Assume $(\Omega, \mathcal{F}, (\mathcal{F})_{n}, \mathbb{P})$ a filtrated probability space and $\tau$ a stopping time such that there is a $\epsilon > 0$ such that for any $n \in \mathbb{N}$
$$ \mathbb{P} \left[ \tau \le n+1 \; | \; \mathcal{F_{n}} \right] \ge \epsilon \; $$
almost sure. Is it right to follow that
$$ \mathbb{P} \left[ \tau \le n+1 \; | \; \tau > n \right] \ge \epsilon $$ holds? My thoughts so far: As $\{\tau > n\} = \complement \{ \tau \le n\} $ and as $\tau$ is a stopping time, i.e.
$$\{\tau \le n \} \in \mathcal{F_{n}} $$
for all $n$, also $\{\tau > n\} \in \mathcal{F_{n}}$. Further I know that
$$ \mathbb{P} \left[ \tau \le n+1 \; | \; \mathcal{F_{n}} \right] = \mathbb{E} \left[ \mathbb{1}_{\{ \tau \le n+1 \}} \; | \; \mathcal{F_{n}}\right] $$
and that
$$ \mathbb{P} \left[ \tau \le n+1 \; | \; \tau > n \right] = \frac{\mathbb{P} \left[ \{ \tau \le n+1\} \cap \{ \tau > n \} \right] } {\mathbb{P} \left[ \tau > n \right] } = \frac{\mathbb{P} \left[ \tau = n+1 \right] } {\mathbb{P} \left[ \tau > n \right] } . $$
I am already sorry in advance if this is trivial - either trivially right or trivially wrong...
$\newcommand{\E}{\mathbb E}\newcommand{\F}{\mathcal F}\newcommand{\P}{\mathbb P}\newcommand{\1}{\mathbf 1}$First of all no need to say sorry. I don't understand what your plan was.
You are given: \begin{align} \E[\1_{\{\tau\leq n+1 \}} |\F_n]\geq \varepsilon \end{align} Let $A:=\{\tau >n \}$. On one hand: \begin{align} \E[\E[\1_{\tau\leq n+1}|\F_n]\ |\ \sigma(A)] \geq \E[\varepsilon|\sigma(A)] = \varepsilon \end{align} Since $\sigma(A)\subset \F_n$, we have on the other hand: \begin{align} \E[\E[\1_{\tau\leq n+1}|\F_n]\ |\ \sigma(A)]=\E[\1_{\tau\leq n+1}|\sigma(A)] \end{align} by the "the smallest $\sigma$-algebra wins" property. That means: \begin{align} \E[\1_{\tau\leq n+1}|\sigma(A)]\geq \varepsilon \end{align} And now we can conclude that: \begin{align} \P(\tau\leq n+1 | \tau > n)\geq \varepsilon \end{align}