(c) Assume
0
≤
i
≤
n
. Using the independence between
X
and
Y
, we can write that
P
(
X
+
Y
=
k

X
=
i
)=
P
(
Y
=
k
−
i

X
=
i
)=
P
(
Y
=
k
−
i
)
.
The conditional PMF of
X
+
Y
given
X
is then given by
P
(
X
+
Y
=
k

X
=
i
)=
braceleftbigg (
n
k
−
i
)
1
2
n
,
if
i
≤
k
≤
i
+
n
0
,
otherwise
(d) Assume
0
≤
k
≤
2
n
. The conditional PMF of
X
given
X
+
Y
can be computed using Bayes’ rule,
P
(
X
=
i

X
+
Y
=
k
)=
P
(
X
+
Y
=
k

X
=
i
)
P
(
X
=
i
)
P
(
X
+
Y
=
k
)
.
Using the results in (b) and (c), we conclude that
P
(
X
=
i

X
+
Y
=
k
)=
(
n
i
)(
n
k

i
)
(
2
n
k
)
,
if
max(0
,k
−
n
)
≤
i
≤
min(
n,k
)
0
,
otherwise
Note that using the results in (b), this can be also written under the following form
P
(
X
=
i

X
+
Y
=
k
)=
(
n
i
)(
n
k

i
)
∑
min(
k,n
)
j
=max(0
,k

n
)
(
n
j
)(
n
k

j
)
,
if
max(0
,k
−
n
)
≤
i
≤
min(
n,k
)
0
,
otherwise
(e) Using symmetry,
E
[
X

X
+
Y
=
k
]=
E
[
Y

X
+
Y
=
k
]
. Now using the linearity of conditional expectation,
E
[
X

X
+
Y
=
k
]+
E
[
Y

X
+
Y
=
k
]=
E
[
X
+
Y

X
+
Y
=
k
]=
E
[
k

X
+
Y
=
k
]=
k,
We conclude therefore that
E
[
X

X
+
Y
=
k
] =
k
2
. This is not surprising as the conditional PMF that we found
in part (d) is symmetric about
k
2
so it’s mean should be
k
2
.
(f) Note that for all choices of joint PMFs of
X
and
Y
we must have
P
(
X
+
Y
=2
n
)=
P
(
X
=
n,Y
=
n
)
≤
P
(
X
=
n
)
.
If we take
Y
=
X
, then
P
(
X
+
Y
=2
n
)=
P
(
X
=
n
)
. Therefore, the joint PMF
P
XY
(
l,k
)=
braceleftbigg
P
X
(
l
)=
(
n
l
)
1
2
n
,
if
0
≤
l
=
k
≤
n
0
,
otherwise
maximizes
P
(
X
+
Y
=2
n
)
.
(g) For all the choices of joint PMFs of
X
and
Y
we must have
P
(
X
+
Y
= 2
n
)
≥
0
. Now take
Y
=
n
−
X
,
then
Y
is also Binomial
(
n,
1
2
)
. But since we always have
X
+
Y
=
n
,
P
(
X
+
Y
=2
n
)=0
. Hence, this choice
minimizes
P
(
X
+
Y
=2
n
)
.
4