Math523-5
-
Upload
rene-barrera -
Category
Documents
-
view
219 -
download
0
description
Transcript of Math523-5
Homework # 5
10.6. (a).
P{
min{X, Y } > i}
= P{X > i}P{Y > i} =
(
∑
k=i+1
1
2k
)2
=1
4i
Thus
P{
min{X, Y } ≤ i}
= 1 −1
4i
(b).
P{X = Y } =
∞∑
i=1
P{X = i, Y = i} =
∞∑
i=1
P{X = i}P{Y = i} =
∞∑
i=1
1
4i=
1
4
1
1 − 4−1=
1
3
(c). Notice that
P{X > Y } + P{Y > X} + P{X = Y } = 1 andP{X > Y } = P{Y > X}
Thus
P{Y > X} =1
2
(
1 − P{X = Y }}
=1
2
(
1 −1
3
)
=1
3
(d).
P{
X devides Y } =
∞∑
i=1
P(
∞⋃
j=1
{X = i, Y = ij})
=
∞∑
i=1
∞∑
j=1
1
2i
1
2ij
=∞∑
i=1
1
2i(1+i)
1
1 − 2−i=
∞∑
i=1
1
2i2
1
2i − 1
(e)
P{X ≥ kY } =∞∑
i=1
P{X ≥ ki, Y = i} =∞∑
i=1
∞∑
j=ki
1
2j
1
2i= 2
∞∑
i=1
1
2i
1
2ki
=2
21+k
1
1 − 2−(1+k)=
2
21+k − 1
10.12.
P{An i.o.} = P(
∞⋂
n=1
∞⋃
k=n
Ak
)
= limn→∞
P(
∞⋃
k=n
Ak
)
Notice that
P(
∞⋃
k=n
Ak
)
≥ P (An)
1
Taking limsup on the both side (notice the limit exists on the left hand side by monotonic-ity),
limn→∞
P(
∞⋃
k=n
Ak
)
≥ lim supn→∞
P (An)
10.13. Assume complete convergence. By the first part of Borel-Cantelli lemma (thepart does not need independence).
P{
|Xn − X | > ǫ i.o.}
= P(
∞⋂
n=1
∞⋃
k=n
{|Xk − X | > ǫ}}
= 0
Or, equivalently,P
{
|Xn − X | ≤ ǫ eventually}
= 1
Thus,lim sup
n→∞
|Xn − X | ≤ ǫ a.s.
Notice that ǫ > 0 can be arbitrarily small, letting ǫ → 0+ on the right leads to
limn→∞
Xn = X a.s. (∗)
On the other hand, assume (*) holds. By 0-1 law, X is equal to a constant almostsurely. Therefore, the sequence {Xn − X} is independent.
For any ǫ > 0,
P(
∞⋂
n=1
∞⋃
k=n
{|Xk − X | > ǫ}}
= P{
|Xn − X | > ǫ i.o.}
≤ P{
limn→∞
Xn 6= X}
= 0
By the second part of Borel-Cantelli lemma (Here the independence is needed), we musthave
∞∑
n=1
P{|Xn − X | > ǫ} < ∞
10. 16* First, A1, · · · , Ak, · · · are independent. By Borel-Cantelli lemma. All we needto show is that
∞∑
k=1
P (Ak)
= ∞ if p ≥ 1/2
< ∞ if p < 1/2
Fix k and write
Tk =The starting time of the first consecutive head-run during
[2k, 2k+1 − 1] that last at least k rounds
2
Then
P (Ak) = P{2k ≤ Tk ≤ 2k+1 − k} =
2k+1−k
∑
j=2k
P{Tk = j}
For j > 2k, P{Tk = j} = (1 − p)pk. For j = 2k, P{Tk = j} = pk. Thus
P (Ak) = pk + (1 − p)(2k − 2)pk ∼ (2p)k (k → ∞)
Therefore, the conclusion follows from the fact that
∞∑
k=1
(2p)k < ∞
if and only if p < 1/2.
10.18. By the relation X = a − Y a.s. and by Theorem 10.1-(c), X is independent ofitself. Consequently, for any number x,
P{X ≤ x} = P{X ≤ x, X ≤ x} = P{X ≤ x}2
Therefore, P{X ≤ x} = 0 or 1. Notice that the distribution function F (x) = P{X ≤ x}is non-decreasing. There is a C such that F (x) = 0 as x < C and F (x) = 1 as x > C.Hence, X = C a.s. Thus, Y = a − C a.s.
3