9. Two Functions of Two Random
Variables
In the spirit of the previous lecture, let us look at an
immediate generalization: Suppose X and Y are two random
variables with joint p.d.f f XY ( x, y). Given two functions g ( x, y )
and h( x, y ), define the new random variables
Z  g ( X ,Y )
W  h ( X , Y ).
(9-1)
(9-2)
How does one determine their joint p.d.f f ZW ( z, w) ? Obviously
with f ZW ( z, w) in hand, the marginal p.d.fs f Z (z) and fW (w)
can be easily determined.
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PILLAI
The procedure is the same as that in (8-3). In fact 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,
(9-3)
where Dz ,w is the region in the xy plane such that the
inequalities g ( x, y )  z and h( x, y )  w are simultaneously
satisfied.
We illustrate this technique in the next example.
y
Dz ,w
Dz ,w
x
2
Fig. 9.1
PILLAI
Example 9.1: Suppose X and Y are independent uniformly
distributed random variables in the interval (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, ).
(9-4)
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 . (9-5)
We must consider two cases: w  z and w  z, since they
give rise to different regions for Dz ,w (see Figs. 9.2 (a)-(b)).
Y
Y
( w, w)
( z, z )
( z, z )
( w, w)
yw
X
X
( b) w  z
(a ) w  z
Fig. 9.2
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PILLAI
For w  z, from Fig. 9.2 (a), the region
by the doubly shaded area. Thus
Dz ,w
is represented
FZW ( z, w)  FXY ( z, w)  FXY ( w, z )  FXY ( z, z ) ,
w  z,
(9-6)
and for w  z, from Fig. 9.2 (b), we obtain
FZW ( z, w)  FXY ( w, w) ,
With
w  z.
x y xy
FXY ( x, y )  FX ( x ) FY ( y )    2 ,
 
(9-7)
(9-8)

we obtain
Thus
(2 w  z ) z /  2 ,
FZW ( z, w)  
2
2
w
/

,

2 /  2 ,
f ZW ( z, w)  
 0,
0  z  w  ,
0  w  z  .
0  z  w ,
otherwise .
(9-9)
(9-10)
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PILLAI
From (9-10), we also obtain
f Z ( z)  

z
2
z
f ZW ( z, w)dw  1  , 0  z   ,
  
(9-11)
2w
(9-12)
and
fW ( w)  
w
0
f ZW ( z, w)dz 

2
,
0  w  .
If g ( x, y ) and h( x, y ) are continuous and differentiable
functions, then as in the case of one random variable (see (530)) it is possible to develop a formula to obtain the joint
p.d.f f ZW ( z, w) directly. Towards this, consider the equations
g ( x, y )  z,
h( x, y )  w.
(9-13)
For a given point (z,w), equation (9-13) can have many
solutions. Let us say
( x1, y1 ), ( x2 , y2 ), , ( xn , yn ),
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PILLAI
represent these multiple solutions such that (see Fig. 9.3)
g ( xi , yi )  z,
h( xi , yi )  w.
y
w
w  w

( x2 , y2 )
( x1, y1 )

i
( xi , yi )
z  z
z
2
1

w
( z , w) 
(9-14)

z
n
x
( xn , yn )
(b)
(a)
Fig. 9.3
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  .
(9-15)
6
PILLAI
Using (7-9) we can rewrite (9-15) as
Pz  Z  z  z, w  W  w  w  f ZW ( z, w)zw.
(9-16)
But 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.
Towards this referring to Fig. 9.4, we observe that the point
A with coordinates (z,w) gets mapped onto the point A with
coordinates ( xi , yi ) (as well as to other points as in Fig. 9.3(b)).
As z changes to z  z to point B in Fig. 9.4 (a), let B 
represent its image in the xy plane. Similarly as w changes
to w  w to C, let C represent its image in the xy plane.
y
w
C
w
C
D
w
A
z
yi
B
z
z
(a)
D
 
A
xi
Fig. 9.4
(b)
B
i
x
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PILLAI
Finally D goes to D, and ABCD represents the equivalent
parallelogram in the XY plane with area  i . Referring back
to Fig. 9.3, the probability in (9-16) can be alternatively
expressed as
 P( X ,Y )     f
i
i
XY
( xi , yi )i .
(9-17)
i
Equating (9-16) and (9-17) we obtain
f ZW ( z, w)   f XY ( xi , yi )
i
i
.
zw
(9-18)
To simplify (9-18), we need to evaluate the area  i of the
parallelograms in Fig. 9.3 (b) in terms of zw. Towards this,
h1
g1 denote
let and
the inverse transformation in (9-14), so
that
xi  g1 ( z, w),
yi  h1 ( z, w).
(9-19)
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PILLAI
As the point (z,w) goes to ( xi , yi )  A, the point ( z  z, w)  B,
the point ( z, w  w)  C , and the point ( z  z, w  w)  D.
Hence the respective x and y coordinates of B  are given by
g1
g
z  xi  1 z,
z
z
(9-20)
h1
h1
h1 ( z  z, w)  h1 ( z, w) 
z  yi 
z.
z
z
(9-21)
g1 ( z  z, w)  g1 ( z, w) 
and
Similarly those of C are given by
xi 
g1
w,
w
yi 
h1
w.
w
(9-22)
The area of the parallelogram ABCD in Fig. 9.4 (b) is
given by
 i   AB AC  sin(    )
(9-23)
  AB cos   AC  sin     AB sin   AC  cos   .
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PILLAI
But from Fig. 9.4 (b), and (9-20) - (9-22)
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
 g1 h1 g1 h1 
i  

