# Almost surely finite stopping time

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 $p\ge 1$

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