Two Functions of Two Random Variables

Two Functions of Two RV.s

X and Y are two random variables with joint p.d.f

and are functions

define the new random variables:

How to determine the joint p.d.f

with in hand, the marginal p.d.fs and can be easily determined.

).,( yxf XY

).,(

),(

YXhW

YXgZ

),( yxg),,( yxh

?),( wzfZW

),( wzfZW )(zfZ)(wfW

Two Functions of Two RV.s

For given z and w,

where is the region in the xy plane such that :

wzDyx XYwz

ZW

dxdyyxfDYXP

wYXhzYXgPwWzZPwzF

,),( , ,),(),(

),(,),()(,)(),(

wzD ,

zyxg ),(wyxh ),(

x

y

wzD ,

wzD ,

X and Y are independent uniformly distributed rv.s in

Define

Determine

Obviously both w and z vary in the interval Thus

two cases:

.0or 0 if ,0),( wzwzFZW

).,( wzfZW

).,0(

).,0(

. ),max( ,),min(,),( wYXzYXPwWzZPwzFZW wzD ,

X

Y

),( ww

),( zz

zwb )(

X

Y

wy

),( ww

),( zz

zwa )(

Example

).,max( ),,min( YXWYXZ

Solution

For

For

With

we obtain

Thus

, , ),(),(),(),( zwzzFzwFwzFwzF XYXYXYZW

,zw

,zw

. , ),(),( zwwwFwzF XYZW

,)( )(),(2

xyyxyFxFyxF YXXY

2

2 2

(2 ) / , 0 ,( , )

/ , 0 .ZW

w z z z wF z w

w w z

.otherwise,0

,0,/2),(

2 wzwzfZW

Example - continued

X

Y

wy

),( ww

),( zz

zwa )(

X

Y

),( ww

),( zz

zwb )(

Also,

and

and are continuous and differentiable functions,

So, it is possible to develop a formula to obtain the joint p.d.f directly.

,0 ,12

),()(

z

zdwwzfzf

z ZWZ

.0 ,2

),()(

0 2

w

wdzwzfwf

w

ZWW

),( yxg ),( yxh

),( wzfZW

Example - continued

Let’s consider:

Let us say these are the solutions to the above equations:

.),( ,),( wyxhzyxg iiii

z

(a)

w

),( wz

z

www

zz

(b)

x

y1

2

i

n

),( 11 yx

),( 22 yx

),( ii yx

),( nn yx

.),( ,),( wyxhzyxg

),,( , ),,( ),,( 2211 nn yxyxyx

Consider the problem of evaluating the probability

This can be rewritten as:

To translate this probability in terms of we need to evaluate the

equivalent region for in the xy plane.

. ),(,),(

,

wwYXhwzzYXgzP

wwWwzzZzP

.),(, wzwzfwwWwzzZzP ZW

),,( yxf XY

wz

The point A with coordinates (z,w) gets mapped onto the point with coordinates

As z changes to to point B in figure (a),

- let represent its image in the xy plane.

As w changes to to C,

- let represent its image in the xy plane.

A

),( ii yx

zz

B

ww

C

(a)

w

Az

ww

B

zz

C D

(b)

y

Aiy B

xix

CD

i

Finally D goes to

represents the equivalent parallelogram in the XY plane with area

The desired probability can be alternatively expressed as

Equating these, we obtain

To simplify this, we need to evaluate the area of the parallelograms in terms of

DCBA

.i

,D

.),(),( i

iiiXYi

i yxfYXP

.),(),(

i

iiiXYZW wz

yxfwzf

.wzi

.i

let and denote the inverse transformations, so that

As the point (z,w) goes to

- the point

- the point

- the point

Hence x and y of are given by:

Similarly, those of are given by:

,),( Ayx ii

,),( Bwzz ,),( Cwwz

.),( Dwwzz

,),(),( 1111 z

z

gxz

z

gwzgwzzg i

.),(),( 1111 z

z

hyz

z

hwzhwzzh i

1g 1h

).,( ),,( 11 wzhywzgx ii

. , 11 ww

hyw

w

gx ii

B

C

The area of the parallelogram is given by

From the figure and these equations,

so that

or

.cos ,sin

,sin ,cos

11

11

ww

gCAz

z

hBA

ww

hCAz

z

gBA

wzz

h

w

g

w

h

z

gi

1111

w

h

z

h

w

g

z

g

z

h

w

g

w

h

z

g

wzi

11

11

1111 det

( , )J z w

DCBA . cos sinsin cos

)sin(

CABACABA

CABAi

This is the Jacobian of the transformation

Using these in , and we get

where represents the Jacobian of the original transformation:

