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One function of two Random variable a.ppt
1. 1
8. One Function of Two Random
Variables
Given two random variables X and Y and a function g(x,y),
we form a new random variable Z as
Given the joint p.d.f how does one obtain
the p.d.f of Z ? Problems of this type are of interest from a
practical standpoint. For example, a receiver output signal
usually consists of the desired signal buried in noise, and
the above formulation in that case reduces to Z = X + Y.
).
,
( Y
X
g
Z
),
,
( y
x
fXY ),
(z
fZ
(8-1)
PILLAI
2. 2
It is important to know the statistics of the incoming signal
for proper receiver design. In this context, we shall analyze
problems of the following type:
Referring back to (8-1), to start with
)
,
( Y
X
g
Z
Y
X
)
/
(
tan 1
Y
X
Y
X
XY
Y
X /
)
,
max( Y
X
)
,
min( Y
X
2
2
Y
X
z
D
y
x
XY
z
Z
dxdy
y
x
f
D
Y
X
P
z
Y
X
g
P
z
Z
P
z
F
,
,
)
,
(
)
,
(
)
,
(
)
(
)
(
(8-2)
(8-3)
PILLAI
3. 3
where in the XY plane represents the region such
that is satisfied. Note that need not be simply
connected (Fig. 8.1). From (8-3), to determine it is
enough to find the region for every z, and then evaluate
the integral there.
We shall illustrate this method through various examples.
z
D
z
y
x
g
)
,
(
)
(z
FZ
z
D
z
D
X
Y
z
D
z
D
Fig. 8.1
PILLAI
4. 4
Example 8.1: Z = X + Y. Find
Solution:
since the region of the xy plane where is the
shaded area in Fig. 8.2 to the left of the line
Integrating over the horizontal strip along the x-axis first
(inner integral) followed by sliding that strip along the y-axis
from to (outer integral) we cover the entire shaded
area.
,
)
,
(
)
(
y
y
z
x
XY
Z dxdy
y
x
f
z
Y
X
P
z
F (8-4)
z
D z
y
x
.
z
y
x
y
z
x
x
y
Fig. 8.2
).
(z
fZ
PILLAI
5. 5
We can find by differentiating directly. In this
context, it is useful to recall the differentiation rule in (7-
15) - (7-16) due to Leibnitz. Suppose
Then
Using (8-6) in (8-4) we get
Alternatively, the integration in (8-4) can be carried out first
along the y-axis followed by the x-axis as in Fig. 8.3.
)
(z
FZ
)
(z
fZ
)
(
)
(
.
)
,
(
)
(
z
b
z
a
dx
z
x
h
z
H (8-5)
)
(
)
(
.
)
,
(
),
(
)
(
),
(
)
(
)
( z
b
z
a
dx
z
z
x
h
z
z
a
h
dz
z
da
z
z
b
h
dz
z
db
dz
z
dH
(8-6)
( , )
( ) ( , ) ( , ) 0
( , ) .
z y z y
XY
Z XY XY
XY
f x y
f z f x y dx dy f z y y dy
z z
f z y y dy
(8-7)
PILLAI
6. 6
In that case
and differentiation of (8-8)
gives
,
)
,
(
)
(
x
x
z
y
XY
Z dxdy
y
x
f
z
F (8-8)
.
)
,
(
)
,
(
)
(
)
(
x
XY
x
x
z
y
XY
Z
Z
dx
x
z
x
f
dx
dy
y
x
f
z
dz
z
dF
z
f
(8-9)
If X and Y are independent, then
and inserting (8-10) into (8-8) and (8-9), we get
)
(
)
(
)
,
( y
f
x
f
y
x
f Y
X
XY
.
)
(
)
(
)
(
)
(
)
(
x
Y
X
y
Y
X
Z dx
x
z
f
x
f
dy
y
f
y
z
f
z
f
(8-10)
(8-11)
x
z
y
x
y
Fig. 8.3
PILLAI
7. 7
The above integral is the standard convolution of the
functions and expressed two different ways. We
thus reach the following conclusion: If two r.vs are
independent, then the density of their sum equals the
convolution of their density functions.
As a special case, suppose that for and
for then we can make use of Fig. 8.4 to determine the
new limits for
)
(z
fX )
(z
fY
0
)
(
x
fX 0
x 0
)
(
y
fY
,
0
y
.
z
D
Fig. 8.4
y
z
x
x
y
)
0
,
(z
)
,
0
( z
PILLAI
8. 8
In that case
or
On the other hand, by considering vertical strips first in
Fig. 8.4, we get
or
if X and Y are independent random variables.
z
y
y
z
x
XY
Z dxdy
y
x
f
z
F
0 0
)
,
(
)
(
.
