PROBABILITY AND STATISTICS
FOR ENGINEERING
Two Functions of Two Random Variables
Hossein Sameti
Department of Computer Engineering
Sharif University of Technology
Two Functions of Two RV.s
 X and Y are two random variables with joint p.d.f f XY ( x, y ).
 h ( x, y ), and g ( x , y ) are functions
 define the new random variables:
Z  g ( X ,Y )
W  h ( X , Y ).
 How to determine the joint p.d.f f ZW ( z, w) ?
 with f ZW ( z, w) in hand, the marginal p.d.fs f Z (z) and fW (w) can be easily
determined.
Two Functions of Two RV.s
 For given z and w,
FZW ( z, w)  P Z ( )  z ,W ( )  w   P g ( X , Y )  z , h( X , Y )  w 
 P ( X , Y )  Dz ,w   

( x , y )Dz , w
f XY ( x, y )dxdy,
g ( x, y )  z
 where D z ,w is the region in the xy plane such that :
h( x, y )  w
y
Dz ,w
Dz ,w
x
Example
 X and Y are independent uniformly distributed rv.s in (0, ).
 Define Z  min( X , Y ), W  max( X , Y ).
 Determine f ZW ( z, w).
Solution
 Obviously both w and z vary in the interval ( 0,  ). Thus
FZW ( z, w)  0, if z  0 or w  0.
FZW ( z, w)  PZ  z,W  w  Pmin( X , Y )  z, max( X , Y )  w .
 two cases:
Dz ,w
Y
Y
( w, w)
( z, z )
( z, z )
yw
( w, w)
X
(a ) w  z
X
( b) w  z
Example - continued
 For w  z,
FZW ( z, w)  FXY ( z, w)  FXY ( w, z )  FXY ( z, z ) ,
w  z,
Y
 For w  z,
( w, w)
FZW ( z, w)  FXY ( w, w) ,
 With
w  z.
yw
( z, z )
X
FXY ( x, y )  FX ( x ) FY ( y ) 
 we obtain
(2 w  z ) z /  2 ,
FZW ( z, w)  
w2 /  2 ,

x y xy
  2,
 

0  z  w  ,
0  w  z  .
 Thus
(a ) w  z
Y
( z, z )
( w, w)
2 /  2 ,
f ZW ( z, w)  
 0,
0  z  w ,
otherwise .
X
( b) w  z
Example - continued
 Also,

f Z ( z )   f ZW ( z, w)dw 
z
 and
fW ( w)  
w
0
2
z
1  , 0  z   ,
  
f ZW ( z, w)dz 
2w

2
,
0  w  .
 g ( x , y ) and h( x, y ) are continuous and differentiable functions,
 So, it is possible to develop a formula to obtain the joint p.d.f f ZW ( z, w)
directly.
 Let’s consider: g ( x, y )  z,
h( x, y )  w.
 Let us say these are the solutions to the above equations:
( x1, y1 ), ( x2 , y2 ), , ( xn , yn ),
g ( xi , yi )  z,
h( xi , yi )  w.
w
y
w  w
( z , w) 
( x1, y1 )
z  z
z
(a)

( x2 , y2 )

w
2
1

( xi , yi )
z
n

( xn , yn )
(b)
i
x
Consider the problem of evaluating the probability
P z  Z  z  z, w  W  w  w 
 P z  g ( X , Y )  z  z, w  h( X , Y )  w  w  .
This can be rewritten as:
Pz  Z  z  z, w  W  w  w  f ZW ( z, w)zw.
To translate this probability in terms of f XY ( x, y ), we need to evaluate the
equivalent region for z w in the xy plane.
 The point A with coordinates (z,w) gets mapped onto the point A with
coordinates
( xi , yi )
 As z changes to z  z to point B in figure (a),
- let
B  represent its image in the xy plane.
 As w changes to w  w to C,
- let C  represent its image in the xy plane.
y
w
C
w
D
w
A
z
yi
B
z
z
(a)
D
C


