Fakultt Informatik Institut fr Systemarchitektur Professur Rechnernetze

A Function of Two Random Variables

Waltenegus g Dargie g
Slides are based on the book: A. Papoulis and S.U. Pillai, "Probability, random variables and stochastic processes", McGraw Hill (4th edition), 2002

Function of Two Variables

Given two random variables X and Y and a g ( X , Y ) , we would f function ti ld like lik to t express a new random variable Z as
Z = g ( X , Y ).

Thi This may signify, i if for f example, l the th energy cost of a server that can be expressed as th cost the t of f processing i (X) and d communication (Y). It is, therefore, important to express the PDF o of Z in terms te s o of the t e jo joint t PDF, , f XY ( x , y )
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Function of Two Variables

Some of the additional relationship we are concerned d are displayed di l d below. b l
X +Y

max( X , Y ) min( X , Y )
Z = g ( X ,Y )

X Y
XY

X 2 +Y2

X /Y

tan 1 ( X / Y )

Function of Two Variables

The distribution of Z is expressed as
F Z ( z ) = P (Z ( ) z ) = P ( g ( X , Y ) z ) = P [( X , Y ) D z ] =

x , y D z

f XY ( x , y ) dxdy

Dz
Dz
X

Addition
Suppose Z = X + Y. Find the distribution and d density d it of f Z. Z The region of D in the xy plane is displayed by the shaded region to the left of the line x + y = z.
z

x + y = z.
x= z y

FZ ( z ) = P( X + Y z ) =

+ y =

z y x =

f XY ( x, y )dxdy,
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Addition
To compute the PDF of Z, we apply L ib it differentiation Leibnitzs diff ti ti rule. l b(z) If H ( z ) = a ( z ) h ( x , z ) dx , then,
b ( z ) h ( x , z ) dH ( z ) db ( z ) da ( z ) = dx . h (b( z ), z ) h (a ( z ), z ) + a ( z ) dz dz dz z

Hence, the PDF of Z can be computed as

+ z y f z y XY ( x, y ) f Z ( z) = f XY ( x, y )dx dy = f XY ( z y, y ) 0 + dy z z +

f XY ( z y, y )dy.

Alternatively, it can also be expressed in terms of the PDF of x

Addition
If X and Y are independent, then
f XY ( x, y ) = f X ( x ) fY ( y )

Inserting g this equation q in the previous p expression yields:

f Z ( z) =
+ y =

f X ( z y ) fY ( y )dy d =

+ x =

f X ( x ) fY ( z x )dx d .

Th The above b expression i is i the th standard t d d convolution of two functions, namely, and d f Y ( z ).

f X (z)

Addition
As a Special case, if f X ( x ) = 0 for x < 0, and f Y ( y ) = 0 for f y > 0, 0 th the new li limit it of f D z is i set t as shown below.
y

FZ ( z ) =
z

z y =0

z y

x =0

f XY ( x, y )dxdy

( z,0)
x=z y

(0, z)

z y f Z ( z ) = f XY ( x, y )dx dy y = 0 z x = 0 z f ( z y, y )dy, z > 0, x = 0 XY 0, z 0.

Alternatively, in the case of independence:

f Z ( z) =
z x =0 z f X ( x ) fY ( z x )dx, z > 0, f XY ( x, z x )dx d = y =0 0, z 0,
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Subtraction
Suppose Z = X - Y. Find the distribution and d density d it of f Z. Z The region of D in the xy plane is displayed by the shaded region to the left of the line x + y = z.
z

y y
x y = z x= y+z x
+ y = z+ y x =

FZ ( z ) = P( X Y z ) =

f XY ( x, y )dxdy
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Subtraction
The density of Z is given by
dF ( z ) fZ ( z) = Z = dz
+ y =

z+ y x =

f XY ( x , y ) dx dy =

f XY ( y + z , y ) dy.

If X and Y are independent, then,

f Z ( z) =
+

f X ( z + y ) fY ( y )dy = f X ( z ) fY ( y ),

As a special case, if f X ( x) = 0, x < 0, and fY ( y) = 0, y < 0, then, Z can be either positive or negative. This requires the analysis of the problem into two separate p conditions, , for z > 0 and z < 0.
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Subtraction
y
x= z+ y
z z

FZ ( z ) =

+ y =0

z+ y x =0

f XY ( x, y )dxdy y

y
x= z+ y
z

FZ ( z ) =
x

+ y = z

z+ y x =0

f XY ( x, y )dxdy

+ 0 f XY ( z + y , y ) dy , fZ ( z ) = + f ( z + y , y ) dy , z XY

z 0, z < 0.
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Multiplication
Suppose Z = X/Y. Find the distribution of Z Our Z. O aim i is i to t find fi d and d expression i for f FZ ( z ) = P ( X / Y z ) . The inequality X / Y z can be expressed as X Yz if Y > 0, and X Yz if Y < 0. Hence Fz ( z ) need to be conditioned by the __ event A = (Y > 0 ) and its compliment A . {X / Y z}= {( X / Y z ) ( A A )}= {( X / Y z ) A} {( X / Y z ) A }
y
x = yz

y
x = yz

x
x

A = (Y > 0 )

A = (Y < 0 )

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Multiplication
Using the property of mutually exclusive: P (X / Y z ) = P (X / Y z,Y > 0) + P (X / Y z,Y < 0) = P ( X Yz , Y > 0 ) + P ( X Yz , Y < 0 ) . Integrating over the two regions results
FZ ( z ) =
+ y =0

yz x =

f XY ( x , y ) dxdy +

0 y =

x = yz

f XY ( x , y ) dxdy .