0
,
0
,
0
,
)
,
(
)
,
(
)
( 0
0 0
z
z
dy
y
y
z
f
dy
dx
y
x
f
z
z
f
z
XY
z
y
y
z
x
XY
Z (8-12)
,
0
,
0
,
0
,
)
(
)
(
)
,
(
)
( 0
0
z
z
dx
x
z
f
x
f
dx
x
z
x
f
z
f
z
y
Y
X
z
x
XY
Z
z
x
x
z
y
XY
Z dydx
y
x
f
z
F
0 0
)
,
(
)
(
(8-13)
PILLAI
9. 9
Example 8.2: Suppose X and Y are independent exponential
r.vs with common parameter , and let Z = X + Y.
Determine
Solution: We have
and we can make use of (13) to obtain the p.d.f of Z = X + Y.
As the next example shows, care should be taken in using
the convolution formula for r.vs with finite range.
Example 8.3: X and Y are independent uniform r.vs in the
common interval (0,1). Determine where Z = X + Y.
Solution: Clearly, here, and as Fig. 8.5
shows there are two cases of z for which the shaded areas are
quite different in shape and they should be considered
separately.
),
(
)
(
),
(
)
( y
U
e
y
f
x
U
e
x
f y
Y
x
X
(8-14)
2
0
z
Y
X
Z
),
(z
fZ
).
(
)
( 2
0
2
0
)
(
2
z
U
e
z
dx
e
dx
e
e
z
f z
z
z
z
x
z
x
Z
(8-15)
).
(z
fZ
PILLAI
10. 10
x
y
y
z
x
1
0
)
(
z
a
x
y
y
z
x
2
1
)
(
z
b
Fig. 8.5
For
For notice that it is easy to deal with the unshaded
region. In that case
,
1
0
z
,
2
1
z
.
1
0
,
2
)
(
1
)
(
2
0
0 0
z
z
dy
y
z
dxdy
z
F
z
y
z
y
y
z
x
Z
(8-16)
.
2
1
,
2
)
2
(
1
)
1
(
1
1
1
1
)
(
2
1
1
z
1
1
1
z
z
dy
y
z
dxdy
z
Z
P
z
F
y
z
y y
z
x
Z
(8-17)
PILLAI
11. 11
Using (8-16) - (8-17), we obtain
By direct convolution of and we obtain the
same result as above. In fact, for (Fig. 8.6(a))
and for (Fig. 8.6(b))
Fig 8.6 (c) shows which agrees with the convolution
of two rectangular waveforms as well.
.
2
1
,
2
,
1
0
)
(
)
(
z
z
z
z
dz
z
dF
z
f Z
Z (8-18)
)
(x
fX ),
( y
fY
1
0
z
2
1
z
.
1
)
(
)
(
)
(
0
z
dx
dx
x
f
x
z
f
z
f
z
Y
X
Z
.
2
1
)
(
1
1
z
dx
z
f
z
Z
(8-19)
(8-20)
)
(z
fZ
PILLAI
12. 12
)
(x
fY
x
1
)
( x
z
fX
x
z
)
(
)
( x
f
x
z
f Y
X
x
z
1
z
1
0
)
(
z
a
)
(x
fY
x
1
)
( x
z
fX
x
)
(
)
( x
f
x
z
f Y
X
x
1
1
z z
1
z
2
1
)
(
z
b
Fig. 8.6 (c)
)
(z
fZ
z
2
0 1
PILLAI
13. 13
Example 8.3: Let Determine its p.d.f
Solution: From (8-3) and Fig. 8.7
and hence
If X and Y are independent, then the above formula reduces
to
which represents the convolution of with
.
Y
X
Z
)
,
(
)
(
y
y
z
x
XY
Z dxdy
y
x
f
z
Y
X
P
z
F
( )
( ) ( , ) ( , ) .
z y
Z
Z XY XY
y x
dF z
f z f x y dx dy f y z y dy
dz z
(8-21)
( ) ( ) ( ) ( ) ( ),
Z X Y X Y
f z f z y f y dy f z f y
(8-22)
)
( z
fX ).
(z
fY
Fig. 8.7
y
x
z
y
x
z
y
x
y
).
