Let $X_n$ be a martingale with $X_0 = 0$ and $EX_n^2<\infty$ for all n. $X_n^2$ is a submartingale and thus $X_n^2 = M_n + A_n$.
I don't understand a step in this theorem: $E(\text{sup}_n |X_n|)\leq 3EA_{\infty}^{1/2}$, where $A_{\infty}=\text{lim}_n A_n$.
Proof: Let $a>0$ and let $N = \text{inf}\{n:A_{n+1}>a^2\}$, then
$$P(\text{sup}_n |X_n|>a)\leq P(N<\infty)+ P(\text{sup}_n |X_{n\wedge N}|>a)$$
I don't understand why {$\text{sup}_n |X_n|>a\} \implies \{N<\infty\}$? Thanks.
Best Answer
The inequality follows from the inclusion $(\sup_n |X_n|>a) \subset (N<\infty) \cup (\text{sup}_n |X_{n\wedge N}|>a)$.
That's because $\begin{aligned}[t](\sup_n |X_n|>a) &= [(\sup_n |X_n|>a)\cap (N<\infty)]\cup [(\sup_n |X_n|>a)\cap (N=\infty)]\\ &=[(\sup_n |X_n|>a)\cap (N<\infty)]\cup [(\sup_n |X_{n\wedge N}|>a)\cap (N=\infty)]\\ &\subset (N<\infty) \cup (\sup_n |X_{n\wedge N}|>a) \end{aligned}$
The union bound then yields $$P(\sup_n |X_n|>a) \leq P(N<\infty) + (\sup_n |X_{n\wedge N}|>a)$$