A
xi
(b)
B
i
x
i .
 Finally D goes to D,
 ABCD represents the equivalent parallelogram in the XY plane with
area  i .
 The desired probability can be alternatively expressed as
 P( X ,Y )     f
i
i
XY
( xi , yi )i .
i
 Equating these, we obtain
f ZW ( z, w)   f XY ( xi , yi )
i
i
.
zw
 To simplify this, we need to evaluate the area  i of the parallelograms in
terms of zw.
 let g1 and h1 denote the inverse transformations, so that
xi  g1 ( z, w),
yi  h1 ( z, w).
 As the point (z,w) goes to ( xi , yi )  A,
- the point ( z  z, w)  B,
- the point ( z, w  w)  C ,
- the point ( z  z, w  w)  D.
 Hence x and y of B  are given by:
g1
g
z  xi  1 z,
z
z
h
h
h1 ( z  z, w)  h1 ( z, w)  1 z  yi  1 z.
z
z
g1 ( z  z, w)  g1 ( z, w) 
 Similarly, those of C are given by:
xi 
g1
w,
w
yi 
h1
w.
w
 The area of the parallelogram ABCD is given by
 i   AB AC  sin(    )
  AB cos   AC  sin     AB sin   AC  cos   .
 From the figure and these equations,
g1
h
z, AC  sin   1 w,
z
w
h
g
AB sin   1 z, AC  cos   1 w.
z
w
AB cos  
 so that
 or
g h 
 g h
 i   1 1  1 1  zw
 z w w z 
 g1

i
 z
 g h1 g1 h1 
 1


det


zw  z w
w z 
 h1

 z
g1 

w 

h1 

w 
J ( z , w)
 This is the Jacobian of the transformation xi  g1 ( z, w),
 g1
 z

J ( z , w)  det 
 h
 1
 z
 Using these in
f ZW ( z, w)   f XY ( xi , yi )
i
g1 
w 

.
h1 

w 
i
. , and
zw
f ZW ( z, w)   | J ( z, w) | f XY ( xi , yi )  
i
yi  h1 ( z, w).
i
| J ( z , w) | 
1
we get
| J ( xi , yi ) |
1
f XY ( xi , yi ),
| J ( xi , yi ) |
( )
*
 where J ( xi , yi ) represents the Jacobian of the original transformation:
 g

 x
J ( xi , yi )  det 
 h
 x

g 

y 

h 
y 
 x x
i
.
, y  yi
Example
 Example 9.2: Suppose X and Y are zero mean independent Gaussian
r.vs with common variance  2 .
where | w |  / 2.
 Define Z  X 2  Y 2 , W  tan 1 (Y / X ),
 Obtain f ZW ( z, w).
Solution
 Here
 Since
f XY ( x, y ) 
1
2
( x
e
2
2
 y 2 ) / 2 2
.
z  g ( x, y )  x 2  y 2 ; w  h( x, y )  tan 1 ( y / x ), | w |  / 2,
 if ( x1, y1 ) is a solution pair so is ( x1, y1 ). Thus
y
 tan w, or
x
y  x tan w.
Example - continued
 Substituting this into z, we get
z  x 2  y 2  x 1  tan 2 w  x sec w, or x  z cos w.
 and
y  x tan w  z sin w.
 Thus there are two solution sets
x1  z cos w, y1  z sin w, x2   z cos w, y2   z sin w.
 J ( z, w) is :
 so that
J ( z , w) 
x
z
y
z
| J ( z, w) | z.
x
w
y
w

cos w
 z sin w
sin w
z cos w
 z,
Example - continued
 Also J ( x, y ) is
x
J ( x, y ) 
y
x2  y2
x2  y2
 y
x2  y2

x
x2  y2
1
x2  y2

1
.
z
 Notice that here also | J ( z, w) | 1 / | J ( xi , yi ) |,
 Using ( ),
*
f ZW ( z, w)  z  f XY ( x1 , y1 )  f XY ( x2 , y2 ) 