Differentiating with respect to Z gives

f Z ( z ) = yf XY ( yz, y )dy + ( y ) f XY ( yz, y )dy
0 + 0

= | y | f XY ( yz, y )dy,

< z < +.
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Multiplication
Note that if X and Y are nonnegative random d variables, i bl then th the th area of f integration reduces to that shown below.
y
x = yz

As a result
FZ ( z ) =

y =0

yz y x =0

f XY ( x, y )dxdy d d

+ y f ( yz, y )dy, z > 0, f Z ( z ) = 0 XY 0, otherwise.

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Circle
Suppose Z = X 2 + Y 2 . Find the distribution and d density d it of f Z. Z The distribution of Z is expressed as
FZ ( z ) = P (X 2 + Y 2 z ) =

But the equation X 2 +Y 2 z represents the area of a circle with a radius z . Therefore,
y

X 2 +Y 2 z

f XY ( x , y ) dxdy .

z
X 2 +Y 2 = z

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Circle
Hence,
FZ ( z ) =

z y= z

z y2 x= z y2

f XY ( x , y ) dxdy .

After applying Leibnitz's distribution:

fZ (z) =

z y= z

1 2 z y
2

(f

XY

( z y 2 , y ) + f XY ( z y 2 , y ) dy .

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Circle
Suppose Z = X 2 + Y 2 . Find the distribution and d density d it of f Z. Z In this case the equation represents a circle of radius z 2 . , Hence,
f ( x , y ) dxdy . Differentiating the above expression results FZ ( z ) =
y= z x= z2 y2 XY z z2 y2

fZ (z) =

z z 2

z z y
2

(f

XY

z 2 y 2 , y ) + f XY (

z 2 y 2 , y ) dy .

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Max and Min

Suppose Z = max( X , Y ). D t Determine i the th distribution di t ib ti and d density d it of f Z. The functions max() and min() are nonlinear functions.
y
X z
X , Z = max( X , Y ) = Y , X >Y, X Y,

x=z

x= y

y
X Y

x= y y=z

( z, z )

x
X >Y
P ( X z, X > Y )

x
Yz
P (Y z , X Y )
Combined
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Max and Min

As a result:
FZ ( z ) = P (max( X , Y ) z ) = P [( X z , X > Y ) (Y z , X Y = P ( X z , X > Y ) + P (Y z , X Y ),

)]

FZ ( z ) = P ( X z , Y z ) = F XY ( z , z ).

If the two random variables are independent, then,

FZ ( z ) = F X ( x ) FY ( y )

In which case the density can be expressed as

f Z ( z ) = F X ( z ) f Y ( z ) + f X ( z ) FY ( z ) ).
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Max and Min

Suppose
Z = min( X , Y ).

In this case
X >Y, X Y.

Y , Z = min( X , Y ) = X ,

The distribution of Z is given as

Fz ( z ) = P(min( X , Y ) z ) = P[(Y z, X > Y ) ( X z, X Y )].
y

y
x= y

x=z

y
x= y

y=z

( z, z )

x
X >Y X Y

x
Combined

x
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Max and Min

The distribution of Z is given by
F z ( z ) = 1 P (Z > z ) = 1 P ( X > z , Y > z ) = F X ( z ) + FY ( z ) F XY ( z , z ) ,

If the two random variables are independent, then the density of Z becomes
f z ( z ) = f X ( z ) + f Y ( z ) f X ( z ) FY ( z ) FX ( z ) f Y ( z ).

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Max and Min

Finally, suppose Z = [min( X , Y ) / max( X , Y ) ]. D fi th Define the distribution di t ib ti and d density d it of f Z. Z Although Z represents a complicated function, by partitioning the whole space as before, it is possible to simplify this function. As before,
X /Y , Z = Y / X , X Y, X >Y.

FZ ( 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 ) .

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Max and Min

If X and Y are both positive random variables, i bl then th 0 < z < 1. The two terms of the above equations are shown by the shaded regions below
y x = yz y x= y x= y y = xz
X >Y

X Y
0

FZ ( z ) =
0

yz x =0

f XY ( x, y )dxdy +
0

xz y =0

f XY ( x, y )dydx.

f Z ( z ) = y f XY ( yz, y )dy + x f XY ( x, xz)dx = y{ f XY ( yz, y ) + f XY ( y, yz ) } dy

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