9. Two Functions of Two Random Variables In the spirit of the previous lecture, let us look at an immediate generalization: Suppose X.
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Transcript 9. Two Functions of Two Random Variables In the spirit of the previous lecture, let us look at an immediate generalization: Suppose X.
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 )
(9-1)
W h( X , Y ).
(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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The procedure is the same as that in (8-3). In fact for given z
and w,
FZW ( z, w) PZ ( ) z,W ( ) w Pg ( 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
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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) PZ z,W w Pmin(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)
yw
X
X
( b) w z
(a ) w z
Fig. 9.2
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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 ,
ot herwise.
(9-9)
(9-10)
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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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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
Pz Z z z, w W w w
Pz g ( X , Y ) z z, w h( X , Y ) w w .
(9-15)
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Using (7-9) we can rewrite (9-15) as
Pz Z z z, w W w w f ZW ( z, w)zw.
(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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Finally D goes to D , and ABC 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
.
zw
(9-18)
To simplify (9-18), we need to evaluate the area i of the
parallelograms in Fig. 9.3 (b) in terms of zw. Towards this,
h1
g1
let and
denote
the inverse transformation in (9-14), so
that
xi g1 ( z, w),
yi h1 ( z, w).
(9-19)
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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 ABC D in Fig. 9.4 (b) is
given by
i AB AC sin( )
(9-23)
AB cos AC sin AB sin AC cos .
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But from Fig. 9.4 (b), and (9-20) - (9-22)
g1
h
z, AC sin 1 w,
z
w
h
g
AB sin 1 z, AC cos 1 w.
z
w
AB cos
so that
g1 h1 g1 h1
i
zw
z w w z
(9-24)
(9-25)
(9-26)
and
g1
i
z
g1 h1 g1 h1
det
zw 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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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)
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, y yi
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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 tan1(Y / X ), where | w | / 2.
Obtain f ZW ( z, w).
Solution: Here
f XY ( x, y )
1
2
2
e
( x 2 y 2 ) / 2 2
.
(9-32)
Since
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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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 tanw z sin w.
Thus there are two solution sets
x1 z cosw, y1 z sin w, x2 z cosw, 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
x
w
y
z
y
w
cos w
sin w
| J ( z, w) | z.
z sin w
z,
z cos w
(9-38)
(9-39)
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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
f Z ( 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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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 tan1(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
uv
,
2
y
uv
.
2
(9-47)
Moreover the Jacobian of the transformation is given by 16
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J ( x, y )
1
1
1 1
2
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
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when only one function is specified.
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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
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by y1 w, x1 z w.
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The Jacobian of the transformation is given by
J ( x, y )
1
1
0
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(2Y ).
Y U (0,1)
be independent.
(9-54)
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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( 2w ) 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 ( 2w) e
2
x1
w
z sec( 2w ) 2 / 2
0
z sec2 ( 2w)e z sec( 2w )
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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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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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 )
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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
mn
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 2q 2q 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
1pq q|n| ,
p 2 q 2 m q|n|
m 0
) p 2 q|n|
n 0, 1, 2,
1
1q2
.
(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 ,
mn
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,,
n0
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),
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
p (1 q) q 2 m ,
n 1
m 0, 1, 2,
(9-69)
2q
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and
p
,
n0
1
q
P( R n ) P{Z m, R n}
2p n
m 0
q , n 1, 2,.
1 q
(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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