We've got the usual filtered stochastic basis $(\Omega, \mathcal F, (\mathcal F_n). \mathbb P), \space \tau : \Omega \to \mathbb{N}\cup \{\infty\}, [\tau \le n] \in \mathcal F_n$

($\tau$ is an $\mathcal F_n$ stopping time).

I'm wondering whether the implication below holds:

$$\tau \in L^1 ( \mathbb P) \Rightarrow \mathbb P [\tau < \infty]=1$$

I think it should hold, because

$$\tau \in L^1 ( \mathbb P) \Rightarrow \mathbb E [\tau] =\int_{\Omega}\tau d \mathbb P < \infty \\ \mathbb P [\tau < \infty] = \int_{[\tau < \infty]} d \mathbb P \le \int_{\Omega} d \mathbb P \le \int_{\Omega}\tau d \mathbb P < \infty $$

The only problem I can see is that $$\int_{\Omega} d \mathbb P \le \int_{\Omega}\tau d \mathbb P $$ might not hold in general. For instance if $X \sim N(0,\sigma^2)$ we've got $\mathbb P (\Omega) =1$ and $\mathbb{E} X =0$. So is this the reason why my proof does not work?

  • 1
    $\begingroup$ You've identified one problem with your reasoning, but even if that inequality were true I don't see how it would prove that $\tau$ is finite with probability one. $\endgroup$
    – Gibarian
    Jun 1 '13 at 10:08

The implication holds for any random variable, not just stopping times. It easily seen by showing the contrapositive: $$ P(X<\infty)<1\;\Rightarrow\;X\notin L^1(P). $$ So suppose $P(X<\infty)<1$, then $P(X=\infty)>0$ and hence $$ {\rm E}[|X|]\geq{\rm E}[|X|\mathbf{1}_{\{X=\infty\}}]=\infty\cdot P(X=\infty)=\infty, $$ and hence $X\notin L^1(P)$. The other implication, however, does not hold in general.

  • $\begingroup$ not sure if I understand what you mean. I elaborated on my reasoning (see above) $\endgroup$
    – shimee
    Jun 1 '13 at 9:29
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    $\begingroup$ @shimee: What is it you don't understand? $\endgroup$ Jun 1 '13 at 9:31
  • $\begingroup$ your proof. You're proving it by contradiction? Also, what do you mean by "The other implication, however, does not hold in general."? Have a look on my proof in the original question. Is it correct? $\endgroup$
    – shimee
    Jun 1 '13 at 9:37
  • $\begingroup$ In general, instead of showing that A implies B you can show that not-A implies not-B. See this. What I mean by "the other implication does not hold" is that you can't say $P(X<\infty)=1\;\Rightarrow\; X\in L^1(P)$. I don't get the first line of your proof. $\endgroup$ Jun 1 '13 at 9:41
  • $\begingroup$ the probability measure of tau being less than infinity is equal to the measure of the entire space. I switched it around (just now) for better clarity. $\endgroup$
    – shimee
    Jun 1 '13 at 9:47

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