Bounded in probability of the running maximum of a stopped process implies almost sure boundedness on the interval $[0,\lim \tau_r)$?

Question

Suppose that $\tau_r$ is a sequence of stopping times increasing to some limit $\tau$ as $r\to \infty$. Moreover, assume that for any time $t\ge 0$, the sequence $X_{\tau_r \wedge t}^* :=\sup_{s\le t} |X^{\tau_r}_s|$ is bounded in probability. In this case, why is $X$ almost surely bounded on the interval $[0,\tau)$ whenever $\tau<\infty$.

Bounded in probability means that for any $\epsilon >0$, we have some $N_\epsilon$ and $B_\epsilon$ such that for $n \ge N_\epsilon$ we have $$P(X_{\tau_r \wedge t}^* > B_\epsilon)<\epsilon.$$

I cannot see how this property leads to almost sure boundedness on $[0,\lim_r \tau_r)$. Why does this hold?

I got this question from reading the proof of Lemma 5 in the following post : https://almostsuremath.com/2010/02/10/existence-of-solutions-to-stochastic-differential-equations/

Answer 1

From the context of the other post, $X_t$ is a cadlag process , $\tau_r$ is a sequence of stopping times increasing to a limit $\tau$. For every $t \geq 0$, we know that the sequence $X^*_{\tau_r \wedge t}$ is bounded in probability for every $t \geq 0$). To write that rigorously :

For every $t \geq 0 , \epsilon>0$, there exists a $B_{\epsilon}>0$ and $r_{\epsilon} \in \mathbb N$ so that $r>r_{\epsilon}$ implies $$ P\left(|X^*_{\tau_{r} \wedge t}| > B_{\epsilon}\right) \leq \epsilon $$

We will prove that $$ P\left(\tau<\infty , \sup_{[0,\tau)} X_t = \infty\right) = 0 $$

That is, if $\tau<\infty$ then $X_t$ is a.s. bounded on $[0,\tau)$. To make this easy , we first write $$ P\left(\tau<\infty , \sup_{[0,\tau)} X_t = \infty\right) = \lim_{K \to \infty} P\left(\tau \leq K, \sup_{[0,\tau)} X_t = \infty\right) $$ which is obvious since the events on the RHS increase to that on the LHS as $K \to \infty$.

Using a similar argument with decreasing events, we may also write for any $K$, $$ P\left(\tau \leq K , \sup_{[0,\tau)} X_t = \infty \right) = \lim_{N \to \infty} P\left(\tau \leq K , \sup_{[0,\tau)} X_t \geq N\right) $$ Now, if $\tau\leq K$ then for every $r$ , we know that $\tau_r \leq K$, so let's apply the boundedness in probability condition with $t = K+1$. In that case, $\tau_r \wedge t = \tau_r$ for all $r$.

Fix $\epsilon>0$. By boundedness in probability and $\tau_{r} \wedge t = \tau_r$, there exists a $B_{\epsilon}>0$ such that for all $r>0$, $$ P\left(\tau \leq K, |X^*_{\tau_r}| > B_{\epsilon}\right) \leq \epsilon $$

Note that we can replace $B_{\epsilon}$ with any $B>B_{\epsilon}$. Also, $$X^*_{\tau_r} = \sup_{[0,\tau_r]} X_t \to \sup_{[0,\tau)} X_t$$ as $r \to \infty$ since $\tau_r \uparrow \tau$ (a.s.). Dragging $r \to \infty$ in the inequality above gives $$ P\left(\tau\leq K ,\sup_{[0,\tau)} X_t \geq B \right) \leq \epsilon $$ for any $B>B_{\epsilon}$. It follows that $$P\left(\tau \leq K , \sup_{[0,\tau)} X_t = \infty \right) = \lim_{B \to \infty} P\left(\tau\leq K ,\sup_{[0,\tau)} X_t \geq B \right) \leq \epsilon $$

for any $\epsilon \geq 0$. The conclusion is that for every $K$, $$ P\left(\tau \leq K , \sup_{[0,\tau)} X_t = \infty \right) = 0 \implies \lim_{K \to \infty}P\left(\tau \leq K , \sup_{[0,\tau)} X_t = \infty \right) = 0 \\ \implies P\left(\tau<\infty, \sup_{[0,\tau)} X_t = \infty\right) = 0 $$

as desired.

Answer 2

If by almost sure boundedness you mean there is a constant $M$ such that $P( \vert X^{\ast}_{\tau^-} \vert>M)=0$ Here is a càdlàg counterexample: Fix $T>0$.

Let $Y_k(t)= \frac{1}{T-t}$ if $t < T(\sum_{m=0}^k (\frac{1}{2})^m ) $, $Y_k (t)$ is zero otherwise.

Let $P(X_t= Y_k (t) , \forall \ t>0)= (\frac{1}{2})^k $.

Define the stopping time $\tau=T$ (you choose the filtration). Let $\tau_r= T-\frac{1}{r}$.