I am reading the proof of the lemma on the equivalent characterizations of a predictable stopping time from George's blog : https://almostsuremath.com/2009/11/30/predictable-stopping-times/
but there are details I cannot figure out. I would greatly appreciate some help figuring these out.
I am having difficulty understanding the proof of the step (2) $\to$ (3) in the Lemma 3 below. So the goal is to construct an adapted continuous process $Y_t$ taking values in $\bar{\mathbb{R}}_{+} $, which is finite for $t< \tau$ and infinite for $t \ge \tau$.
He starts with defining
$$Y_t = \sum \lambda_n \max(t-\tau_n,0)$$ on the set $\tau >0$.
My first question is : why does the fact that this reduces to a finite sum when $t<\tau$ only leave us to show that for large enough $\lambda_n$, the left limit at $\tau$ will be almost surely infinite whenever $\tau>0$? We still need to show that $Y$ is adapted and continuous, and infinite for $t \ge \tau$. I cannot see how all these are taken care of by showing that $Y_{\tau_{-}}$ is almost surely infinite when $\tau>0$.
Secondly: how is $$Y_{\tau_{-}}= \sum \lambda_n \max(\tau-\tau_n,0)?$$ In the case $t< \tau$, $Y$ is just a finite sum so this makes sense, but when $\tau>0$, it could be an infinite sum, so how can we pass the left limit inside the sum?
Third: How do we use monotone convergence to show that $P(\lambda_n(\tau – \tau_n)<1 \;\text{and} \; \tau>0)<2^{-n}$ for large $\lambda_n$, from equation (2)?
Finally: what is the point of replacing $X_t$ by $\tau-t$ on any zero probability set? I don't see why this is necessary from the construction of $Y$ and defining $X=1/(1+Y)$ here.
Best Answer
$Y$ is adapted: Since $\tau_n$ are stopping times $\{\tau_n\le t\}\in{\cal F}_t\,.$ Clearly, $$ \max(t-\tau_n,0)=t1_{\{\tau_n\le t\}}-\tau_n1_{\{\tau_n\le t\}}\,. $$ The first term on the RHS is clearly ${\cal F}_t$-measurable. To see that for the second term observe that $\tau_n1_{\{\tau_n\le t\}}>a$ if and only if $a<\tau_n\le t\,.$ Since every set $\{a<\tau_n\le t\}$ is in ${\cal F}_t$ we are done.
$t\mapsto Y_t$ is continuous: Clearly, every $\max(t-\tau_n,0)$ is a continuous function of $t\,.$ Because on $\{t<\tau\}$ the sum is finite, $t\mapsto Y_t$ is continuous on $\{t<\tau\}\,.$ Since $\{t>\tau\}\subset \{Y_t=\infty\}$ it is clear that $t\mapsto Y_t$ is continuous on $\{t>\tau \}\,.$ The not so trivial question is the continuity at $\tau$ but that's handled by George Lowther who shows that $Y_{\tau-}=\infty$ a.s.
The limit can be passed inside the sum because of monotonicity and nonnegativity $$\lim\limits_{k\to\infty}Y_{\tau_k}=\lim\limits_{k\to\infty}\sum\limits_{n=1}^\infty\lambda_n\max(\tau_k-\tau_n,0)=\sup_k\sup_m\sum_{n=1}^m\lambda_n\max(\tau_k-\tau_n,0)$$ and the two suprema can be swapped.
Regarding your Third question: $$ \{\tau_n<\tau\}=\bigcup_m\Big\{\tau_n<\tau-\textstyle\frac{1}{m}\Big\} $$ where the sets on the RHS are increasing in $m\,.$ By monotone convergence $$ \lim_{m\to\infty}\mathbb P\Big\{\tau_n<\tau-\textstyle\frac{1}{m}\Big\}=\mathbb P\{\tau_n<\tau\}\,. $$ Using monotone convergence on (2), we have for large enough $m$ $$ \mathbb P\Bigg(\Big(n\wedge\tau-2^{-n}<\tau_n<\tau-\textstyle\frac{1}{m}\Big)\text{ or }\tau_n=\tau=0\Bigg)>1-2^{-n}\,. $$ This implies $$ \mathbb P\Big(\tau_n<\tau-\textstyle\frac{1}{m}\text{ or }\tau_n=\tau=0\Big)>1-2^{-n}\,. $$ Then, $$ \mathbb P\Big(\tau-\tau_n\le\textstyle\frac{1}{m}\text{ and }(\tau_n>0\text{ or }\tau>0\text{ or }\tau_n\not=\tau)\Big)< 2^{-n}\,. $$ If we drop the condition $\tau_n>0$ or $\tau_n\not=\tau$ then the set on the LHS gets only smaller. Therefore, $$ \mathbb P\Big(\tau-\tau_n\le\textstyle\frac{1}{m}\text{ and }\tau>0\Big)< 2^{-n}\,. $$ Clearly, $m$ depends on $n$ and one can choose $\lambda_n$ to be $m(n)\,.$ This implies $$ \mathbb P\Big(\lambda_n(\tau-\tau_n)\le 1\text{ and }\tau>0\Big)<2^{-n}\,. $$
Finally, $X$ is replaced by $\tau-t$ on any zero probability set simply because all properties shown for $Y$ only hold almost surely, and we want $$ \tau=\inf\{t\ge 0:X_t=0\} $$ to hold surely.