Supermartingale vanishing at some stopping time

Question

Let $\left\{X_t\right\}_{t\in[0, T]}$ be a continuous and non-negative supermartingale. We define the stopping time $$\tau_0:=\inf\{t\in[0,T]:X(t)=0\}\wedge T$$ and immediately obtain by continuity of $X$ that $\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)=0$ a.s.

I want to prove that $X(t)=0$ for almost every $\omega\in\{\tau_0<T\}$ where $t\in[\tau_0(\omega),T]$.

What I established so far (using optional stopping) is the following: $$E[\mathbf{1}_{\{\tau_0<T\}}X(T)]=E[\mathbf{1}_{\{\tau_0<T\}}E[X(T)\mid F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0$$ Hence, as $X$ is non-negative, the desired result follows for the terminal time $t=T$. Can someone please help me to prove the whole result for all the other relevant times? Thanks a lot in advance!

Answer 1

You just need to look at more stopping times. For any $t\geq 0$, you have $${E}[\mathbf{1}_{\{\tau_0<T\}}X(T \wedge (\tau_0+t))] ={E}[\mathbf{1}_{\{\tau_0<T\}}E[X(T\wedge (\tau_0+t))|F_{\tau_0}]] \le {E}[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0.$$ This gives
$$P\biggl(X(T\wedge(\tau_0+t))=0\mbox{ for all }t\in\mathbb{Q}\cap[0,T]; \tau_0<T\biggr) =P(\tau_0<T),$$ and sample path continuity will give you the desired conclusion.

Answer 2

You can generalize your derivation from: $$E[\mathbf{1}_{\{\tau_0<T\}}X(T)]=E[\mathbf{1}_{\{\tau_0<T\}}E[X(T)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0$$ to: $$E[\mathbf{1}_{\{\tau_0=s\}}X(t)]=E[\mathbf{1}_{\{\tau_0=s\}}E[X(t)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0=s\}}X(\tau_0)]=0$$ where $t > s$.

As you mentioned, using the fact that $X(t)$ is non-negative, you can imply state from there that: $$E[\mathbf{1}_{\{\tau_0=s\}}X(t)] = \int_0^\infty x p(x_t = x \wedge \tau_0=s) dx \leq 0$$ This can only hold if $p(x_t = x \wedge \tau_0 = s) = 0$ for all $x > 0$. Therefore, for any path that has a stopping time $s$, $p(x_t > x) = 0$ for all $x > 0$.

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For future readers: How to finish the proof - For a fixed $s$ I have shown that $P(X(t) = 0|\tau_0 = s) = 1$ for all $t > s$. Now take a countable dense subset $t_i$ of $(s, T]$. Then $P(X(t_i) = 0 \,\,\,\ \forall t_i | \tau_0 = s) = 1$. A continuous path that is zero at a countable dense set of points must be zero everywhere.

Now to finish: $$P(X_t \neq 0 \,\,\, \textrm{for some} \,\,t > \tau_0 \wedge \tau_0 < T) = \int_0^T p(X_t \neq 0 \,\,\, \textrm{for some} \,\,t > s \wedge \tau_0 = s) ds = 0$$

Answer 3

Here are two approaches, one more elementary than the other.

1. Modify your argument to show that $X(r) = 0$ a.s on the event $\{\tau_0< r\}$ for each $r\in (0,T]$. Deduce from this that $X(r)=0$ for every rational in $(\tau_0,T)$, a.s. on the event $\{\tau_0<T\}$. Now invoke the continuity of $X$.

2. Let $B=\{\omega: \tau_0(\omega)<T, X(t,\omega)>0$ for some $t\in(\tau_0,T)\}$, and suppose that $P[B]>0$. Then by Meyer's Optional Section Theorem, there is a stopping time $S\ge\tau_0$ such that $P[S<\infty]>0$ and $X(S(\omega),\omega)>0$ for each $\omega\in\{S<\infty\}$. By optional stopping at time $S$, $E[X(S)]=0$ because $S\ge\tau_0$. On the other hand $$ E[X(S)] \ge E[X(S); S<\infty] >0. $$ This contradiction shows that $P[B]=0$.

Answer 4

Let $\epsilon > 0$ and consider the stopping time $$\tau_\epsilon = \inf\{t \in [\tau_0, T] : X_t \ge \epsilon\} \wedge T.$$ Since $\tau_0 \le \tau_\epsilon$ almost surely, optional stopping gives $E[X_{\tau_\epsilon}] \le E[X_{\tau_0}]$. Now let consider the events $\{\tau_0 < T\}$ and $\{\tau_0 = T\}$ and write $$E[X_{\tau_\epsilon} ; \tau_0 < T] + E[X_{\tau_\epsilon} ; \tau_0 =T] \le E[X_{\tau_0} ; \tau_0 < T] + E[X_{\tau_0} ; \tau_0 =T].$$ Now $X_{\tau_0} = 0$ on the event $\{\tau_0 < T\}$. On the event $\{\tau_0 = T\}$ we have $X_{\tau_0} = X_T$, and moreover, since $\tau_\epsilon \ge \tau_0$, we also have $\tau_\epsilon = T$ and $X_{\tau_\epsilon} = X_T$. So our inequality becomes $$E[X_{\tau_\epsilon} ; \tau_0 < T] + E[X_{T} ; \tau_0 =T] \le 0 + E[X_{T} ; \tau_0 =T]$$ showing that $E[X_{\tau_\epsilon} ; \tau_0 < T] \le 0$. Since $X_{\tau_\epsilon}$ is nonnegative we have $X_{\tau_\epsilon} = 0$ almost everywhere on $\{\tau_0 < T\}$. On the other hand, $\{\tau_\epsilon < T\} \subset \{\tau_0 < T\}$, and by continuity $X_{\tau_\epsilon} = \epsilon$ almost everywhere on $\{\tau_\epsilon < T\}$. So we conclude $P(\tau_\epsilon < T) = 0$, i.e. almost surely on $\{\tau_0 < T\}$, there does not exist $t$ in $[\tau_0, T)$ with $X_t \ge \epsilon$. Letting $\epsilon \downarrow 0$ along a sequence, we get that almost surely on $\{\tau_0 < T\}$, there does not exist $t \in [t_0, T)$ with $X_t > 0$. This is the desired conclusion.