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1.1
Discrete
Case
Suppose
X
1
, . . . , X
n
are discrete with joint p.d.f.
f
X
1
,...,X
n
(
x
1
, . . . , x
n
),
and
Y
1
, . . . , Y
m
are given by
m
functions
Y
1
=
u
1
(
X
1
, . . . , X
n
)
.
.
.
Y
m
=
u
m
(
X
1
, . . . , X
n
)
If we
let
A
y
:=
{
(
x
1
, . . . , x
n
) :
r
1
(
x
1
, . . . , x
n
) =
y
1
, . . . , u
m
(
x
1
, . . . , x
n
) =
y
m
}
then the joint p.d.f. of
Y
1
, . . . , Y
m
is given by
f
Y
1
,...,Y
m
(
y
1
, . . . , y
m
) =
f
X
1
,...,X
n
(
x
1
, . . . , x
n
)
(
x
1
,...,x
n
)
∈
A
y
Example
1
(Sum
of
Binomial
Random
Variables)
Suppose
X
∼
B
(
m, p
)
and
Y
∼
B
(
n, p
)
are
independent binomial random variables with
p.d.f.s
f
X
(
k
)
=
m
p
k
(1
−
p
)
m
−
k
k
f
Y
(
k
)
=
n
p
k
(1
−
p
)
n
−
k
k
If we
define
Z
=
X
+
Y
,
what is the
p.d.f.
f
Z
(
z
)
?
Since
X
is the
number of successes in
a
sequence
of
m
independent trials,
and
Y
the
number of
successes in
n
trials, both
with
the
same
success probability,
a
first guess would be
that then
Z
should
just be
the
number of successes in
m
+
n
trials with
success
probability
p
, i.e.
Z
∼
B
(
m
+
n, p
)
.
This turns out to
be
true, but we’ll
first have
to
check
this formally:
P
(
Z
=
z
)
=
P
(
{
X
= 0
, Y
=
z
}
or
{
X
= 1
, Y
=
z
−
1
}
. . .
or
{
X
=
z, Y
= 0
}
)
z
z
=
P
(
X
=
k, Y
=
z
−
k
) =
P
(
X
=
k
)
P
(
Y
=
z
−
k
)
k
=0
k
=0
z
�
�
�
�
=
m
p
k
(1
−
p
)
m
−
k
n
p
z
−
k
(1
−
p
)
n
−
z
+
k
k
z
−
k
k
=0
z
�
� �
�
=
�
m
n
p
z
(1
−
p
)
n
−
z
k
z
−
k
k
=0
Now,
the
term
p
z
(1