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
, we have
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
(
2
).
(b) Let
C
=
{
(
x, y
)

x
+
y
≤
z
}
and note that
C
y
=
{
x

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

Y

=
∞
.
Since
E

Y
 ≤
E

x
+
Y

+
x
, then
g
(
x
) =
def
E

x
+
Y

=
∞
for all
x
. But then part (a) implies that
E

X
+
Y

=
Eg
(
X
) =
∞
.
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 '07
 RomanVershynin
 Probability, Trigraph, Xn, subsequence, xn xn, Infinite monkey theorem, Ynij

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