MATH 587
Solutions to Assignment 4
December 2, 2003
1. (a) Consider a subsequence
f
(
X
n
i
,Y
n
i
,i
≥
1. Since
X
n
i
→
X
in
P
, there is a further subsequence
X
n
i
j
,j
≥
1 such that
X
n
i
j
→
X
a.s. Since
Y
n
i
j
→
Y
in
P
, there is a further subsequence
Y
n
i
j
k
,j
≥
1 such that
Y
n
i
j
k
→
Y
a.s. Since also
X
n
i
j
k
→
X
a.s. and
f
is continuous, then
f
(
X
n
i
j
k
,Y
n
i
j
k
)
→
f
(
X, Y
) a.s. Hence
f
(
X
n
,Y
n
)
→
f
(
X, Y
)in
P
.
(b) Apply part (a) with
f
(
x, y
)=
x
+
y
and
f
(
x, y
)=
xy
.
2. (a) For a measurable rectangle
C
=
A
×
B
,wehave
µ
X,Y
(
A
×
B
)=
P
{
X
∈
A, Y
∈
B
}
=
P
{
X
∈
A
}
P
{
Y
∈
B
}
=
µ
X
(
A
)
µ
Y
(
B
)=
µ
X
⊗
µ
Y
(
A
×
B
). Hence
µ
X,Y
and
µ
X
⊗
µ
Y
coincide on the
semialgebra
S
of measurable rectangles, and therefore on the
σ
algebra
B
(
R
2
).
(b) Let
C
=
{
(
x, y
)

x
+
y
≤
z
}
and note that
C
y
=
{
x

x
≤
z
−
y
}
. Then
F
Z
(
z
)=
RR
I
C
(
x, y
)
dµ
X
⊗
µ
Y
=
R £ R
I
C
y
(
x
)
dµ
X
¤
dµ
Y
=
R
µ
X
(
C
y
)
dµ
Y
=
R
F
X
(
z
−
y
)
dF
Y
(
y
).
3. By Fubini,
Eφ
(
X, Y
)=
R £ R
φ
(
x, y
)
dP
Y
¤
dP
X
=
R
g
(
x
)
dP
X
=
Eg
(
X
). Next, suppose
E

Y

=
∞
.
Since