I am reading the proof that the jump times of a Poisson process are totally inaccessible from the following post : https://almostsuremath.com/2010/06/24/poisson-processes/#scn_pp_def1
From definition 9 in https://almostsuremath.com/2009/11/30/predictable-stopping-times/#scn_pst_def1
a stopping time $\tau$ is totally inaccessible if $P(\sigma=\tau<\infty)=0$ for all predictable stopping times $\sigma$.
In the proof below, the author shows that, using the compensated process $M_t=X_t – \lambda t$ being a martingale, that the jump times of a Poisson process $X$ are totally inaccessible, by showing that $P(\Delta X_\tau \neq 0)=0$ for each predictable stopping time $\tau$.
My questions are:
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In the proof, how are we able to replace $\tau$ by $\tau \wedge t$? How does this not hurt generality here?
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In the final equalities, how are we able to take the limit outside of the expectation in computing $E[M_\tau – M_{\tau_n}]$?
Best Answer
For 1: Suppose that $\tau \wedge t$ is totally inaccessible for all $t>0$ and let $\sigma$ be a predictable stopping time. By the monotone convergence theorem for sets, $P(\sigma = \tau < \infty) = \lim_{t \rightarrow \infty} P(\sigma = \tau \wedge t < \infty)$. But for all $t > 0$ we have $\tau \wedge t$ is totally inaccessible, so $P(\sigma = \tau \wedge t < \infty) = 0$, and hence $P(\sigma = \tau < \infty) = 0$ as well.
For 2: This follows from the fact that $M_{\tau_n} = \mathbb{E}[M_\tau | \mathcal F_{\tau_n}]$ and $\{\mathbb{E}[X|\mathcal B]: \mathcal B \text{ is a sub-$\sigma$ algebra}\}$ is a uniformly integrable class of random variables for any integrable $X$.