(z
fZ
PILLAI
14. 14
As a special case, suppose
In this case, Z can be negative as well as positive, and that
gives rise to two situations that should be analyzed
separately, since the region of integration for and
are quite different. For from Fig. 8.8 (a)
and for from Fig 8.8 (b)
After differentiation, this gives
0 0
)
,
(
)
(
y
y
z
x
XY
Z dxdy
y
x
f
z
F
0
)
,
(
)
(
z
y
y
z
x
XY
Z dxdy
y
x
f
z
F
.
0
,
0
)
(
and
,
0
,
0
)
(
y
y
f
x
x
f Y
X
0
z 0
z
,
0
z
,
0
z
.
0
,
)
,
(
,
0
,
)
,
(
)
( 0
z
dy
y
y
z
f
z
dy
y
y
z
f
z
f
z
XY
XY
Z (8-23) Fig. 8.8 (b)
y
x
y
z
x
z
y
x
y
z
x
z
z
(a)
PILLAI
15. 15
Example 8.4: Given Z = X / Y, obtain its density function.
Solution: We have
The inequality can be rewritten as if
and if Hence the event in (8-24) need
to be conditioned by the event and its compliment
Since by the partition theorem, we have
and hence by the mutually exclusive property of the later
two events
Fig. 8.9(a) shows the area corresponding to the first term,
and Fig. 8.9(b) shows that corresponding to the second term
in (8-25).
.
/
)
( z
Y
X
P
z
FZ
(8-24)
z
Y
X
/ Yz
X ,
0
Y
Yz
X .
0
Y
z
Y
X
/
0
Y
A .
__
A
__
,
A A
.
0
,
0
,
0
,
/
0
,
/
/
Y
Yz
X
P
Y
Yz
X
P
Y
z
Y
X
P
Y
z
Y
X
P
z
Y
X
P
(8-25)
Fig. 8.9
y
x
yz
x
(a)
y
x
yz
x
(b)
A
z
Y
X
A
z
Y
X
A
A
z
Y
X
z
Y
X
)
/
(
)
/
(
)
(
)
/
(
/
PILLAI
16. 16
Integrating over these two regions, we get
Differentiation with respect to z gives
Note that if X and Y are nonnegative random variables, then
the area of integration reduces to that shown in Fig. 8.10.
.
)
,
(
)
,
(
)
(
0
0
y yz
x
XY
y
yz
x
XY
Z dxdy
y
x
f
dxdy
y
x
f
z
F (8-26)
.
,
)
,
(
|
|
)
,
(
)
(
)
,
(
)
(
0
0
z
dy
y
yz
f
y
dy
y
yz
f
y
dy
y
yz
yf
z
f
XY
XY
XY
Z
(8-27)
y
x
yz
x
Fig. 8.10
PILLAI
17. 17
This gives
or
Example 8.5: X and Y are jointly normal random variables
with zero mean so that
Show that the ratio Z = X / Y has a Cauchy density function
centered at
Solution: Inserting (8-29) into (8-27) and using the fact
that we obtain
0 0
)
,
(
)
(
y
yz
x
XY
Z dxdy
y
x
f
z
F
otherwise.
,
0
,
0
,
)
,
(
)
( 0
z
dy
y
yz
f
y
z
f XY
Z (8-28)
.
1
2
1
)
,
(
2
2
2
2
1
2
1
2
2
2
)
1
(
2
1
2
2
1
y
rxy
x
r
XY e
r
y
x
f (8-29)
.
/ 2
1
r
,
1
)
(
1
2
2
)
(
2
2
1
2
0
0
2
/
2
2
1
2
0
2
r
z
dy
ye
r
z
f y
Z
),
,
(
)
,
( y
x
f
y
x
f XY
XY
PILLAI
18. 18
where
Thus
which represents a Cauchy r.v centered at Integrating
(8-30) from to z, we obtain the corresponding
distribution function to be
Example 8.6: Obtain
Solution: We have
.
1
2
1
)
(
2
2
2
1
2
1
2
2
2
0
rz
z
r
z
,
)
1
(
)
/
(
/
1
)
( 2
2
1
2
2
1
2
2
2
2
1
r
r
z
r
z
fZ
(8-30)
.
/ 2
1
r
.
1
arctan
1
2
1
)
(
2
1
1
2
r
r
z
z
FZ
(8-31)
.
)
,
(
)
( 2
2
2
2
z
Y
X
XY
Z dxdy
y
x
f
z
Y
X
P
z
F
.