 zw
 z w w z 
(9-24)
(9-25)
(9-26)
and
 g1

i
 z
 g1 h1 g1 h1 


  det 
zw  z w
w z 
 h1

 z
g1 

w 

h1 

w 
The right side of (9-27) represents the Jacobian
the transformation in (9-19). Thus
(9-27)
J ( z , w) of
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PILLAI
 g1
 z

J ( z , w)  det 
 h
 1
 z
g1 
w 

.
h1 

w 
(9-28)
Substituting (9-27) - (9-28) into (9-18), we get
1
f ZW ( z, w)   | J ( z, w) | f XY ( xi , yi )  
f XY ( xi , yi ),
i
i | J ( xi , yi ) |
(9-29)
since
1
| J ( z , w) | 
| J ( xi , yi ) |
(9-30)
where J ( xi , yi ) represents the Jacobian of the original
transformation in (9-13) given by
 g

 x
J ( xi , yi )  det 
 h
 x

g 

y 

h 
y 
 x x
i
.
(9-31)
11
, y  yi
PILLAI
Next we shall illustrate the usefulness of the formula in
(9-29) through various examples:
Example 9.2: Suppose X and Y are zero mean independent
Gaussian r.vs with common variance  2 .
Define Z  X 2  Y 2 , W  tan 1 (Y / X ), where | w |  / 2.
Obtain f ZW ( z, w).
Solution: Here
f XY ( x, y ) 
Since
1
2
2
e
( x 2  y 2 ) / 2 2
.
(9-32)
z  g ( x, y )  x 2  y 2 ; w  h( x, y )  tan 1 ( y / x ), | w |  / 2, (9-33)
if ( x1, y1 ) is a solution pair so is ( x1, y1 ). From (9-33)
y
 tan w, or
x
y  x tan w.
(9-34)
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PILLAI
Substituting this into z, we get
z  x 2  y 2  x 1  tan 2 w  x sec w, or x  z cos w.
(9-35)
and
(9-36)
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. (9-37)
We can use (9-35) - (9-37) to obtain J ( z, w). From (9-28)
J ( z , w) 
so that
x
z
y
z
x
w
y
w

cos w
 z sin w
sin w
z cos w
| J ( z, w) | z.
 z,
(9-38)
(9-39)
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PILLAI
We can also compute J ( x, y ) using (9-31). From (9-33),
x
J ( x, y ) 
y
x2  y2
 y
x2  y2
x2  y2

x
x2  y2
1
x2  y2

1
.
z
(9-40)
Notice that | J ( z, w) | 1 / | J ( xi , yi ) |, agreeing with (9-30).
Substituting (9-37) and (9-39) or (9-40) into (9-29), we get
f ZW ( z, w)  z  f XY ( x1 , y1 )  f XY ( x2 , y2 ) 

z

e
2
 z 2 / 2 2
0  z  , | w |
,

2
(9-41)
.
Thus
fZ ( z)  
 /2
 / 2
f ZW ( z, w)dw 
z

2
e
 z 2 / 2 2
0  z  ,
,
which represents a Rayleigh r.v with parameter

1
0

fW ( w)   f ZW ( z, w)dz 
, | w |

2
,
 2,
(9-42)
and
(9-43)
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PILLAI
which represents a uniform r.v in the interval (  / 2,  / 2).
Moreover by direct computation
f ZW ( z, w)  f Z ( z )  fW ( w)
(9-44)
implying that Z and W are independent. We summarize these
results in the following statement: If X and Y are zero mean
independent Gaussian random variables with common
variance, then X 2  Y 2 has a Rayleigh distribution and tan 1(Y / X )
has a uniform distribution. Moreover these two derived r.vs
are statistically independent. Alternatively, with X and Y as
independent zero mean r.vs as in (9-32), X + jY represents a
complex Gaussian r.v. But
X  jY  Ze jW ,
(9-45)
where Z and W are as in (9-33), except that for (9-45) to hold
good on the entire complex plane we must have    W  15 ,
and hence it follows that the magnitude and phase of PILLAI
a complex Gaussian r.v are independent with Rayleigh
and uniform distributions U ~ ( , ) respectively. The
statistical independence of these derived r.vs is an interesting
observation.
Example 9.3: 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.
(9-46)
Now since u = x + y, v = x - y, always | v | u, and there is
only one solution given by
x
uv
,
2
y
uv
.
2
(9-47)
Moreover the Jacobian of the transformation is given by 16
PILLAI
1
1
J ( x, y ) 
 2
1 1
and hence
1 u / 
fUV (u, v )  2 e ,
2
0  | v |  u  ,
(9-48)
represents the joint p.d.f of U and V. This gives
fU (u )  
u
u
1
fUV (u, v )dv  2
2

