Solutions of Selected Problems from
Probability
Essentials, Second Edition
Solutions to selected problems of Chapter 2
2.1 Let’s first prove by induction that #(2
Ω
n
) = 2
n
if Ω =
{
x
1
, . . . , x
n
}
.
For
n
= 1
it is clear that #(2
Ω
1
) = #(
{∅
,
{
x
1
}}
) = 2.
Suppose #(2
Ω
n

1
) = 2
n

1
.
Observe that
2
Ω
n
=
{{
x
n
} ∪
A, A
∈
2
Ω
n

1
} ∪
2
Ω
n

1
}
hence #(2
Ω
n
) = 2#(2
Ω
n

1
) = 2
n
.
This proves
finiteness. To show that 2
Ω
is a
σ
algebra we check:
1.
∅ ⊂
Ω hence
∅ ∈
2
Ω
.
2. If
A
∈
2
Ω
then
A
⊂
Ω and
A
c
⊂
Ω hence
A
c
∈
2
Ω
.
3. Let (
A
n
)
n
≥
1
be a sequence of subsets of Ω. Then
∞
n
=1
A
n
is also a subset of Ω hence
in 2
Ω
.
Therefore 2
Ω
is a
σ
algebra.
2.2 We check if
H
=
∩
α
∈
A
G
α
has the three properties of a
σ
algebra:
1.
∅ ∈ G
α
∀
α
∈
A
hence
∅ ∈ ∩
α
∈
A
G
α
.
2. If
B
∈ ∩
α
∈
A
G
α
then
B
∈ G
α
∀
α
∈
A
. This implies that
B
c
∈ G
α
∀
α
∈
A
since each
G
α
is a
σ
algebra. So
B
c
∈ ∩
α
∈
A
G
α
.
3. Let (
A
n
)
n
≥
1
be a sequence in
H
. Since each
A
n
∈ G
α
,
∞
n
=1
A
n
is in
G
α
since
G
α
is a
σ
algebra for each
α
∈
A
. Hence
∞
n
=1
A
n
∈ ∩
α
∈
A
G
α
.
Therefore
H
=
∩
α
∈
A
G
α
is a
σ
algebra.
2.3 a. Let
x
∈
(
∪
∞
n
=1
A
n
)
c
. Then
x
∈
A
c
n
for all
n
, hence
x
∈ ∩
∞
n
=1
A
c
n
. So (
∪
∞
n
=1
A
n
)
c
⊂
∩
∞
n
=1
A
c
n
.
Similarly if
x
∈ ∩
∞
n
=1
A
c
n
then
x
∈
A
c
n
for any
n
hence
x
∈
(
∪
∞
n
=1
A
n
)
c
.
So
(
∪
∞
n
=1
A
n
)
c
=
∩
∞
n
=1
A
c
n
.
b. By parta
∩
∞
n
=1
A
n
= (
∪
∞
n
=1
A
c
n
)
c
, hence (
∩
∞
n
=1
A
n
)
c
=
∪
∞
n
=1
A
c
n
.
2.4 lim inf
n
→∞
A
n
=
∪
∞
n
=1
B
n
where
B
n
=
∩
m
≥
n
A
m
∈ A ∀
n
since
A
is closed under taking
countable intersections.
Therefore lim inf
n
→∞
A
n
∈ A
since
A
is closed under taking
countable unions.
By De Morgan’s Law it is easy to see that lim sup
A
n
= (lim inf
n
→∞
A
c
n
)
c
, hence lim sup
n
→∞
A
n
∈
A
since lim inf
n
→∞
A
c
n
∈ A
and
A
is closed under taking complements.
Note that
x
∈
lim inf
n
→∞
A
n
⇒ ∃
n
*
s.t
x
∈ ∩
m
≥
n
*
A
m
⇒
x
∈ ∩
m
≥
n
A
m
∀
n
⇒
x
∈
lim sup
n
→∞
A
n
. Therefore lim inf
n
→∞
A
n
⊂
lim sup
n
→∞
A
n
.
2.8 Let
L
=
{
B
⊂
R
:
f

1
(
B
)
∈ B}
. It is easy to check that
L
is a
σ
algebra. Since
f
is continuous
f

1
(
B
) is open (hence Borel) if
B
is open. Therefore
L
contains the open
sets which implies
L ⊃ B
since
B
is generated by the open sets of
R
. This proves that
f

1
(
B
)
∈ B
if
B
∈ B
and that
A
=
{
A
⊂
R
:
∃
B
∈ B
with
A
=
f

1
(
B
)
∈ B} ⊂ B
.
1
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Solutions to selected problems of Chapter 3
3.7 a. Since
P
(
B
)
>
0
P
(
.

B
) defines a probability measure on
A
, therefore by Theorem
2.4 lim
n
→∞
P
(
A
n

B
) =
P
(
A

B
).
b. We have that
A
∩
B
n
→
A
∩
B
since
1
A
∩
B
n
(
w
) =
1
A
(
w
)
1
B
n
(
w
)
→
1
A
(
w
)
1
B
(
w
).
Hence
P
(
A
∩
B
n
)
→
P
(
A
∩
B
). Also
P
(
B
n
)
→
P
(
B
). Hence
P
(
A

B
n
) =
P
(
A
∩
B
n
)
P
(
B
n
)
→
P
(
A
∩
B
)
P
(
B
)
=
P
(
A

B
)
.
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 Spring '11
 Probability, lim fn, Dominated convergence theorem, Selected Problems, Monotone convergence theorem, Fnk

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