 Thus
fZ ( z)  
 /2
 / 2
z

z
e
2
2
/ 2 2
f ZW ( z, w)dw 
0  z  , | w |
,
z

2
ez
2
/ 2 2
,

2
.
0  z  ,
2
 which represents a Rayleigh r.v with parameter  ,
Example - continued
 Also,

1
0

fW ( w)   f ZW ( z, w)dz 
, | w |

2
,
 which represents a uniform r.v in (  / 2,  / 2).
 Moreover,
f ZW ( z, w)  f Z ( z )  fW ( w)
 So Z and W are independent.
 To summarize, If X and Y are zero mean independent Gaussian random
variables with common variance, then
-
X 2  Y 2 has a Rayleigh distribution
- tan 1 (Y / X ) has a uniform distribution.
- These two derived r.vs are statistically independent.
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
X  jY  Ze jW ,
 where Z and W are as in
z  g ( x, y )  x 2  y 2 ; w  h( x, y )  tan 1 ( y / x), | w |  / 2,
except that for the abovementioned, to hold good on the entire complex
plane we must have    W   ,
 The magnitude and phase of a complex Gaussian r.v are independent
with Rayleigh and uniform distributions U ~ ( ,  ) respectively.
Example
 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.
Solution
 It is given that
f XY ( x, y ) 
1
( x  y ) / 
e
, x  0,
2

y  0.
 Now since u = x + y, v = x - y, always | v | u,and there is only one
solution given by
uv
uv
x
2
,
y
2
.
Example - continued
 Moreover the Jacobian of the transformation is given by
1
1
J ( x, y ) 
 2
1 1
 and hence
1
fUV (u, v )  2 e u /  ,
0  | v |  u  ,
2
represents the joint p.d.f of U and V. This gives
fU (u )  
 and
fV ( v )  
u
u

|v|
1
fUV (u, v )dv  2
2
fUV (u, v )du 
1
2 2


u
e u /  dv 
u

|v|
u
u / 
e
, 0  u  ,
2

e  u /  du 
1 |v|/ 
e
,    v  .
2
 Notice that in this case the r.vs U and V are not independent.
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
Z  g ( X , Y ),
 X and Y : two random variables.
 To determine f Z (z ) by making use of the above formulation in ( ), we
*
can define an auxiliary variable W  X or W  Y
 and the p.d.f of Z can be obtained from f ZW ( z, w) by proper integration.
Example
Z = X + Y
 Let W = Y so that the transformation is one-to-one and the solution is
given by y1  w, x1  z  w.
 The Jacobian of the transformation is given by
J ( x, y ) 
 and hence
 or
1
0
1
1
1
f ZW ( x, y )  f XY ( x1, y1 )  f XY ( z  w, w)
f Z ( z )   f ZW ( z, w)dw  


f XY ( z  w, w)dw,
 This reduces to the convolution of f X (z ) and fY (z ) if X and Y are
independent random variables.
Example
 Let X
U (0,1) and Y
 Define Z   2 ln X 
1/ 2
U (0,1) be independent.
cos( 2Y ).
 Find the density function of Z.
Solution
 Making use of the auxiliary variable W = Y,
x1  e
  z sec( 2w )  2 / 2
,
y1  w,
J ( z , w) 
x1
z
x1
w
y1
z
y1
w
 z sec 2 ( 2w) e  z sec( 2w ) 

  z sec 2 ( 2w)e  z sec( 2w ) 
0
2
/2
.
2
/2
x1
w
1
Example - continued
 Using these in ( ), we obtain
*
f ZW ( z , w)  z sec (2 w) e
2
   z  ,
  z sec( 2 w ) 
2
/2
,
0  w  1,
 and
1
f Z ( z )   f ZW ( z, w)dw  e
0
 z2 / 2

1
0
z sec (2 w) e
2
 z tan(2 w )  / 2
 Let u  z tan(2 w) so that du  2 z sec2 (2 w)dw.
 Notice that as w varies from 0 to 1, u varies from   to  .
2
dw.
Example - continued
 Using this in the previous formula, we get
f Z ( z) 

