1 8. one function of two random variables given two random variables x and y and a function g(x,y),...
TRANSCRIPT
1
8. One Function of Two Random Variables
Given two random variables X and Y and a function g(x,y),
we form a new random variable Z as
Given the joint p.d.f how does one obtain
the p.d.f 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 ),(zfZ
(8-1)
2
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:
Referring back to (8-1), to start with
),( YXgZ
YX
)/(tan 1 YX
YX
XY
YX /
),max( YX
),min( YX
22 YX
zDyx XY
zZ
dxdyyxf
DYXPzYXgPzZPzF
, ,),(
),(),()()(
(8-2)
(8-3)
3
where in the XY plane represents the region such that is satisfied. Note that need not be simply connected (Fig. 8.1). From (8-3), to determine it is enough to find the region for every z, and then evaluate the integral there.
We shall illustrate this method through various examples.
zD
zyxg ),(
)(zFZ
zD
zD
X
Y
zD
zD
Fig. 8.1
4
Example 8.1: Z = X + Y. Find Solution:
since the region of the xy plane where is the shaded area in Fig. 8.2 to 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 (8-4)
zD zyx .zyx
yzx
x
y
Fig. 8.2
).(zfZ
5
We can find by differentiating directly. In this context, it is useful to recall the differentiation rule in (7-15) - (7-16) due to Leibnitz. Suppose
Then
Using (8-6) in (8-4) we get
Alternatively, the integration in (8-4) can be carried out first along the y-axis followed by the x-axis as in Fig. 8.3.
)(zFZ)(zfZ
)(
)( .),()(
zb
zadxzxhzH (8-5)
)(
)( .
),(),(
)(),(
)()( zb
zadx
z
zxhzzah
dz
zdazzbh
dz
zdb
dz
zdH(8-6)
.),(
),(0),(1 ),( )(
dyyyzf
dyz
yxfyyzfdydxyxf
zzf
XY
XYXY
yz
XYZ
(8-7)
6
In that case
and differentiation of (8-8) gives
,),()(
x
xz
y XYZ dxdyyxfzF (8-8)
.),(
),( )(
)(
x XY
x
xz
y XYZ
Z
dxxzxf
dxdyyxfzdz
zdFzf
(8-9)
If X and Y are independent, then
and inserting (8-10) into (8-8) and (8-9), we get
)()(),( yfxfyxf YXXY
.)()()()()(
x YXy YXZ dxxzfxfdyyfyzfzf
(8-10)
(8-11)
xzy
x
y
Fig. 8.3
7
The above integral is the standard convolution of the functions and expressed two different ways. We thus reach the following conclusion: 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 and for then we can make use of Fig. 8.4 to determine the new limits for
)(zf X )(zfY
0)( xf X 0x 0)( yfY
,0y
.zD
Fig. 8.4
yzx
x
y
)0,(z
),0( z
8
In that case
or
On the other hand, by considering vertical strips first in Fig. 8.4, 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 z
zdyyyzfdydxyxfz
zfz
XYz
y
yz
x XYZ (8-12)
,0,0
,0,)()(),()(
0
0 z
zdxxzfxfdxxzxfzf
z
y YXz
x XYZ
z
x
xz
y XYZ dydxyxfzF
0
0 ),()(
(8-13)
9
Example 8.2: Suppose X and Y are 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 p.d.f 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 8.3: X and Y are independent uniform r.vs in the common interval (0,1). Determine where Z = X + Y. Solution: Clearly, here, and as Fig. 8.5 shows there are two cases of z for which the shaded areas are quite different in shape and they should be considered separately.
),()( ),()( yUeyfxUexf yY
xX
(8-14)
20 zYXZ
),(zfZ
).( )( 2
0
2
0
)(2 zUezdxedxeezf zzzz xzxZ
(8-15)
).(zfZ
10
x
y
yzx
10 )( za
x
y
yzx
21 )( zb
Fig. 8.5
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
zdyyzdxdyzF
z
y
z
y
yz
xZ(8-16)
.21 ,2
)2(1)1(1
1 11)(
21
1z
1
1
1
zz
dyyz
dxdyzZPzF
y
zy yzxZ
(8-17)
11
Using (8-16) - (8-17), we obtain
By direct convolution of and we obtain the same result as above. In fact, for (Fig. 8.6(a))
and for (Fig. 8.6(b))
Fig 8.6 (c) shows which agrees with the convolution of two rectangular waveforms as well.
