Homework # 3
23.1. For any G-measurable, non-negative bounded random variable Z, |EY Z|
E|Y |Z. This gives that E(Y )Z| E|Y |Z. So we have
E(Y |G) Z
E
E E |Y |G Z
Or
E
E(|Y |G) E(Y |G) Z
0
For any > 0, take Z = 1cfw_E(|Y |G)E(Y |G) . We have that
E(|Y
Homework # 5
26.2
p
EX = p
t
p1
P cfw_X tdt p
0
0
X
tp2 dt
= pE Y
tp2 E(Y 1cfw_Xt dt
0
=
p
E Y X p1
p1
26.3 By the conclusion from Problem 26.2
EX p qE Y X p1 EY p
1/p
EX p
1/q
where the second step follows from the Hlder inequality.
o
26.4. We may assume
Homework # 2
20.3
n
1
n
1/n
log
=
Yj
j=1
n
Xj EX1
j=1
a.s.
Hence,
n
1/n
lim
Yj
n
20.7.
= eEX1
a.s.
j=1
2
2
X1 + + Xn
X1 + + Xn X1 + + Xn
=
2
2
X1 + + Xn
n
n
1
By the law of the large numbers,
X1 + + Xn
=1
n
n
lim
a.s.
2
2
X1 + + Xn
2
= EX1 = V ar(X1 ) + E
Homework # 4
24.2. For each n,
cfw_S T n = cfw_S n cfw_T n Fn
24.4. For each n,
n
cfw_S + T n =
cfw_S = kcfw_T n k Fn
k=0
24.9. All we need is to show that for any integer m 0, cfw_T m FT . By denition
of FT we need to show for all integer n 0, cfw_T m cf
Homework # 1
18.6. Notice that for any m > 0
m
n
P |Xn | > m = 1
when n > m. Consequently, Xn is not uniformly tight. Therefore, Xn is not pre-compact
in the sense of weak convergence. Thus, Xn is not convergent in distribution. Since, the
convergence in