
)(zfZThis figure shows which agrees with the convolution of two rectangular waveforms as well.
Let Determine its p.d.f
and hence
If X and Y are independent,
which represents the convolution of with
Example
Solution
.YXZ
),( )(
y
yz
x XYZ dxdyyxfzYXPzF
( )( ) ( , ) ( , ) .z y
ZZ XY XYy x
dF zf z f x y dx dy f y z y dydz z
)( zf X ).(zfY
y
x
zyx zyx
y
).(zfZ
)()()()()( zfzfdyyfyzfzf YXYXZ
Suppose
For
and for
After differentiation, this gives
0
0 ),( )(
y
yz
x XYZ dxdyyxfzF
0 ),( )(
zy
yz
x XYZ dxdyyxfzF
.0 ,0)( and ,0 ,0)( yyfxxf YX
,0z
,0z
.0,),(
,0,),()(
0
zdyyyzf
zdyyyzfzf
z XY
XYZ
y
x
yzx
z
y
xyzx
zz
Example – a special case
Given Z = X / Y, obtain its density function.
- if - if Since by the partition theorem, we have
and hence by the mutually exclusive property of the later two events
Example
Solution
zYX /YzX ,0YYzX .0Y
zYX / 0 YA
__
,A A
. 0,0,
0,/0,/ /
YYzXPYYzXP
YzYXPYzYXPzYXP
AzYXAzYX
AAzYXzYX
)/()/(
)()/( /
Integrating over these two regions, we get
Differentiation with respect to z gives
y
x
yzx
y
xyzx
Example – Continued
.),( ),( )(0
0
y yzx XYy
yz
x XYZ dxdyyxfdxdyyxfzF
. ,),(||
),()(),()(
0
0
zdyyyzfy
dyyyzfydyyyzyfzf
XY
XYXYZ
0, YYzXP 0, YYzXP
If X and Y are nonnegative random variables, then the area of integration reduces to that shown in this figure:
y
x
yzx
Example – Continued
0
0 ),( )(
y
yz
x XYZ dxdyyxfzF
otherwise.,0,0,),( )(
0 zdyyyzfyzf XY
Z
This gives
or
X and Y are jointly normal random variables with zero mean so that
Show that the ratio Z = X / Y has a Cauchy density function centered at
Using
and the fact that we obtain
Example
Solution
.12
1),(22
2
2121
2
22
)1(21
221
yrxyxr
XY er
yxf
./ 21 r
,1)(
122)(
221
20
0
2/
221
20
2
rzdyye
rzf y
Z
),,(),( yxfyxf XYXY
. ,),(||
),()(),()(
0
0
zdyyyzfy
dyyyzfydyyyzyfzf
XY
XYXYZ
where
Thus
which represents a Cauchy r.v centered at
Integrating the above from to z, we obtain the corresponding distribution function to be
Example – Continued
.12
1)(
2221
21
2
220
rzzrz
,)1()/(
/1)( 221
221
22
221
rrzrzfZ
./ 21 r
.1
arctan121)(
21
12
rrzzFZ
Obtain
We have
But, represents the area of a circle with radius and hence
This gives after repeated differentiation
.),()(22
22
zYX XYZ dxdyyxfzYXPzF
.22 YXZ ).(zfZ
Example
Solution
zYX 22 ,z
.),()(
2
2
z
zy
yz
yzx XYZ dxdyyxfzF
. ),(),(2
1)(
22
2
z
zy XYXYZ dyyyzfyyzfyz
zf
x
y
zzYX 22
z
z
Result - If X and Y are independent zero mean Gaussian r.vs with common variance then is an exponential r.vs with parameter
X and Y are independent normal r.vs with zero Mean and common variance
Determine for
Direct substitution with gives
where we have used the substitution
)(zfZ .22 YXZ
.2
21 ,0r2
2 2 2
22
/ 2 ( ) / 22 22 2 0
/ 2 /2 / 22 2 0
1 1 1( ) 2 22
cos 1 ( ),2cos
zz zz y yZ y z
zz
ef z e dy dyz y z y
e z d e U zz
.sinzy
,2 22 YX .2 2
.z
Example
Solution
Let Find
This corresponds to a circle with radius Thus
If X and Y are independent Gaussian as in the previous example,
.22 YXZ ).(zfZ
.z
Example
Solution
. ),(),()(
2222
22
z
z XYXYZ dyyyzfyyzfyz
zzf
.),()(
22
22
z
zy
yz
yzx XYZ dxdyyxfzF
),( coscos2
122
12)(
2222
222222
2/2
/2
0
2/2
0 22
2/2
0
2/)(222
zUezdzzez
dyyz
ezdyeyz
zzf
zz
zzz yyzZ
Rayleigh distribution
(*)
If where X and Y are real, independent normal r.vs with zero mean and equal variance, then the r.v has a Rayleigh density.
W is said to be a complex Gaussian r.v with zero mean, whose real and imaginary parts are independent r.vs.
As we saw its magnitude has Rayleigh distribution.
What about its phase
Example – Conclusion
,iYXW
22 YXW
?tan 1
YX
Let
U has a Cauchy distribution with
As a result
The magnitude and phase of a zero mean complex Gaussian r.v has Rayleigh and uniform distributions respectively. We will show later, these two derived r.vs are also independent of each
other!