.21,2
,10)()(
zz
zz
dz
zdFzf Z
Z (8-18)
)(xf X ),( yfY
10 z
21 z
. 1 )()()(
0 zdxdxxfxzfzf
z
YXZ
.2 1 )(1
1 zdxzf
zZ
(8-19)
(8-20)
)(zfZ
12
)(xfY
x1
)( xzf X
xz
)()( xfxzf YX
xz1z
10 )( za
)(xfY
x1
)( xzf X
x
)()( xfxzf YX
x11z z
1z
21 )( zb
Fig. 8.6 (c)
)(zfZ
z20 1
13
Example 8.3: Let Determine its p.d.f
Solution: From (8-3) and Fig. 8.7
and hence
If X and Y are independent, then the above formula reduces to
which represents the convolution of with
.YXZ
),( )(
y
yz
x XYZ dxdyyxfzYXPzF
.),(),(
)()( dyyzyfdydxyxf
zdz
zdFzf XYy
xz
x XYZ
Z (8-21)
),()()()()(
yfzfdxyfyzfzf YXYXZ
(8-22)
)( zf X ).(zfY
Fig. 8.7
y
x
zyx zyx
y
).(zfZ
14
As a special case, suppose
In this case, Z can 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 Fig. 8.8 (a)
and for from Fig 8.8 (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
0z 0z,0z
,0z
.0,),(
,0,),()(
0
zdyyyzf
zdyyyzfzf
z XY
XY
Z (8-23) Fig. 8.8 (b)
y
x
yzx
z
y
x
yzx
zz
(a)
15
Example 8.4: Given Z = X / Y, obtain its density function. Solution: We have The inequality can be rewritten as if and if Hence the event in (24) need to be conditioned by the event and its compliment Since by the partition theorem, we have
and hence by the mutually exclusive property of the later two events
Fig. 8.9(a) shows the area corresponding to the first term, and Fig. 8.9(b) shows that corresponding to the second term in (8-25).
. /)( zYXPzFZ (8-24)zYX / YzX ,0Y
YzX .0Y zYX / 0 YA .
__
A,
__
SAA
. 0,0,
0,/0,/ /
YYzXPYYzXP
YzYXPYzYXPzYXP
(8-25)
Fig. 8.9
y
x
yzx
(a)
y
xyzx
(b)
AzYXAzYXAAzYXzYX )/()/()()/( /
16
Integrating over these two regions, we get
Differentiation with respect to z gives
Note that if X and Y are nonnegative random variables, then the area of integration reduces to that shown in Fig. 8.10.
.),( ),( )(0
0
y yzx XYy
yz
x XYZ dxdyyxfdxdyyxfzF (8-26)
. ,),(||
),()(),()(
0
0
zdyyyzfy
dyyyzfydyyyzyfzf
XY
XYXYZ
(8-27)
y
x
yzx
Fig. 8.10
17
This gives
or
Example 8.5: 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 Solution: Inserting (8-29) into (8-27) and using the fact that we obtain
0
0 ),( )(
y
yz
x XYZ dxdyyxfzF
otherwise.,0
,0,),( )(
0 zdyyyzfyzf XY
Z (8-28)
.12
1),(
22
2
2121
2
2
2
)1(2
1
221
yrxyx
rXY e
ryxf (8-29)
./ 21 r
,1
)(
12
2)(
221
20
0
2/
221
20
2
r
zdyye
rzf y
Z
),,(),( yxfyxf XYXY
18
where
Thus
which represents a Cauchy r.v centered at Integrating (8-30) from to z, we obtain the corresponding distribution function to be
Example 8.6: Obtain Solution: We have
.12
1)(
2221
21
2
220
rzzr
z
,)1()/(
/1)(
221
221
22
221
rrz
rzfZ
(8-30)
./ 21 r
.1
arctan1
2
1)(
21
12
r
rzzFZ
(8-31)
.),()(22
22
zYX XYZ dxdyyxfzYXPzF
.22 YXZ
(8-32)
).(zfZ
19
But, represents the area of a circle with radius and hence from Fig. 8.11,
This gives after repeated differentiation
As an illustration, consider the next example.
