Continuous nonnegative supermartingale vanishes after it hits zero

Question

Let $(X_t)$ be a continuous nonnegative supermartingale and $T = \inf\{t\geq 0 \colon X_t = 0 \}$ then $X_t = 0$ for every $t\geq T$.

Idea of solution:

Since $T$ is stopping time, by Doob theorem: $$E(X_{T+q} 1_{T < \infty} | F_T) \leq X_T 1_{T < \infty} =0 $$ for every $q \in \mathbb{Q}$

Then taking expectation we have that $E(X_{T+q}1_{T < \infty}) \leq 0$ . Since $X_{T+q}1_{T < \infty} $ is positive $X_{T+q}1_{T < \infty} = 0$ a.e. (on a set $\Omega_q$ with $\mathbb{P}(\Omega_q)=1$). Taking $\Omega'=\cap_{q \in \mathbb{Q}^+} \Omega_q$ we will have that $$X_{T+t}1_{T < \infty} = 0$$ on $\Omega'$ for every $t\geq 0$.

The only problem is we are not allow to use Doob theorem since $T$ is not bounded and $X$ is not U.I. I try to use $T \wedge k $ to make the stoping time bounded but I couldn't take limit properly.

Answer 1

We can prove this result just using conclusions in Karatzas&Shreve Brownian Motion and Stochastic Calculus.

By Problem 1.3.16, we have $X_{\infty}\left(\omega\right)={\displaystyle \lim_{t\rightarrow\infty}X_{t}\left(\omega\right)}$ exists for a.e. $\omega\in\Omega$, and $E\left(\left|X_{\infty}\right|\right)<\infty$ and $\left\{ X_{t},\mathscr{F}_{t}:0\le t\le\infty\right\}$ is a supermartingale. Then clearly we have $\left\{ -X_{t},\mathscr{F}_{t}:0\le t\le\infty\right\}$ is a right-continuous submartingale.

Firstly, using Problem 1.2.7, since $X_{t}$ is continuous and $\left\{ 0\right\} $ is closed, $T$ is a stopping time. Then using Lemma 1.2.8, for all $0\le t<\infty$, $T+t$ is also a stopping time. Note that $T$ is $\mathscr{F}_{T}$-measurable.

Then using Theorem 1.3.22, we have $E\left[-X_{T+t}\chi_{\left\{ T<\infty\right\} }|\mathscr{F}_{T}\right]=\chi_{\left\{ T<\infty\right\} }E\left[-X_{T+t}|\mathscr{F}_{T}\right]\geq-\chi_{\left\{ T<\infty\right\} }X_{T}=0$

hence $E\left[X_{T+t}\chi_{\left\{ T<\infty\right\} }|\mathscr{F}_{T}\right]\le0 $for all $0\le t<\infty$. Then we have $E\left[X_{T+t}\chi_{\left\{ T<\infty\right\} }\right]\le0$. Since $X_{T+t}\chi_{\left\{ T<\infty\right\} }$ is nonnegative, we have $E\left[X_{T+t}\chi_{\left\{ T<\infty\right\} }\right]=0$. Hence $X_{T+t}\chi_{\left\{ T<\infty\right\} }=0$ a.e on $\left\{ T<\infty\right\} $ for all $0\le t<\infty$.

Then $X_{T+q}\chi_{\left\{ T<\infty\right\} }=0$ for all $q\in\left[0,\infty\right)\cap\mathbb{Q} $ holds a.s. on $\left\{ T<\infty\right\} $, i.e. there exists $A\subset\Omega$ with $P\left(A\right)=1$ such that for all $\omega\in A$ and $q\in\left[0,\infty\right)\cap\mathbb{Q}$, $X_{T+q}\chi_{\left\{ T<\infty\right\} }\left(\omega\right)=0$. Since $X$ has continuous path, for all $\omega\in A$ and $t\in\left[0,\infty\right)$, $X_{T+t}\chi_{\left\{ T<\infty\right\} }\left(\omega\right)=0$, that is $X_{T+t}=0$; $0\le t<\infty$ holds a.s. on $\left\{ T<\infty\right\}$.

Answer 2

I will give simple proof for discrete time case. Someone else might extend it for contiunous time super-martingales.

First note that for any $A \in \mathcal{F}_k$ we have $$ E(M_t I_A)\le E(M_k I_A)$$

Now write $$ \begin{align} E(M_{T+x}) &= E(\lim_{n \rightarrow \infty}\sum_{k}^{n} M_{k+x} I_{T=k}) \\ &\le \liminf_{n\rightarrow \infty} \sum_{k}^{n} E(M_{k+x} I_{T=k}) \\ &\le \liminf_{n\rightarrow \infty} \sum_{k}^{n} E(M_{k} I_{T=k}) = 0 \end{align} $$ Here we used our first inequality with $A=I_{T=k}$, and Fatou’s lemma.

Hence, $0\le E(M_{T+k}) \le 0$, therefore $M_{T+k}=0$ almost surely for every k.