Chapter 4

[not yet completed]

Martingales

4.1 Conditional Expectation
 page 224 At the end of the proof following example 1.5, the final integrand should be g(X) instead of g(Y).
 page 226 For the centered equation in the first proof on this page, you can think of the inequality as saying that the p-norm of E(X|F) <= the p-norm of X.

4.2 Martingales, Almost Sure Convergence
 page 231 Thinking about the definition Durrett gives for a martingale: in part (iii), why not just replace Fn by sigma(X1,..., Xn) ? This would be just as useful (as far as I can tell), and not change the object being defined. In fact, Breiman uses this alternative definition.
 page 234 About half-way down the page, the function Hn is incorrectly defined. It should be 1{N
 page 235 It is intuitively helpful to notice that a + (Xm-a)+ is just a euphemism for max(Xm,a).

4.3 Examples
 page 240 The equality of corollary (3.2) is almost surely (not an exact equality). If you examine the proof, you'll notice that it only shows the probability of the equality holding is 1.
 page 244 Kakutani dichotomy This introductory paragraph before theorem (3.5) can be a little confusing. About halfway through, we see that "(3.3) implies that Xn ->X nu-a.s." If you look at (3.3), it is not entirely clear how it says this. It is actually more pertinent to use (3.4) to see that Xn is a martingale, notice that Xn is nonnegative, and apply (2.11) to see that the sequence does in fact have a nu-a.s. limit. Next Durrett mentions "the convergence of the infinite product," and it is not entirely clear that we have seen any infinite products here. But we have: Xn is actually the product of qm for m=1..n (this is later officially stated as one of the displayed equations in the proof of (3.5).)
 page 245 Near the end of the proof of (3.5), it is stated that P(X=infty)=0. But we don't have a prob measure P! The measure should be nu instead.
 page 246 In the statement of theorem (3.7), it should be: almost surely, (Zn=0 for all n sufficiently large). Notice that how large n needs to be may very well depend on omega. That is, the proof does not show that there is an N so that for all n>N, Zn=0 almost surely.