2
2
1
du
e z / 2 
e u / 2



2
2


2
1
e  z / 2 ,    z  ,
2
1
 As you can see, Z ~ N (0,1).
 A practical procedure to generate Gaussian random variables is from
two independent uniformly distributed random sequences, based on
1/ 2
Z   2 ln X  cos( 2Y ).
Example
 Let X and Y be independent identically distributed Geometric random
variables with
P( X  k )  P(Y  K )  pq k ,
 (a) Show that min (X , Y ) and X
–Y
k  0, 1, 2,.
are independent random variables.
 (b) Show that min (X , Y ) and max (X , Y ) – min (X , Y ) are also independent
Solution
 (a) Let
Z = min (X , Y ) , and W = X – Y.
 Note that Z takes only nonnegative values {0, 1, 2, }, while W takes
both positive, zero and negative values {0,  1,  2, }.
Example - continued
 We have
P(Z = m, W = n) = P{min (X , Y ) = m, X – Y = n}.
 But
Y
Z  min( X ,Y )  
X
X  Y  W  X  Y is nonnegativ e
X  Y  W  X  Y is negative.
 Thus
P( Z  m,W  n )  P{min( X ,Y )  m, X  Y  n, ( X  Y  X  Y )}
 P(min( X , Y )  m, X  Y  n, X  Y )
 P(min( X , Y )  m, X  Y  n, X  Y )
Example - continued
P( Z  m,W  n )  P(Y  m, X  m  n, X  Y )
 P ( X  m, Y  m  n , X  Y )
 P( X  m  n ) P(Y  m)  pq m  n pq m , m  0, n  0

m
mn
 P( X  m) P(Y  m  n )  pq pq , m  0, n  0
 p 2 q 2 m |n| , m  0, 1, 2, n  0,  1,  2,
represents the joint probability mass function of the random variables Z
and W. Also
P ( Z  m )   P ( Z  m,W  n )   p 2 q 2 m q |n|
n
n
 p 2 q 2 m (1  2 q  2 q 2  )
2q
 p 2 q 2 m (1  1 q )  pq 2 m (1  q)
 p (1  q) q 2 m ,
m  0, 1, 2,.
Example - continued
 Thus Z represents a Geometric random variable since 1  q 2  p(1  q),
and


P(W  n ) 
 P( Z  m,W  n )   p 2 q 2 m q|n|
m 0
m 0
 p 2 q|n| (1  q 2  q 4 
 1pq q|n| ,
)  p 2 q|n|
n  0,  1,  2,
1
1q2
.
 Note that
P( Z  m,W  n)  P( Z  m) P(W  n),
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.
Example - continued
 (b) Let
Z = min (X , Y ) , R = max (X , Y ) – min (X , Y ).
 In this case both Z and R take nonnegative integer values 0, 1, 2, .
 we get
P{Z  m, R  n}  P{min( X , Y )  m, max( X , Y )  min( X , Y )  n, X  Y }
 P{min( X , Y )  m, max( X , Y )  min( X , Y )  n, X  Y }
 P{Y  m, X  m  n, X  Y )  P( X  m, Y  m  n, X  Y }
 P{ X  m  n, Y  m, X  Y )  P( X  m, Y  m  n, X  Y }
 pq m  n pq m  pq m pq m  n ,

pq m  n pq m ,

2 p 2 q 2 m  n ,
  2 2m
 p q ,
m  0, 1, 2,,
n  1, 2,
m  0, 1, 2,,
n0
m  0, 1, 2,,
n  1, 2,
m  0, 1, 2,,
n  0.
Example - continued
 This equation is the joint probability mass function of Z and R.
 Also we can obtain:



P( Z  m)   P{Z  m, R  n}  p q (1  2 q n )  p 2 q 2 m 1  p
2
2m
n 0
 and
 p(1  q) q 2 m ,
n 1
2q

m  0, 1, 2,
p

,
n0

 1 q
P( R  n )   P{Z  m, R  n}  
2p n
m 0
1 q q , n  1, 2,.
 From (9-68)-(9-70), we get
P( Z  m, R  n)  P( Z  m) P( R  n)
 This proves the independence of the random variables Z and R as well.