$(X_{S}, X_{T})$ 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 $$\mathbb{E}(X_T \mid \mathcal{F}_S) = X_S.$$

Answer 2

Let $\left(\Omega, \mathcal{F},(\mathcal{F}_{n}), \mathbb P\right)$ be a filtered probability space.

If $(X_{n})$ is a sub-martingale and $S \leqslant T$ are bounded stopping times. Then $\mathbb E [ X_T | \mathcal F_S ] \ge X_S$.

First, we need the following lemmas:

$\textbf{Lemma 1}$ If $(X_{n})$ is a sub-martingale and $T$ is a stopping time. Then $X_{n \wedge T}$ is $\mathcal F_n$-measurable and $\mathbb E [ X_{(n+1) \wedge T} | \mathcal F_n ] \le X_{n \wedge T}$.

$\textbf{Proof}$ We have $X_{n \wedge T}$ is $\mathcal F_{n \wedge T}$-measurable and $\mathcal F_{n \wedge T} \subseteq \mathcal F_n$. Then $X_{n \wedge T}$ is $\mathcal F_n$-measurable. We have $$\begin{aligned} \mathbb E [ X_{(n+1) \wedge T} | \mathcal F_n ] &= \mathbb E [ X_{(n+1) \wedge T} \mathbf{1}_{\{T \le n\}} + X_{(n+1) \wedge T} \mathbf{1}_{\{T > n\}} | \mathcal F_n ] \\ &= \mathbb E [ X_{(n+1) \wedge T} \mathbf{1}_{\{T \le n\}} | \mathcal F_n ] +\mathbb E [ X_{(n+1) \wedge T} \mathbf{1}_{\{T > n\}} | \mathcal F_n ] \\ &= \mathbb E [ X_{n \wedge T} \mathbf{1}_{\{T \le n\}} | \mathcal F_n ] + \mathbb E [ X_{n+1} \mathbf{1}_{\{T > n\}} | \mathcal F_n ] \\ \end{aligned}$$

It follows from $X_{n \wedge T}$ is $\mathcal F_n$-measurable and $\{T \le n\} \in \mathcal F_n$ that $X_{n \wedge T} \mathbf{1}_{\{T \le n\}}$ is $\mathcal F_n$-measurable. Hence $ \mathbb E [ X_{n \wedge T} \mathbf{1}_{\{T \le n\}} | \mathcal F_n ] = X_{n \wedge T} \mathbf{1}_{\{T \le n\}}$.

It follows from $\{T > n\} \in \mathcal F_n$ that $\mathbf{1}_{\{T > n\}}$ is $\mathcal F_n$-measurable. Hence $\mathbb E [ X_{n+1} \mathbf{1}_{\{T > n\}} | \mathcal F_n ] = \mathbf{1}_{\{T > n\}} \mathbb E [ X_{n+1} | \mathcal F_n ]$. On the other hand, $(X_n)_{n \in \mathbb N}$ is a sub-martingale. So $\mathbb E [ X_{n+1} | \mathcal F_n ] \le X_n$.

As such, we get $$\begin{aligned} \mathbb E [ X_{(n+1) \wedge T} | \mathcal F_n ] &\le X_{n \wedge T} \mathbf{1}_{\{T \le n\}} + \mathbf{1}_{\{T > n\}} X_n \\ &= X_{n \wedge T} \mathbf{1}_{\{T \le n\}} + \mathbf{1}_{\{T > n\}} X_{n \wedge T} \\ &= X_{n \wedge T} \end{aligned}$$ $\blacksquare$

$\textbf{Lemma 2}$ If $\mathcal G$ is a sub $\sigma$-algebra of $\mathcal F$ and $X:\Omega \to \mathbb R$ is $\mathcal G$-measurable. Then $$X \ge 0 \quad \text{a.s} \iff \forall \Lambda \in \mathcal G: \mathbb E [\mathbf{1}_\Lambda X] \ge 0$$

$\textbf{Proof}$

$\Longrightarrow$ Because $X \ge 0$ a.s, $\mathbf{1}_\Lambda X \ge 0$ a.s. The claim then follows.

$\Longleftarrow$ Assume the contrary that there exists $\Lambda \in \mathcal A$ such that $\mu (\Lambda) >0$ and $X(\omega) <0$ for all $\omega \in \Lambda$. Then $\mathbb E (\mathbf{1}_\Lambda X) = \int_\Omega \mathbf{1}_\Lambda X \, d \mu <0$, which is impossible. $\blacksquare$

$\textbf{Lemma 3}$ If $(X_{n})$ is a sub-martingale and $S \le T$ are bounded stopping times. Then $\mathbb E [ X_T -X_S ] \ge 0$.

$\textbf{Proof}$ We have $$\begin{aligned} \mathbb E [ X_T -X_S ] &= \mathbb E \left [ (X_T -X_S) \sum_{m=0}^N \mathbf{1}_{\{S=m\}} \right ] \\ &= \sum_{m=0}^N \mathbb E \left [ \mathbf{1}_{\{S=m\}} (X_T -X_m) \right ] \\ &\overset{(1)}{=}{} \sum_{m=0}^N \mathbb E \left [\mathbf{1}_{\{S=m\}} \sum_{n=m}^N (X_{T \wedge (n+1)} -X_{T \wedge n}) \right ] \\ &= \sum_{m=0}^N \sum_{n=m}^N \mathbb E \left [\mathbf{1}_{\{S=m\}} (X_{T \wedge (n+1)} -X_{T \wedge n}) \right ] \\ &\overset{(2)}{=}{} \sum_{m=0}^N \sum_{n=m}^N \mathbb E \left [\mathbf{1}_{\{S=m\}} \mathbb E \left [X_{T \wedge (n+1)} - X_{T \wedge n} | \mathcal F_{n} \right ] \right ] \\ \end{aligned}$$

(1) In case $n=N$, we have $X_{T \wedge (n+1)} = X_T$. In case $n=m$, we have $T \ge S = m$ and thus $X_{T \wedge n} = X_m$.

(2) Because $n \ge m$, $\{S=m\}$ is $\mathcal F_n$-measurable.

By $\textbf{Lemma 1}$, $\mathbb E \left [X_{T \wedge (n+1)} - X_{T \wedge n} | \mathcal F_{n} \right ] = \mathbb E \left [X_{T \wedge (n+1)}| \mathcal F_{n} \right ] - X_{T \wedge n} \ge 0$. Hence $\mathbb E [ X_T -X_S ] \ge 0$. $\blacksquare$

Now come back to our main theorem. By $\textbf{Lemma 2}$, it suffices to prove that $\forall \Lambda \in \mathcal F_S: \mathbb E [\mathbf{1}_\Lambda \mathbb E [ X_T -X_S | \mathcal F_X]] \ge 0$ or equivalently $\forall \Lambda \in \mathcal F_S: \mathbb E [\mathbf{1}_\Lambda ( X_T -X_S )] \ge 0$.

We define $S'$ by $$S'(\omega) =\begin{cases} S (\omega) &\text{if} \quad \omega \in \Lambda \\ T (\omega) &\text{if} \quad \omega \notin \Lambda \end{cases}$$

It's easy to verify that $S'$ is a bounded stopping time such that $S' \le T$ and $\mathbb E [\mathbf{1}_\Lambda ( X_T -X_S )] = \mathbb E [ X_T -X_{S'} ]$. We have $\mathbb E [ X_T -X_{S'}] \ge 0$ by $\textbf{Lemma 3}$ and thus $\mathbb E [\mathbf{1}_\Lambda ( X_T -X_S )] \ge 0$. This completes the proof. $\blacksquare$