2
2
Y
X
Z
(8-32)
).
(z
fZ
PILLAI
19. 19
But, represents the area of a circle with radius
and hence from Fig. 8.11,
This gives after repeated differentiation
As an illustration, consider the next example.
z
Y
X
2
2
,
z
.
)
,
(
)
(
2
2
z
z
y
y
z
y
z
x
XY
Z dxdy
y
x
f
z
F (8-33)
.
)
,
(
)
,
(
2
1
)
( 2
2
2
z
z
y
XY
XY
Z dy
y
y
z
f
y
y
z
f
y
z
z
f (8-34)
Fig. 8.11
x
y
z
z
Y
X
2
2
z
z
PILLAI
20. 20
Example 8.7 : X and Y are independent normal r.vs with zero
Mean and common variance Determine for
Solution: Direct substitution of (8-29) with
Into (8-34) gives
where we have used the substitution From (8-35)
we have the following result: If X and Y are independent zero
mean Gaussian r.vs with common variance then
is an exponential r.vs with parameter
Example 8.8 : Let Find
Solution: From Fig. 8.11, the present case corresponds to a
circle with radius Thus
)
(z
fZ .
2
2
Y
X
Z
.
2
2
1
,
0
r
2
2 2 2
2
2
/ 2
( ) / 2
2 2
2 2
0
/ 2
/2
/ 2
2 2
0
1 1 1
( ) 2
2
2
cos 1
( ),
2
cos
z
z z
z y y
Z
y z
z
z
e
f z e dy dy
z y z y
e z
d e U z
z
(8-35)
.
sin
z
y
,
2
2
2
Y
X .
2 2
.
2
2
Y
X
Z
).
(z
fZ
.
2
z PILLAI
21. 21
.
)
,
(
)
,
(
)
( 2
2
2
2
2
2
z
z
XY
XY
Z dy
y
y
z
f
y
y
z
f
y
z
z
z
f
.
)
,
(
)
(
2
2
2
2
z
z
y
y
z
y
z
x
XY
Z dxdy
y
x
f
z
F
),
(
cos
cos
2
1
2
2
1
2
)
(
2
2
2
2
2
2
2
2
2
2
2
/
2
/2
0
2
/
2
0 2
2
2
/
2
0
2
/
)
(
2
2
2
z
U
e
z
d
z
z
e
z
dy
y
z
e
z
dy
e
y
z
z
z
f
z
z
z
z
z
y
y
z
Z
And by repeated differentiation, we obtain
Now suppose X and Y are independent Gaussian as in
Example 8.7. In that case, (8-36) simplifies to
which represents a Rayleigh distribution. Thus, if
where X and Y are real, independent normal r.vs with zero
mean and equal variance, then the r.v has a
Rayleigh density. W is said to be a complex Gaussian r.v
with zero mean, whose real and imaginary parts are
independent r.vs. From (8-37), we have seen that its
magnitude has Rayleigh distribution.
(8-36)
(8-37)
,
iY
X
W
2
2
Y
X
W
PILLAI
22. 22
What about its phase
Clearly, the principal value of lies in the interval
If we let then from example 8.5, U has a
Cauchy distribution with (see (8-30) with )
As a result
To summarize, the magnitude and phase of a zero mean
complex Gaussian r.v has Rayleigh and uniform distributions
respectively. Interestingly, as we will show later, these two
derived r.vs are also independent of each other!
?
tan 1
Y
X
(8-38)
.
,
1
/
1
)
( 2
u
u
u
fU
(8-39)
).
2
/
,
2
/
(
tan / ,
U X Y
0
,
2
1
r
.
otherwise
,
0
,
2
/
2
/
,
/
1
1
tan
/
1
)
sec
/
1
(
1
)
(tan
|
/
|
1
)
( 2
2
U
f
du
d
f (8-40)
PILLAI
23. 23
Let us reconsider example 8.8 where X and Y have nonzero
means and respectively. Then is said to
be a Rician r.v. Such a scene arises in fading multipath
situation where there is a dominant constant component
(mean) in addition to a zero mean Gaussian r.v. The constant
component may be the line of sight signal and the zero mean
Gaussian r.v part could be due to random multipath
components adding up incoherently (see diagram below).
The envelope of such a signal is said to have a Rician p.d.f.
Example 8.9: Redo example 8.8, where X and Y have
nonzero means and respectively.
