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Student.


Feb
25
revised Martingale and bounded stopping time
added 73 characters in body
Feb
24
asked Martingale and bounded stopping time
Feb
24
accepted Generalized Second Borel-Cantelli lemma
Feb
23
asked Generalized Second Borel-Cantelli lemma
Feb
22
awarded  Critic
Feb
21
revised Proving existence of limit by Martingale.
added 660 characters in body
Feb
21
asked Convergence of Martingale.
Feb
21
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.
Feb
21
answered A fair coin flip, which of the events are independent
Feb
21
comment Proving existence of limit by Martingale.
@ByronSchmuland Yeah, you are right. There is a problem if $M > 1$.
Feb
21
asked Proving existence of limit by Martingale.
Feb
20
accepted Martingale that converges almost surely to $-\infty$.
Feb
20
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.
Feb
20
comment Martingale that converges to zero
@ColinMcQuillan But couldn't there exists a set $A = \{\omega:Y_n(\omega) \to 1\}$ such that $\mathbb P\{A\} > 0$?
Feb
20
accepted Martingale that converges to zero
Feb
20
asked Martingale that converges to zero
Feb
20
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 $\sum_{m=1}^n \xi_n \to -\infty$.
Feb
20
asked Martingale that converges almost surely to $-\infty$.
Feb
14
comment Superharmonic function and super martingale.
I got it thanks!
Feb
14
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?