Example – Conclusion
tan / ,U X Y
. ,1
/1)( 2
uu
ufU
. otherwise,0
,2/2/,/11tan
/1)sec/1(
1)(tan|/|
1)( 22
Ufdud
f
What if X and Y have nonzero means and respectively?
Since
Example
X Y
,2
1),(222 2/])()[(
2
YX yx
XY eyxf
Solution
2 2cos , sin , , cos , sin ,X Y X Yx z y z
,2
2
2
)(
202
2/)(
/23
/2
/)cos(/2
/2
/)cos(2
2/)(
/2
/2
/)cos(/)cos(2
2/)(
222
22222
22222
zIze
dedeze
deezezf
z
zzz
zzz
Z
0
cos2
0
)cos(0
121)( dedeI
substituting this into (*), and letting
we get
where
Rician p.d.f.
the modified Bessel function of the first kind and zeroth order
Application
Fading multipath situation where there is
- a dominant constant component (mean)
- a zero mean Gaussian r.v.
The constant component may be the line of sight signal and the zero mean Gaussian r.v part could be due to random multipath components adding up incoherently (see diagram below). The envelope of such a signal is said to have a Rician p.d.f.
Line of sight signal (constant)
a
Multipath/Gaussian noise
Determine
nonlinear operators special cases of the more general order statistics. We can arrange any n-tuple such that:
Example
Solution
).,min( ),,max( YXWYXZ ).(zfZ
, )()2()1( nXXX
, , , , min 21)1( nXXXX
. , , ,max 21)( nn XXXX
, , , , 21 nXXX
represent r.vs, The function that takes on the value in each possible
sequence is known as the k-th order statistic.
represent the set of order statistics among n random variables.
represents the range, and when n = 2, we have the max and min statistics.
Example - continued
nXXX , , , 21
)(kX )(kx nxxx , , , 21
)1()( XXR n
(1) (2) ( )( , , , )nX X X
Since we have
since and form a partition.
Example - continued
x
yzx yx
zX
YX
),( )( YXzXPa
x
y
zY
YX yx
zy
),( )( YXzYPb
x
y
),( zz
)(c
,,,,
),max(YXYYXX
YXZ
,,,
,,),max()(YXzYPYXzXP
YXzYYXzXPzYXPzFZ
)( YX )( YX
From the rightmost figure,
If X and Y are independent, then
and hence
Similarly
Thus
).,(,)( zzFzYzXPzF XYZ
Example - continued
)()()( zFzFzF YXZ
).()()()()( zFzfzfzFzf YXYXZ
.,,,
),min(YXXYXY
YXW
. ,,),min()( YXwXYXwYPwYXPwFW
x
yyx
wy
(a)
x
y
yx wx
(b)
x
y
),( ww
(c)
Example - continued
, ),()()( ,11)(
wwFwFwFwYwXPwWPwF
XYYX
W
X and Y are independent exponential r.vs with common parameter .
Determine for
We have and hence
But and so that
Thus min ( X, Y ) is also exponential with parameter 2.
Example
Solution
).,min( YXW )(wfW
)()()()( )( wFwFwFwFwF YXYXW
).()()()()()( )( wfwFwFwfwfwfwf YXYXYXW
,)( )( wYX ewfwf ,1)( )( w
YX ewFwF
).(2)1(22 )( 2 wUeeeewf wwwwW
X and Y are independent exponential r.vs with common parameter .
Define Determine
Example
Solution
. ),max(/),min( YXYXZ ).(zfZ
.,/,,/
YXXYYXYX
Z
( ) / , / ,
, , .ZF z P Z z P X Y z X Y P Y X z X Y
P X Yz X Y P Y Xz X Y
We solve it by partitioning the whole space.
Since X and Y are both positive random variables in this case, we have .10 z
Example - continued
x
y
yx yzx
(a)
x
yyx
xzy
(b)
.),( ),( )(
0
z
0
0
0
x
y XY
yz
x XYZ dydxyxfdxdyyxfzF
. otherwise,0
,10,)1(
2
)1(22
),(),(
),( ),( )(
2
0 2
0
)1(2
0
)()(2
0
0
0
zz
dyuez
dyye
dyeey
dyyzyfyyzfy
dxxzxfxdyyyzfyzf
uyz
yzyyyz
XYXY
XYXYZ
)(zfZ
z1
2
Let X and Y be independent Poisson random variables with parameters and respectively. Let Determine the p.m.f of Z.
Z takes integer values For any gives only a finite number of
options for X and Y. The event is the union of (n + 1) mutually exclusive
events given by
Example – Discrete Case
1 2.YXZ
Solution
nYXn , ,2 ,1 ,0
, ,kA X k Y n k
}{ nYX kA
As a result
If X and Y are also independent, then
and hence
. ) ,(
,)()(
0
0
n
k
n
k
knYkXP
knYkXPnYXPnZP
Example – continued
)()( , knYPkXPknYkXP
. , ,2 ,1 ,0 ,!
)(
)!(!!
!)!(!
) ,()(
21)(
021
)(2
0
1
0
21
2121
nn
e
knkn
ne
kne
ke
knYkXPnZP
n
n
k
knkknn
k
k
n
k
Thus Z represents a Poisson random variable with parameter
Sum of independent Poisson random variables is also a Poisson random variable whose parameter is the sum of the parameters of the original random variables.
Such a procedure for determining the p.m.f of functions of discrete random variables is somewhat tedious.
As we shall see, the joint characteristic function can be used in this context to solve problems of this type in an easier fashion.
Example – conclusion
.21