Solution: Since
substituting this into (8-36) and letting
X
Y
2
2
Y
X
Z
X
Y
,
2
1
)
,
(
2
2
2
2
/
]
)
(
)
[(
2
Y
X y
x
XY e
y
x
f
Rician
Output
Line of sight
signal (constant)
a
Multipath/Gaussian
noise
PILLAI
24. 24
we get the Rician probability density function to be
where
is the modified Bessel function of the first kind and zeroth
order.
Example 8.10: Determine
Solution: The functions max and min are nonlinear
2 2
cos , sin , , cos , sin ,
X Y X Y
x z y z
,
2
2
2
)
(
2
0
2
2
/
)
(
/2
3
/2
/
)
cos(
/2
/2
/
)
cos(
2
2
/
)
(
/2
/2
/
)
cos(
/
)
cos(
2
2
/
)
(
2
2
2
2
2
2
2
2
2
2
2
2
2
z
I
ze
d
e
d
e
ze
d
e
e
ze
z
f
z
z
z
z
z
z
z
Z
0
cos
2
0
)
cos(
0
1
2
1
)
( d
e
d
e
I
(8-41)
).
,
min(
),
,
max( Y
X
W
Y
X
Z
).
(z
fZ
(8-42)
PILLAI
25. 25
operators and represent special cases of the more general
order statistics. In general, given any n-tuple
we can arrange them in an increasing order of magnitude
such that
where and is the second smallest
value among and finally
If represent r.vs, the function that takes on
the value in each possible sequence is
known as the k-th order statistic. represent
the set of order statistics among n random variables. In this
context
represents the range, and when n = 2, we have the max and
min statistics.
,
,
,
, 2
1 n
X
X
X
,
)
(
)
2
(
)
1
( n
X
X
X
,
,
,
,
min 2
1
)
1
( n
X
X
X
X
)
2
(
X
,
,
,
, 2
1 n
X
X
X .
,
,
,
max 2
1
)
( n
n X
X
X
X
n
X
X
X ,
,
, 2
1 )
(k
X
)
(k
x
n
x
x
x ,
,
, 2
1
)
1
(
)
( X
X
R n
(8-44)
(8-43)
PILLAI
(1) (2) ( )
( , , , )
n
X X X
26. 26
Returning back to that problem, since
we have (see also (8-25))
since and are mutually exclusive sets that
form a partition. Figs 8.12 (a)-(b) show the regions
satisfying the corresponding inequalities in each term
above.
,
,
,
,
)
,
max(
Y
X
Y
Y
X
X
Y
X
Z (8-45)
,
,
,
,
,
)
,
max(
)
(
Y
X
z
Y
P
Y
X
z
X
P
Y
X
z
Y
Y
X
z
X
P
z
Y
X
P
z
FZ
)
( Y
X )
( Y
X
x
y
z
x y
x
z
X
Y
X
)
,
(
)
( Y
X
z
X
P
a
Fig. 8.12
x
y
z
Y
Y
X
y
x
z
y
)
,
(
)
( Y
X
z
Y
P
b
x
y
)
,
( z
z
)
(c
PILLAI
27. 27
(8-46)
Fig. 8.12 (c) represents the total region, and from there
If X and Y are independent, then
and hence
Similarly
Thus
).
,
(
,
)
( z
z
F
z
Y
z
X
P
z
F XY
Z
)
(
)
(
)
( y
F
x
F
z
F Y
X
Z
).
(
)
(
)
(
)
(
)
( z
F
z
f
z
f
z
F
z
f Y
X
Y
X
Z
(8-47)
.
,
,
,
)
,
min(
Y
X
X
Y
X
Y
Y
X
W (8-48)
.
,
,
)
,
min(
)
( Y
X
w
X
Y
X
w
Y
P
w
Y
X
P
w
FW
PILLAI
28. 28
Once again, the shaded areas in Fig. 8.13 (a)-(b) show the
regions satisfying the above inequalities and Fig 8.13 (c)
shows the overall region.
From Fig. 8.13 (c),
where we have made use of (7-5) and (7-12) with
and
,
)
,
(
)
(
)
(
,
1
1
)
(
w
w
F
w
F
w
F
w
Y
w
X
P
w
W
P
w
F
XY
Y
X
W
(8-49)
,
2
2
y
x
.
