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1
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.
zD
zyxg ),(
)(zFZ
zD
zD
X
Y
zD
zD
Fig. 8.1
4
5
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
6
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
( , ) .
z y z yXY
Z XY XY
XY
f x yf z f x y dx dy f z y y dy
z z
f z y y dy
(8-7)
7
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
8
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
9
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
zdxxzfxfdxxzxfzfz
x YXz
x XYZ
z
x
xz
y XYZ dydxyxfzF
0
0 ),()(
(8-13)
10
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
( )( ) ( , ) ( , ) .
z yZ
Z XY XYy x
dF zf z f x y dx dy f y z y dy
dz z
(8-21)
( ) ( ) ( ) ( ) ( ),Z X Y X Yf z f z y f y dy f z f y
(8-22)
)( zf X ).(zfY
Fig. 8.7
y
x
zyx zyx
y
).(zfZ
11
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)
12
Example 8.6: Obtain Solution: We have
.),()(22
22
zYX XYZ dxdyyxfzYXPzF
.22 YXZ ).(zfZ
13
But, represents the area of a circle with radius and hence from Fig. 8.11,
This gives after repeated differentiation
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
14
Example 8.10:
Determine:
).,min( ),,max( YXWYXZ
)()( wfzf WZ and
15
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
16
(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
17
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)
18
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.
).,min( YXW ?)(wfW
)()()()( )( wFwFwFwFwF YXYXW
).()()()()()( )( wfwFwFwfwfwfwf YXYXYXW
,)( )( wYX ewfwf ,1)( )( w
YX ewFwF
).(2)1(22 )( 2 wUeeeewf wwwwW
(8-50)
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