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)