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 Feb25 revised Martingale and bounded stopping time added 73 characters in body Feb24 asked Martingale and bounded stopping time Feb24 accepted Generalized Second Borel-Cantelli lemma Feb23 asked Generalized Second Borel-Cantelli lemma Feb22 awarded Critic Feb21 revised Proving existence of limit by Martingale. added 660 characters in body Feb21 asked Convergence of Martingale. Feb21 comment Proving existence of limit by Martingale. @Ilya You're right. But this question appears as an exercise for martingales. So there must be some way that we can relate this fixed sequence to a super or sub martingale. I'm thinking that could there be $X_n$ such that $X_n$ converges almost surely, and $\prod_m (1+y_m)$ is one of $X_n$'s realization. Feb21 answered A fair coin flip, which of the events are independent Feb21 comment Proving existence of limit by Martingale. @ByronSchmuland Yeah, you are right. There is a problem if $M > 1$. Feb21 asked Proving existence of limit by Martingale. Feb20 accepted Martingale that converges almost surely to $-\infty$. Feb20 comment Martingale that converges to zero @ColinMcQuillan I see! Because $Y_1,Y_2,..$ are i.i.d., so we can find an $\epsilon > 0$ such that $\mathbb P[|Y_1 - 1| ~> \epsilon] < 1$ for all $n$. Therefore $\mathbb P[Y_n = 1] = 0$. So here $i.i.d.$ is a necessary condition. Feb20 comment Martingale that converges to zero @ColinMcQuillan But couldn't there exists a set $A = \{\omega:Y_n(\omega) \to 1\}$ such that $\mathbb P\{A\} > 0$? Feb20 accepted Martingale that converges to zero Feb20 asked Martingale that converges to zero Feb20 comment Martingale that converges almost surely to $-\infty$. I see. How about let $\xi_{n}$ be $-1$ with probability $1-\frac{1}{2^{n}}$, and be $2^{n}-1$with probability $1/2^{n}$.Since $\sum_{n=1}^{\infty}1/2^{n}<\infty,$we have $\mathbb{P}\{\xi_{n}>0 \, \text{i.o.}\}=0$. In other words, $\mathbb{P}\{\xi_{n}<0 \, \text{i.o.}\}=1$. So we have $\sum_{m=1}^n \xi_n \to -\infty$. Feb20 asked Martingale that converges almost surely to $-\infty$. Feb14 comment Superharmonic function and super martingale. I got it thanks! Feb14 comment Superharmonic function and super martingale. Hi, I guess by "integrate that which is independent and keep that which is measurable", you mean something like $E(X + Y|\mathcal{F})=E(X) + Y$ is $X$ is independent of $\mathcal{F}$ and $Y$ is measurable w.r.p. to $\mathcal{F}$. Is that right?