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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
May 28, 2015 at 7:38 answer added Michael timeline score: 3
May 27, 2015 at 22:02 comment added Michael By differentiating Wald we get $E[(S_J - J\gamma'(r))e^{rS_J-J\gamma(r)}] = 0$, which for $r=0$ gives $E[S_J] = E[J]E[X]$ (this recovers a more common version of Wald). Differentiating again gives $E[J]Var(X)=E[(S_J-JE[X])^2]$. If $E[X]=0$ then we get $E[J]Var(X)=E[S_J^2]$, which relates to second moments as you ask about.
May 27, 2015 at 21:59 comment added Michael Note $\gamma'(r) = \frac{E[Xe^{rX}]}{E[e^{rX}]}$ and $\gamma''(r) = \frac{-E[Xe^{rX}]^2}{E[e^{rX}]^2} + \frac{E[X^2e^{rX}]}{E[e^{rX}]}$, so for $r=0$ we get $\gamma(0)=0$, $\gamma'(0)=E[X]$, $\gamma''(0)=Var(X)$.
May 27, 2015 at 21:55 comment added Michael The general Wald equality might be useful for problems of this type: If $J$ is a stopping time for a sum $S_k = \sum_{i=1}^k X_i$ of iid random variables $\{X_i\}_{i=1}^{\infty}$, then $E[e^{rS_J - J\gamma(r)}]=1$, where $\gamma(r) = \log(E[e^{rX}])$.
May 27, 2015 at 21:52 comment added Michael The previous question you linked to started at the origin and asked for the time to exit the box. This one seems different: Where do we start? And why do you consider both positive and negative integers if 0 absorbs (so we cannot go from positive to negative)? I do not see much relationship between the previous question and this question.
May 26, 2015 at 18:13 history edited John Lotos CC BY-SA 3.0
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May 26, 2015 at 18:02 history asked John Lotos CC BY-SA 3.0