1 1

1 1

( , ) det .

g g

z wJ z w

h h

z w

,),(|),(|

1),(|),(|),(

iiiXY

iiiiiXYZW yxf

yxJyxfwzJwzf

|),(|

1 |),(|

ii yxJwzJ

( , )i iJ x y

.det),(

, ii yyxx

ii

y

h

x

h

y

g

x

g

yxJ

).,( ),,( 11 wzhywzgx ii

.),(),(

i

iiiXYZW wz

yxfwzf

(*)

Example 9.2: Suppose X and Y are zero mean independent Gaussian r.vs with common variance

Define where

Obtain

Here

Since

if is a solution pair so is Thus

),/(tan , 122 XYWYXZ

.2

).,( wzfZW

.2

1),(

222 2/)(2

yx

XY eyxf

,2/||),/(tan),(;),( 122 wxyyxhwyxyxgz

),( 11 yx ).,( 11 yx

.tanor ,tan wxywx

y

.2/|| w

Example

Solution

Substituting this into z, we get

and

Thus there are two solution sets

so that

.cos or ,sec tan1 222 wzxwxwxyxz

.sintan wzwxy

.sin ,cos ,sin ,cos 2211 wzywzxwzywzx

:is ),( wzJ,

cossin

sincos),( z

wzw

wzw

w

y

z

y

w

x

z

x

wzJ

.|),(| zwzJ

Example - continued

Also is

Notice that here also

Using (*),

Thus

which represents a Rayleigh r.v with parameter

),( yxJ

.11

),(22

2222

2222

zyx

yx

x

yx

y

yx

y

yx

x

yxJ

|,),(|/1|),(| ii yxJwzJ

.2

|| ,0 ,

),(),(),(

22 2/2

2211

wzez

yxfyxfzwzf

z

XYXYZW

,0 ,),()(22 2/

2

2/

2/

zez

dwwzfzf zZWZ

,2

Example - continued

Also,

which represents a uniform r.v in

Moreover,

So Z and W are independent.

To summarize, If X and Y are zero mean independent Gaussian random variables with common variance, then

- has a Rayleigh distribution

- has a uniform distribution.

- These two derived r.vs are statistically independent.

).2/,2/(

22 YX )/(tan 1 XY

)()(),( wfzfwzf WZZW

,2

|| ,1

),()(

0

wdzwzfwf ZWW

Example - continued

Alternatively, with X and Y as independent zero mean Gaussian r.vs with

common variance, X + jY represents a complex Gaussian r.v. But

where Z and W are as in

except that for the abovementioned, to hold good on the entire complex plane we must have

The magnitude and phase of a complex Gaussian r.v are independent with Rayleigh and uniform distributions respectively.

,jWZejYX

, W

),( ~ U

,2/||),/(tan),(;),( 122 wxyyxhwyxyxgz

Example - continued

Let X and Y be independent exponential random variables with common parameter .

Define U = X + Y, V = X - Y.

Find the joint and marginal p.d.f of U and V.

It is given that

Now since u = x + y, v = x - y, always and there is only one solution given by

.0 ,0 ,1

),( /)(2

yxeyxf yxXY

.2

,2

vuy

vux

,|| uv

Example

Solution

Moreover the Jacobian of the transformation is given by

and hence

represents the joint p.d.f of U and V. This gives

and

Notice that in this case the r.vs U and V are not independent.

2 11

1 1 ),(

yxJ

, || 0 ,2

1),( /

2 uvevuf u

UV

,0 ,2

1),()( /

2

/2

ueu

dvedvvufuf uu

u

uu

u UVU

/ | |/2 | | | |

1 1( ) ( , ) , .

2 2u v

V UVv vf v f u v du e du e v

Example - continued

As we will show, the general transformation formula in (*) making use of two functions

can be made useful even when only one function is specified.

Auxiliary Variables

Suppose

X and Y : two random variables.

To determine by making use of the above formulation in (*), we

can define an auxiliary variable

and the p.d.f of Z can be obtained from by proper integration.

),,( YXgZ

YWXW or

),( wzfZW

)(zfZ

Z = X + Y

Let W = Y so that the transformation is one-to-one and the solution is given by

The Jacobian of the transformation is given by

and hence

or

This reduces to the convolution of and if X and Y are independent random variables.

1 1 0

1 1 ),( yxJ

),(),(),( 11 wwzfyxfyxf XYXYZW

,),(),()( dwwwzfdwwzfzf XYZWZ

)(zf X )(zfY

. , 11 wzxwy

Example

Let and be independent.

