Let $(Y_n)_{n \geq 1} $ be a sequence of independent r.v.'s s.t. $$P(Y_n=y) = {n \choose k } \left(\frac1n\right)^y \left(1-\frac1n\right)^{n-y}\quad {\rm if }\;y \in \{0,1,\dots,n\}$$
How to show that
- $\tau := \inf{\{n : Y_n>1\}}$ is a stopping time wrt the natural filtration $\mathcal{F}_n=\sigma(Y_1,\dots,Y_n)$
- $\tau$ is a.s. finite?
We have $$\{\tau\leqslant n\}=\bigcup_{k=1}^n\{Y_k>1 \}\in\mathcal F_n, $$ so $\tau$ is a stopping time. Moreover, \begin{align} \mathbb P(\tau=\infty) &= \mathbb P\left(\bigcap_{n=2}^\infty\{Y_n\leqslant1\}\right) \\ &=\prod_{n=2}^\infty \mathbb P(Y_n\leqslant 1)\\ &=\prod_{n=2}^\infty\left(1-\frac1n\right)^{n-1}\left(2-\frac1n\right)\\ &\leqslant 2\prod_{n=2}^\infty\left(1-\frac1n\right)^{n-1}. \end{align} Since $$\lim_{n\to\infty}\left(1-\frac1n\right)^{n-1}=\frac1e<1, $$ it follows that $$\prod_{n=2}^\infty\left(1-\frac1n\right)^{n-1}=0$$ and hence $\tau$ is almost surely finite.