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