1
1 w
y
x
x
y
y
x
w
y
(a)
Fig. 8.13
x
y
y
x
w
x
(b)
x
y
)
,
( w
w
(c)
PILLAI
29. 29
Example 8.11: Let X and Y be independent exponential r.vs
with common parameter . Define Find
Solution: From (8-49)
and hence
But and so that
Thus min ( X, Y ) is also exponential with parameter 2.
Example 8.12: Suppose X and Y are as give in the above
example. Define Determine
).
,
min( Y
X
W ?
)
(w
fW
)
(
)
(
)
(
)
(
)
( w
F
w
F
w
F
w
F
w
F Y
X
Y
X
W
).
(
)
(
)
(
)
(
)
(
)
(
)
( w
f
w
F
w
F
w
f
w
f
w
f
w
f Y
X
Y
X
Y
X
W
,
)
(
)
( w
Y
X e
w
f
w
f
,
1
)
(
)
( w
Y
X e
w
F
w
F
).
(
2
)
1
(
2
2
)
( 2
w
U
e
e
e
e
w
f w
w
w
w
W
(8-50)
.
)
,
max(
/
)
,
min( Y
X
Y
X
Z ).
(z
fZ
PILLAI
30. 30
Solution: Although represents a complicated
function, by partitioning the whole space as before, it is
possible to simplify this function. In fact
As before, this gives
Since X and Y are both positive random variables in this case,
we have The shaded regions in Figs 8.14 (a)-(b)
represent the two terms in the above sum.
Fig. 8.14
)
max(
/
)
min(
.
,
/
,
,
/
Y
X
X
Y
Y
X
Y
X
Z (8-51)
( ) / , / ,
, , .
Z
F 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
.
1
0
z
x
y
y
x
yz
x
(a)
x
y
y
x
xz
y
(b)
(8-52)
PILLAI
31. 31
From Fig. 8.14
Hence
Example 8.13 (Discrete Case): Let X and Y be independent
Poisson random variables with parameters and
respectively. Let Determine the p.m.f of Z.
.
)
,
(
)
,
(
)
(
0
z
0
0 0
x
y
XY
yz
x
XY
Z dydx
y
x
f
dxdy
y
x
f
z
F
.
otherwise
,
0
,
1
0
,
)
1
(
2
)
1
(
2
2
)
,
(
)
,
(
)
,
(
)
,
(
)
(
2
0
2
0
)
1
(
2
0
)
(
)
(
2
0
0
0
z
z
dy
ue
z
dy
ye
dy
e
e
y
dy
yz
y
f
y
yz
f
y
dx
xz
x
f
x
dy
y
yz
f
y
z
f
u
y
z
yz
y
y
yz
XY
XY
XY
XY
Z
(8-54)
1
2
.
Y
X
Z
(8-53)
)
(z
fZ
z
Fig. 8.15
1
2
PILLAI
32. 32
Solution: Since X and Y both take integer values
the same is true for Z. For any gives
only a finite number of options for X and Y. In fact, if X = 0,
then Y must be n; if X = 1, then Y must be n-1, etc. Thus the
event is the union of (n + 1) mutually exclusive
events given by
As a result
If X and Y are also independent, then
and hence
,
,
2
,
1
,
0
n
Y
X
n
,
,
2
,
1
,
0
.
)
,
(
,
)
(
)
(
0
0
n
k
n
k
k
n
Y
k
X
P
k
n
Y
k
X
P
n
Y
X
P
n
Z
P
(8-55)
, ,
k
A X k Y n k
)
(
)
(
, k
n
Y
P
k
X
P
k
n
Y
k
X
P
(8-56)
}
{ n
Y
X
k
A
.
,
,
2
,
1
,
0 n
k
PILLAI
33. 33
Thus Z represents a Poisson random variable with
parameter indicating that sum of independent Poisson
random variables is also a Poisson random variable whose
parameter is the sum of the parameters of the original
random variables.
As the last example illustrates, the above procedure for
determining the p.m.f of functions of discrete random
variables is somewhat tedious. As we shall see in Lecture 10,
the joint characteristic function can be used in this context
to solve problems of this type in an easier fashion.
,
2
1
.
,
,
2
,
1
,
0
,
!
)
(
)!
(
!
!
!
)!
(
!
)
,
(
)
(
2
1
)
(
0
2
1
)
(
2
0
1
0
2
1
2
1
2
1
n
n
e
k
n
k
n
n
e
k
n
e
k
e
k
n
Y
k
X
P
n
Z
P
n
n
k
k
n
k
k
n
n
k
k
n
k
(8-57)
PILLAI
