$(X_{S}, X_{T})$ is a sub-martingale with respect to the filtration $(\mathcal{F}_{S}, \mathcal{F}_{T})$

Question

I'm reading a theorem about stopping time in my lecture note:

Let $\left(\Omega, \mathcal{F},(\mathcal{F}_{n})_{n \in \mathbb{N}}, \mathbb P\right)$ be a filtered probability space, $(X_{n}, \mathcal{F}_{n})_{n \in \mathbb N}$ be a sub-martingale, and $S \le T$ be bounded stopping times. Then $(X_{S}, X_{T})$ is a sub-martingale with respect to the filtration $(\mathcal{F}_{S}, \mathcal{F}_{T})$.

The usual definition of a sub-martingale is a sequence of random variables, but $(X_{S}, X_{T})$ is one pair of random variables. Similarly, the filtration $(\mathcal{F}_{S}, \mathcal{F}_{T})$ is one pair of $\sigma$-algebra. This notation is very different than what I've seen so far.

Could you please elaborate in this point? Many thanks!

Answer 1

You need to verify the following properties:

• $X_S$ (resp. $X_T$) is measurable w.r.t to $\mathcal{F}_S$ (resp. $\mathcal{F}_T$),
• $X_T$ is integrable,
• $\mathbb{E}(X_T \mid \mathcal{F}_S) = X_S$.

Hints: Since $S \leq T$ are bounded stopping times, there exists some $N \in \mathbb{N}$ such that $S \leq T \leq N$.

1. To prove measurability, you need to verify that $$\{X_S \in B\} \cap \{S \leq n\} \in \mathcal{F}_n$$ for all $n \in \mathbb{N}_0$ and Borel sets $B$. To this end, write $$\{X_S \in B\} \cap \{S \leq n\} = \bigcup_{k=0}^n \{X_k \in B\} \cap \{S=k\}.$$ (Clearly, an analogous statement holds for $T$.)
2. To prove integrability, use that $$X_T = \sum_{k=0}^N X_k 1_{\{T=k\}}$$ and the fact that $\mathbb{E}(|X_k|)<\infty$ for each $k$.

It remains to prove the third property (i.e. to compute the conditional expectation). We will use the following statement:

Lemma: Let $(M_n)_{n \in \mathbb{N}}$ be a martingale. If $T$ a bounded stopping time, then $\mathbb{E}(M_T) = \mathbb{E}(M_0)$.

3. Use the Doob decomposition of the submartingale $(X_n)_{n \in \mathbb{N}}$ and the above lemma to show that $\mathbb{E}(X_T) \geq \mathbb{E}(X_S)$.
4. Fix $F \in \mathcal{F}_S \subseteq \mathcal{F}_T$. Show that $\varrho := S 1_F + T 1_{F^c}$ defines a stopping time satisfying $\varrho \leq T$. Apply the previous step to the stopping times $\varrho$, $T$ to obtain that $$\mathbb{E}(X_\varrho) = \mathbb{E}(X_T).$$ Rearrange the terms on both sides to conclude that $$\mathbb{E}(X_S 1_F) = \mathbb{E}(X_T 1_F).$$ Since $F \in \mathcal{F}_S$ is arbitrary and $X_S$ is $\mathcal{F}_S$-measurable, this proves $$\mathbb{E}(X_T \mid \mathcal{F}_S) = X_S.$$