344
Problem Hints and Solutions
Solution for Problem 2.10. To calculate E (Mn+1 |Fn ), rst note that
Mn+1 is equal to (An +1)/(An + Bn +1) with probability Mn = An /(An + Bn )
and Mn+1 equals An /(An + Bn + 1) with probability 1 Mn . Thus,
An + 1
An
An
Bn
Appendix I: Problem Hints and Solutions
Chapter 1
Solution for Problem 1.1. Let Ti,j denote the expected time to go from
level i to level j , and note by formula (1.16) that T25,20 = 15 and T21,20 = 3.
By rst-step analysis we also have
T20,19 =
1
9
1+
cf
362
Problem Hints and Solutions
sup |gn (, t) g (, t)|
0tT
sup
s,t:|st|1/n
|g (, s) g (, t)| n ( ).
By the uniform continuity of t g (, t) on [0, T ], one has for each that
n ( ) 0 as n . Two applications of the DCT then show that gn g
in L2 (dP dt).
Cha
356
Problem Hints and Solutions
so Mt converges with probability one. If M denotes the value of this limit,
then by Fatous lemma and the bound E (|Mt |) B we have E (|M |) B.
Incidentally, a sequence such as cfw_n is relate to the notion of a a localizin