A Càdlàg non-negative submartingale is class DL

Question

Let X = $\{X_t\}_t$ be a right-continuous sub-martingale w.r.t. its natural filtration, and assume $X \geq 0 $ a.s. I am trying to show that X is of class DL.

According to my definition, this means the family $\{X_T\}_{T \in J_a}$ is uniformly continuous, where $J_a = \{T \text{ is a stopping time} : \mathbb{P}(T \leq a) = 1 \}$.

I proceeded via the optional stopping theorem for bounded stopping times. So,

$\mathbb{E}(X_T1_{\{X_T>\lambda\}}) \leq \mathbb{E}(X_a1_{\{X_T>\lambda\}})$ for any $ a> 0, \lambda >0, T \in J_a$.

And by Markov's inequality, $\mathbb{P}(X_T > \lambda) \leq \frac{\mathbb{E}(X_T)}{\lambda} \leq \frac{\mathbb{E}(X_a)}{\lambda} $

How can I then conclude $ \lim_{\lambda\to\infty} \sup_{T \in J_a} \int_{\{X_T > \lambda\} }X_T d\mathbb{P} = 0 $ ?

Answer 1

You want to show $$\lim_{\lambda \rightarrow \infty} \sup_{T \in J_a} \mathbb{E}[X_T 1_{X_T > \lambda}] = 0.$$

Using that $\mathbb{E}[X_T1_{X_T > \lambda}] \le \mathbb{E}[X_a 1_{X_T > \lambda}]$, we have

$$\lim_{\lambda \rightarrow \infty} \sup_{T \in J_a} \mathbb{E}[X_T 1_{X_T > \lambda}] \le \lim_{\lambda \rightarrow \infty} \sup_{T \in J_a} \mathbb{E}[X_a 1_{X_T > \lambda}].$$

Since $\mathbb{P}(X_T > \lambda) \le \frac{\mathbb{E}[X_a]}{\lambda}$, we have that $\lim_{\lambda \rightarrow \infty}\sup_{T \in J_a}\mathbb{P}(X_T > \lambda) = 0$. Combined with the fact that the singleton $\{X_a\}$ is uniformly integrable (an alternative form of the uniformly integrable condition is that for all $\varepsilon > 0$ there exists $\delta > 0$ such that $\mathbb{E}[X_a1_A] < \epsilon$ for all $A$ with $\mathbb{P}(A) < \delta$), we conclude $$\lim_{\lambda \rightarrow \infty} \sup_{T \in J_a} \mathbb{E}[X_a 1_{X_T > \lambda}] = 0.$$

More specifically, given $\varepsilon > 0$, let $\delta > 0$ be such that $\mathbb{E}[X_a1_A] < \varepsilon$ for all $A$ with $\mathbb{P}(A) < \delta$. Then for $\lambda > K := \frac{\mathbb{E}[X_a]}{\delta},$ we have $\mathbb{P}(X_T > \lambda) < \delta$ for all $T \in J_a$. Hence $\mathbb{E}[X_a1_{X_T > \lambda}] < \varepsilon$, and since this holds for all $T \in J_a$ we have $\sup_{T \in J_a}\mathbb{E}[X_a1_{X_T > \lambda}] \le \varepsilon$ for all $\lambda > K$. Since $\varepsilon$ was arbitrary, we conclude $\lim_{\lambda \rightarrow \infty} \sup_{T \in J_a} \mathbb{E}[X_a 1_{X_T > \lambda}] = 0$.

Answer 2

Let $X^*_a:=\sup_{0\le t\le a}X_t$. Doob's maximal inequality states $$ \Bbb P[X^*_a>\lambda]\le\lambda^{-1}\Bbb E[X_a],\qquad \lambda>0.\qquad\qquad (1) $$ If $T\in J_a$ then $$ X_T\le\Bbb E[X_a\mid\mathcal F_T], $$ and so $$ \Bbb E[X_T; X_T>\lambda]\le\Bbb E[X_a; X_T>\lambda]\le\Bbb E[X_a; X^*_a>\lambda].\qquad (2) $$ Given $\epsilon>0$ the integrability of $X_a$ implies the existence of $\delta>0$ such that if $\Bbb P[B]<\delta$ then $\Bbb E[X_a; B]<\epsilon$. Using (1), if $\lambda>\Bbb E[X_a]/\delta$ then $\Bbb P[X^*_a>\lambda]<\delta$ and so by (2) $$ \Bbb E[X_T; X_T>\lambda]\le\Bbb E[X_a; X^*_a>\lambda]<\epsilon, $$ for all $T\in J_a$.