Show $\sup_{s \in [0,t]} \mathbb{E}|X_s|<\infty$ for any submartingale $(X_t)_{t \geq 0}$

Question

My problem:

For any submartingale $(X_s)_{s\geq0}$ and for all $t\geq0$ show that $\sup_{s\in[0,t]}\mathbb{E}[|X_s]|]$ is finite.

What I have until now: I know that $\mathbb{E}[X_s]$ is increasing hence the supremum of the expected values is bounded by $\mathbb{E}[X_t]=\mathbb{E}[X_t^+]-\mathbb{E}[X_t^-]$. I also know that therefore $\mathbb{E}[X_t]=\mathbb{E}[X_t^+]-\mathbb{E}[X_t^-]\geq \mathbb{E}[X_s^+]-\mathbb{E}[X_s^-]\geq0$, where the last inequality uses that the martingale may start wlog at $X_0=0$.

But how can i prevent $X_s^+$ and $X_s^-$ from both diverging to infinity s.t. their difference remains "nice"? Any attemt to estimate $\mathbb{E}[|X_s|]=\mathbb{E}[X_s^+]+\mathbb{E}[X_s^-]$ without terms depending on $s$ failed so far...

Thanks!

Answer 1

From

$$|X_s| = 2X_s^+ - X_s, \qquad s \leq t$$

we conclude

$$\mathbb{E}(|X_s|) \leq 2 \sup_{s \leq t}\mathbb{E}(X_{s}^+) - \underbrace{\mathbb{E}(X_s)}_{\geq \mathbb{E}X_0}$$

Note that $(X_t^+)_{t \geq 0}$ is a sub-martingale (since $(X_t)_{t \geq 0}$ is a sub-martingale) and therefore

$$\sup_{s \leq t} \mathbb{E}(X_{s}^+) \leq \mathbb{E}(X_t^+)< \infty$$

Hence

$$\sup_{s \leq t} \mathbb{E}(|X_s|) \leq \mathbb{E}(X_t^+) - \mathbb{E}X_0<\infty$$

Roughly speaking: Since the expectations are increasing, the expectations of the negative part cannot diverge. On the other hand, the boundedness of the expectations (on compact $t$-intervalls) implies the boundedness of the expectations of the positive parts.