Define

Find the density function of Z.

Making use of the auxiliary variable W = Y,

,

,

1

2/)2sec(1

2

wy

ex wz

.)2(sec

10

)2(sec

),(

2/)2sec(2

12/)2sec(2

11

11

2

2

wz

wz

ewz

w

xewz

w

y

z

y

w

x

z

x

wzJ

)1,0( UX )1,0( UY

).2cos(ln2 2/1 YXZ

Example

Solution

Using these in (*), we obtain

and

Let so that

Notice that as w varies from 0 to 1, u varies from to

2sec(2 ) / 22( , ) sec (2 ) ,

, 0 1,

z wZWf z w z w e

z w

Example - continued

22 1 1 tan(2 ) / 2/ 2 2

0 0( ) ( , ) sec (2 ) .z wz

Z ZWf z f z w dw e z w e dw

tan(2 )u z w 22 sec (2 ) .du z w dw

.

Using this in the previous formula, we get

As you can see,

A practical procedure to generate Gaussian random variables is from two independent uniformly distributed random sequences, based on

, ,2

1

2

2

1)( 2/

1

2/2/ 222

zedu

eezf zuzZ

).1,0( ~ NZ

Example - continued

).2cos(ln2 2/1 YXZ

Let X and Y be independent identically distributed Geometric random variables with

(a) Show that min (X , Y ) and X – Y are independent random variables.

(b) Show that min (X , Y ) and max (X , Y ) – min (X , Y ) are also independent

(a) Let

Z = min (X , Y ) , and W = X – Y.

Note that Z takes only nonnegative values while W takes both positive, zero and negative values

Example

Solution

.,2 ,1 ,0 ,)()( kpqKYPkXP k

}, ,2 1, 0,{ }. ,2 1, 0,{

We have

P(Z = m, W = n) = P{min (X , Y ) = m, X – Y = n}.

But

Thus

negative. is

enonnegativ is ),min(

YXWYXX

YXWYXYYXZ

),,),(min(

),,),(min(

)} (,,),{min(),(

YXnYXmYXP

YXnYXmYXP

YXYXnYXmYXPnWmZP

Example - continued

represents the joint probability mass function of the random variables Z

and W. Also

,2 ,1 ,0 ,2 ,1 ,0 ,

0,0,)()(

0,0,)()(

),,(

),,( ),(

||22

nmqp

nmpqpqnmYPmXP

nmpqpqmYPnmXP

YXnmYmXP

YXnmXmYPnWmZP

nm

nmm

mnm

.,2 ,1 ,0 ,)1(

)1()1(

)221(

),()(

2

222

222

||22

12

mqqp

qpqqp

qqqp

qqpnWmZPmZP

m

mm

m

n n

nm

qq

Example - continued

Thus Z represents a Geometric random variable since and

Note that

establishing the independence of the random variables Z and W.

The independence of X – Y and min (X , Y ) when X and Y are

independent Geometric random variables is an interesting observation.

),1(1 2 qpq

Example - continued

2 2 | |

0 0

2 | | 2 4 2 | | 12

| |

1

1

( ) ( , )

(1 )

, 0, 1, 2, .

m n

m m

n n

n

qpq

P W n P Z m W n p q q

p q q q p q

q n

),()(),( nWPmZPnWmZP

(b) Let

Z = min (X , Y ) , R = max (X , Y ) – min (X , Y ).

In this case both Z and R take nonnegative integer values

we get

. ,2 1, 0,

Example - continued

.0 ,,2 ,1 ,0 ,

,2 ,1 ,,2 ,1 ,0 ,2

0 ,,2 ,1 ,0 ,

,2 ,1 ,,2 ,1 ,0 ,

} ,,() ,,{

} ,,() ,,{

} ,),min(),max( ,),{min(

} ,),min(),max( ,),{min(},{

22

22

nmqp

nmqp

nmpqpq

nmpqpqpqpq

YXnmYmXPYXmYnmXP

YXnmYmXPYXnmXmYP

YXnYXYXmYXP

YXnYXYXmYXPnRmZP

m

nm

mnm

nmmmnm

This equation is the joint probability mass function of Z and R. Also we can obtain:

and

From (9-68)-(9-70), we get

This proves the independence of the random variables Z and R as well.

,2 ,1 ,0 ,)1(

1)21(},{)(

2

0

22

1

22 2

mqqp

qpqqpnRmZPmZP

m

n

m

n

nmpq

)()(),( nRPmZPnRmZP

.,2 ,1 ,

0 , },{)(

12

1

0 nq

nnRmZPnRP

nm

qp

qp

Example - continued