u
e u /  dv 
u
u
u / 
e
, 0  u  ,
2

(9-49)
and
fV ( v )  

|v|
fUV (u, v )du 
1
2 2


|v|
e  u /  du 
1 |v|/ 
e
,    v  . (9-50)
2
Notice that in this case the r.vs U and V are not independent.
As we show below, the general transformation formula in
(9-29) making use of two functions can be made useful even
17
when only one function is specified.
PILLAI
Auxiliary Variables:
Suppose
Z  g ( X , Y ),
(9-51)
where X and Y are two random variables. To determine f Z (z)
by making use of the above formulation in (9-29), we can
define an auxiliary variable
W X
or W  Y
(9-52)
and the p.d.f of Z can be obtained from f ZW ( z, w) by proper
integration.
Example 9.4: Suppose Z = X + Y and let W = Y so that the
transformation is one-to-one and the solution is given
18
by y1  w, x1  z  w.
PILLAI
The Jacobian of the transformation is given by
1
J ( x, y ) 
0
1
1
1
and hence
f ZW ( x, y )  f XY ( x1, y1 )  f XY ( z  w, w)
or
f Z ( z )   f ZW ( z, w)dw  


f XY ( z  w, w)dw,
(9-53)
which agrees with (8.7). Note that (9-53) reduces to the
convolution of f X (z ) and fY (z) if X and Y are independent
random variables. Next, we consider a less trivial example.
Example 9.5: Let X  U (0,1) and
Define Z   2 ln X 1 / 2 cos( 2Y ).
Y  U (0,1)
be independent.
(9-54)
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PILLAI
Find the density function of Z.
Solution: We can make use of the auxiliary variable W = Y
in this case. This gives the only solution to be
x1  e
  z sec( 2w )  2 / 2
(9-55)
,
y1  w,
(9-56)
and using (9-28)
J ( z , w) 
x1
z
x1
w
y1
z
y1
w
 z sec ( 2w) e
2

  z sec ( 2w)e
2
x1
w
  z sec( 2w )  2 / 2
0
  z sec( 2w )  2 / 2
(9-57)
1
.
Substituting (9-55) - (9-57) into (9-29), we obtain
f ZW ( z , w)  z sec (2 w) e
2
   z  ,
  z sec( 2 w ) 
2
/2
0  w  1,
,
(9-58)
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PILLAI
and
1
f Z ( z )   f ZW ( z, w)dw  e
 z2 / 2
0

1
0
z sec (2 w) e
2
 z tan(2 w )  / 2
2
dw. (9-59)
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  .
Using this in (9-59), we get
f Z ( z) 

1
z2 / 2
u 2 / 2 du
e
e




2
2


1
z2 / 2
e
,    z  ,
2
(9-60)
1
which represents a zero mean Gaussian r.v with unit
variance. Thus Z  N (0,1). Equation (9-54) can be used as
a practical procedure to generate Gaussian random variables
from two independent uniformly distributed random
sequences.
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PILLAI
Example 9.6 : Let X and Y be independent identically distributed
Geometric random variables with
P( X  k )  P(Y  K )  pq k ,
k  0, 1, 2,.
(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 random variables.
Solution: (a) Let
(9-61)
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, }. We have
P(Z = m, W = n) = P{min (X , Y ) = m, X – Y = n}. But
X  Y  W  X  Y is nonnegativ e
Y
Z  min( X ,Y )  
X 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 )
22
 P(min( X , Y )  m, X  Y  n, X  Y ) (9-62) PILLAI
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, (9-63)
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,.
Thus Z represents a Geometric random variable since
1  q 2  p(1  q), and
(9-64)
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P(W  n ) 

 P ( Z  m, W  n ) 
m 0
 p 2 q|n| (1  q 2  q 4 
 1pq q|n| ,

2 2 m |n|
p
 q q
m 0
)  p 2 q|n|
n  0,  1,  2,
1
1q2
.
(9-65)
Note that
P( Z  m,W  n)  P( Z  m) P(W  n),
(9-66)
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.
(b) Let
Z = min (X , Y ) , R = max (X , Y ) – min (X , Y ).
(9-67)
In this case both Z and R take nonnegative integer values 0, 1, 2, .
Proceeding as in (9-62)-(9-63) we get
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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 ,

mn
m
pq
pq
,

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.
(9-68)
Eq. (9-68) represents the joint probability mass function of Z and R
in (9-67). From (9-68),


n 0
n 1

P( Z  m)   P{Z  m, R  n}  p 2 q 2 m (1  2 q n )  p 2 q 2 m 1  p
 p(1  q) q 2 m ,
m  0, 1, 2,
(9-69)
2q

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and
p

,
n0

1

q

P( R  n )   P{Z  m, R  n}  
2p n
m 0
1 q q , n  1, 2,.
(9-70)
From (9-68)-(9-70), we get
P( Z  m, R  n)  P( Z  m) P( R  n)
(9-71)
which proves the independence of the random variables Z and R
defined in (9-67) as well.
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