Let $\tau$ be a stopping time for the filtration $\{\mathcal{F}_n\}$ and suppose there is a constant N s.t. for every $n\ge 0$, $\mathcal{P}(\tau\le n+N|\mathcal{F}_n)\ge \epsilon \gt 0$ for some $\epsilon$.

Show that $\mathcal{P}(\tau< \infty)=1$ and $\mathbb{E}[\tau^p]< \infty$ for every $p\ge 1$

  • $\begingroup$ I changed your $\le\infty$ to $<\infty$, which I presume is what you intended. $\endgroup$ – user940 Nov 14 '11 at 17:50
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    $\begingroup$ Are you sure that you copied the problem correctly? As written, I don't think it is true. $\endgroup$ – user940 Nov 14 '11 at 18:57
  • $\begingroup$ pretty sure its $<\infty$. When I did this question I used the tower property on indicator functions. $\endgroup$ – Pk.yd Oct 15 '12 at 11:27

For every $n\geqslant0$, consider $A_n=[\tau\gt n]$, then the event $A_n$ is in $\mathcal F_n$ because $\tau$ is a stopping time, hence $\mathbb P(\tau\leqslant n+N,A_n)\geqslant\varepsilon\mathbb P(A_n)$, that is, $\mathbb P(n\lt\tau\leqslant n+N)\geqslant\varepsilon\mathbb P(n\lt\tau)$.

Since $\mathbb P(n\lt\tau\leqslant n+N)=\mathbb P(\tau\gt n)-\mathbb P(\tau\gt n+N)$, this yields $$ \mathbb P(\tau\gt n+N)\leqslant(1-\varepsilon)\mathbb P(\tau\gt n). $$ Iterating this on the multiples of $N$ yields, for every $k\geqslant0$, $$ \mathbb P(\tau\gt kN)\leqslant(1-\varepsilon)^k. $$ In particular, $\mathbb P(\tau=\infty)\leqslant\mathbb P(\tau\gt kN)$ for every $k\geqslant0$ hence $\mathbb P(\tau=\infty)=0$, which shows that $\tau$ is almost surely finite.

Likewise, for any $p\gt0$, $$ \tau^p\leqslant\sum\limits_{k\geqslant0}(k+1)^pN^p\mathbf 1_{kN\lt \tau\leqslant(k+1)N}\leqslant N^p\sum\limits_{k\geqslant0}(k+1)^p\mathbf 1_{\tau\gt kN}, $$ hence, integrating this almost sure inequality, $$ \mathbb E(\tau^p)\leqslant N^p\sum\limits_{k\geqslant0}(k+1)^p\mathbb P(\tau\gt kN)\leqslant N^p\sum\limits_{k\geqslant0}(k+1)^p(1-\varepsilon)^k. $$ The last series converges, hence $\tau^p$ is integrable.

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  • $\begingroup$ when you take the expectation of the terms across the last but one inequality, wouldn't you have to make sure that the sum on the rightmost side of the inequality converges before moving the expectation inside the sum? $\endgroup$ – Calculon Feb 28 '15 at 11:57
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    $\begingroup$ @Calculon No, because everything is nonnegative hence the sums, whether finite or infinite, always coincide. $\endgroup$ – Did Feb 28 '15 at 15:22
  • $\begingroup$ Thanks, that clears it for me. $\endgroup$ – Calculon Feb 28 '15 at 19:08
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    $\begingroup$ @perlman $P(\tau\leqslant n+N,A_n)=P(\tau\leqslant n+N\mid A_n)P(A_n)\geqslant\varepsilon P(A_n)$ because $P(\tau\leqslant n+N\mid \mathcal F_n)\geqslant\varepsilon$ and $A_n\in\mathcal F_n$. $\endgroup$ – Did Nov 19 '17 at 10:00
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    $\begingroup$ @perlman If $P(B\mid\mathcal G)\geqslant\epsilon$ almost surely and $A\in\mathcal G$, one has $$P(B\mid A)=E(\mathbf 1_B\mid A)=E(E(\mathbf 1_B\mid\mathcal G)\mid A)=E(P(B\mid\mathcal G)\mid A)\geqslant E(\epsilon\mid A)=\epsilon$$ $\endgroup$ – Did Nov 20 '17 at 21:21

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