zYX 22,z
.),()(
2
2
z
zy
yz
yzx XYZ dxdyyxfzF (8-33)
. ),(),(2
1)(
22
2
z
zy XYXYZ dyyyzfyyzfyz
zf (8-34)
Fig. 8.11
x
y
zzYX 22
z
z
20
Example 8.7 : X and Y are independent normal r.vs with zeroMean and common variance Determine for Solution: Direct substitution of (8-29) with Into (8-34) gives
where we have used the substitution From (8-35)we have the following result: If X and Y are independent zeromean Gaussian r.vs with common variance then is an exponential r.vs with parameter
Example 8.8 : Let Find Solution: From Fig. 8.11, the present case corresponds to a circle with radius Thus
)(zfZ .22 YXZ .2 21 ,0r
),(2
1
cos
cos
1
2
12
2
1)(
2
2
2
222
2/2
/2
0 2
2/
0 22
2/
2/)(22
zUedz
ze
dyyz
edye
yzzf
zz
zzz
zy
yyzZ
(8-35)
.sinzy
,222 YX .2 2
.22 YXZ ).(zfZ
.2z
21
. ),(),()(
2222
22
z
z XYXYZ dyyyzfyyzfyz
zzf
.),()(
22
22
z
zy
yz
yzx XYZ dxdyyxfzF
),( cos
cos2
12
2
12)(
2222
222222
2/2
/2
0
2/2
0 22
2/2
0
2/)(222
zUez
dz
ze
z
dyyz
ez
dyeyz
zzf
zz
zzz yyzZ
And by repeated differentiation, we obtain
Now suppose X and Y are independent Gaussian as in Example 8.7. In that case, (8-36) simplifies to
which represents a Rayleigh distribution. Thus, 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. From (8-37), we have seen that its magnitude has Rayleigh distribution.
(8-36)
(8-37)
,iYXW
22 YXW
22
What about its phase
Clearly, the principal value of lies in the interval If we let then from example 8.5, U has a Cauchy distribution with (see (8-30) with )
As a result
To summarize, the magnitude and phase of a zero mean complex Gaussian r.v has Rayleigh and uniform distributions respectively. Interestingly, as we will show later, these two derived r.vs are also independent of each other!
?tan 1
Y
X (8-38)
. ,1
/1)(
2
u
uufU
(8-39)
).2/,2/(
,/tan YXU 0 ,21 r
. otherwise,0
,2/2/,/1
1tan
/1
)sec/1(
1)(tan
|/|
1)(
22
Ufdud
f (8-40)
23
Let us reconsider example 8.8 where X and Y have nonzero means and respectively. Then is said to be a Rician r.v. Such a scene arises in fading multipath situation where there is a dominant constant component (mean) in addition to 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.
Example 8.9: Redo example 8.8, where X and Y have nonzero means and respectively. Solution: Since
substituting this into (8-36) and letting
X Y 22 YXZ
X Y
,2
1),(
222 2/])()[(2
YX yx
XY eyxf Rician Output
Line of sight signal (constant)
a
Multipath/Gaussian noise
24
we get the Rician probability density function to be
where
is the modified Bessel function of the first kind and zeroth order.
Example 8.10: Determine Solution: The functions max and min are nonlinear
,sin ,cos , ,sin 22 YXYXzy
,2
2
2
)(
202
2/)(
/23
/2
/)cos(/2
/2
/)cos(2
2/)(
/2
/2
/)cos(/)cos(2
2/)(
222
22
222
22
222
zI
ze
dedeze
deeze
zf
z
zzz
zzz
Z
0
cos2
0
)cos(0
1
2
1)( dedeI
(8-41)
).,min( ),,max( YXWYXZ
).(zfZ
(8-42)
25
operators and represent special cases of the more general order statistics. In general, given any n-tuple we can arrange them in an increasing order of magnitude such that
where and is the second smallest value among and finally If 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. In this context
represents the range, and when n = 2, we have the max and min statistics.
, , , , 21 nXXX
, )()2()1( nXXX
, , , , min 21)1( nXXXX )2(X
, , , , 21 nXXX . , , ,max 21)( nn XXXX
nXXX , , , 21 )(kX
)(kx nxxx , , , 21
)1()( XXR n (8-44)
(8-43)
(1) (2) ( )( , , , )nX X X
26
Returning back to that problem, since
we have (see also (8-25))
since and are mutually exclusive sets that form a partition. Figs 8.12 (a)-(b) show the regions satisfying the corresponding inequalities in each term above.
