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Let $X_1, X_2, X_3, \ldots$ be i.i.d. random variables with zero mean and let $S_n := X_1 + \ldots + X_n$. Does $T := \inf\{n: S_n > 0\}$ always have infinite first moment?

In the trivial case, where $X_i = 0$, we have $T = \infty$. For random walk, in which $X_i = \pm 1$ with probability $1/2$, it can be shown that $\mathbb{E}[T]=\infty$. I am wondering if it is always the case that $\mathbb{E}[T]=\infty$.

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The answer is yes. This follows indirectly from the optional stopping theorem. The relevant formulation of the theorem says the following. If $X_n$ is a martingale, the increments of $X_n$ are "conditionally bounded", and $\mathbb{E}(\tau) < \infty$, then $\mathbb{E}[X_\tau]=\mathbb{E}[X_0]$. "Conditionally bounded" means that $\mathbb{E}(|X_{n+1}-X_n| \, | \mathcal{F}_n) \leq C$ independent of $n$.

Here $S_n$ is a martingale because it is a sum of iid variables and its increments are conditionally bounded by $\mathbb{E}[|X_i|]$. So the first two hypotheses hold. Now by contraposition, since the conclusion fails, the third hypothesis must also fail. That is, since $\mathbb{E}[S_\tau] > 0$ while $\mathbb{E}[S_0] = 0$, we conclude that $\mathbb{E}(\tau)=\infty$.

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  • $\begingroup$ I am not sure what you meant by 'on average you are moving to the right'. In the problem, we assume $E[X_i]=0$, which means the expected moving distance is $0$. Can you please elaborate on this? $\endgroup$ – Zilin J. Mar 5 '15 at 21:34
  • $\begingroup$ @roy_24601 Sorry, I missed the zero mean requirement. $\endgroup$ – Ian Mar 5 '15 at 21:36
  • $\begingroup$ @roy_24601 I've corrected my post now. $\endgroup$ – Ian Mar 5 '15 at 21:39
  • $\begingroup$ In my case, the increment in the martingale might not be bounded. The problem didn't assume anything on $X_i$ other than being mean zero. $\endgroup$ – Zilin J. Mar 5 '15 at 21:47
  • $\begingroup$ @roy_24601 Sorry, what you actually need is that they are "conditionally bounded", meaning that $\mathbb{E}(|X_{n+1}| | \mathcal{F}_n) \leq C$ for a fixed $C$ independent of $n$. I think this follows from integrability of the $X_n$. $\endgroup$ – Ian Mar 5 '15 at 21:49

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