Homework # 5
10.6. (a).
X
2
o
1
1
P min{X, Y } > i = P {X > i}P {Y > i} =
= i
k
2
4
n
k=i+1
Thus
o
1
P min{X, Y } ≤ i = 1 − i
4
n
(b).
P {X = Y } =
∞
X
P {X = i, Y = i} =
i=1
∞
X
P {X = i}P {Y = i} =
i=1
∞
X
1
1
1
1
=
=
i
−1
4
41−4
3
i=1
(c). Notice that
P {X > Y } + P {Y > X} + P {X = Y } = 1 andP {X > Y } = P {Y > X}
Thus
P {Y > X} =
o 1
1
1 1
1 − P {X = Y } =
1−
=
2
2
3
3
(d).
∞
∞
∞ X
∞
n
[
X
X
1 1
P X devides Y } =
P
{X = i, Y = ij} =
2i 2ij
i=1
j=1
i=1 j=1
=
∞
X
i=1
∞
1
2i(1+i)
X 1
1
1
=
2
1 − 2−i
2i 2i − 1
i=1
(e)
P {X ≥ kY } =
∞
X
i=1
=
2
21+k
∞
∞ X
∞
X
X
1 1
1 1
=2
P {X ≥ ki, Y = i} =
j
i
2 2
2i 2ki
1
1 − 2−(1+k)
i=1
i=1 j=ki
=
2
21+k
10.12.
P {An i.o.} = P
−1
∞ [
∞
\
n=1 k=n
Notice that
P
∞
[
k=n
Ak = lim P
n→∞
Ak ≥ P (An )
1
∞
[
k=n
Ak
Taking limsup on the both side (notice the limit exists on the left hand side by monotonicity),
∞
[
lim P
Ak ≥ lim sup P (An )
n→∞
n→∞
k=n
10.13. Assume complete convergence. By the first part of Borel-Cantelli lemma (the
part does not need independence).
∞ [
∞
\
o
P |Xn − X| > ǫ i.o. = P
{|Xk − X| > ǫ} = 0
n=1 k=n
Or, equivalently,
P |Xn − X| ≤ ǫ eventually = 1
Thus,
lim sup |Xn − X| ≤ ǫ
a.s.
n→∞
Notice that ǫ > 0 can be arbitrarily small, letting ǫ → 0+ on the right leads to
lim Xn = X
n→∞
a.s.
(∗)
On the other hand, assume (*) holds. By 0-1 law, X is equal to a constant almost
surely. Therefore, the sequence {Xn − X} is independent.
For any ǫ > 0,
P
∞ [
∞
\
n=1 k=n
o
n
o
{|Xk − X| > ǫ} = P |Xn − X| > ǫ i.o. ≤ P lim Xn 6= X = 0
n→∞
By the second part of Borel-Cantelli lemma (Here the independence is needed), we must
have
∞
X
P {|Xn − X| > ǫ} < ∞
n=1
10. 16* First, A1 , · · · , Ak , · · · are independent. By Borel-Cantelli lemma. All we need
to show is that

∞
if p ≥ 1/2
= ∞
X
P (Ak )

k=1
<∞
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
X−k
P {Tk = j}
j=2k
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
∞
X
(2p)k < ∞
k=1
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 of
itself. 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