,,
,,),max(
YXY
YXXYXZ (8-45)
,,,
,,),max()(
YXzYPYXzXP
YXzYYXzXPzYXPzFZ
)( YX )( YX
x
yzx yx
zX
YX
),( )( YXzXPa Fig. 8.12
x
y
zY
YX yx
zy
),( )( YXzYPb
x
y
),( zz
)(c
27
(8-46)
Fig. 8.12 (c) represents the total region, and from there
If X and Y are independent, then
and hence
Similarly
Thus
).,(,)( zzFzYzXPzF XYZ
)()()( yFxFzF YXZ
).()()()()( zFzfzfzFzf YXYXZ (8-47)
.,
,,),min(
YXX
YXYYXW (8-48)
. ,,),min()( YXwXYXwYPwYXPwFW
28
Once again, the shaded areas in Fig. 8.13 (a)-(b) show the regions satisfying the above inequalities and Fig 8.13 (c) shows the overall region.
From Fig. 8.13 (c),
where we have made use of (7-5) and (7-12) with and
, ),()()(
,11)(
wwFwFwF
wYwXPwWPwF
XYYX
W
(8-49),22 yx
.11 wyx
x
yyx
wy
(a)
Fig. 8.13
x
y
yx wx
(b)
x
y
),( ww
(c)
29
Example 8.11: Let X and Y be independent exponential r.vs with common parameter . Define Find Solution: From (8-49)
and hence
But and so that
Thus min ( X, Y ) is also exponential with parameter 2.
Example 8.12: Suppose X and Y are as give in the above example. Define Determine
).,min( YXW ?)(wfW
)()()()( )( wFwFwFwFwF YXYXW
).()()()()()( )( wfwFwFwfwfwfwf YXYXYXW
,)( )( wYX ewfwf ,1)( )( w
YX ewFwF
).(2)1(22 )( 2 wUeeeewf wwwwW
(8-50)
. ),max(/),min( YXYXZ ).(zfZ
30
Solution: Although represents a complicated function, by partitioning the whole space as before, it is possible to simplify this function. In fact
As before, this gives
Since X and Y are both positive random variables in this case, we have The shaded regions in Figs 8.14 (a)-(b) represent the two terms in the above sum.
Fig. 8.14
)max(/)min(
.,/
,,/
YXXY
YXYXZ (8-51)
( ) / , / ,
, , .
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
.10 z
x
y
yx yzx
(a)x
y
yx
xzy
(b)
(8-52)
31
From Fig. 8.14
Hence
Example 8.13 (Discrete Case): Let X and Y be independent Poisson random variables with parameters and respectively. Let Determine the p.m.f of Z.
.),( ),( )(
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
dyyedyeey
dyyzyfyyzfydxxzxfxdyyyzfyzf
uyzyzyyyz
XYXYXYXYZ
(8-54)
1 2.YXZ
(8-53)
)(zfZ
z
Fig. 8.151
2
32
Solution: Since X and Y both take integer values the same is true for Z. For any gives only a finite number of options for X and Y. In fact, if X = 0, then Y must be n; if X = 1, then Y must be n-1, etc. Thus the event is the union of (n + 1) mutually exclusive events given by
As a result
If X and Y are also independent, then
and hence
, ,2 ,1 ,0 nYXn , ,2 ,1 ,0
. ) ,(
,)()(
0
0
n
k
n
k
knYkXP
knYkXPnYXPnZP
(8-55) , , knYkXAk
)()( , knYPkXPknYkXP
(8-56)
}{ nYX
kA
.,,2,1,0 nk
33
Thus Z represents a Poisson random variable with parameter indicating that 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.
As the last example illustrates, the above procedure for determining the p.m.f of functions of discrete random variables is somewhat tedious. As we shall see in Lecture 10, the joint characteristic function can be used in this context to solve problems of this type in an easier fashion.
,21
. , ,2 ,1 ,0 ,!
)(
)!(!
!
!)!(!
) ,()(
21)(
021
)(2
0
1
0
21
21
21
nn
e
knk
n
n
e
kne
ke
knYkXPnZP
n
n
k
knkknn
k
k
n
k
(8-57)