If $X_n$ is a submartingale, and $M,N$ are stopping times with $M \leq N$, then $X_M \leq E(X_{N} \mid \mathcal{F}_M)$

Question

I'm trying to show that if $X_n$ is a submartingale, with stopping times $M \leq N$, $P(N\leq k) =1$, then $E(X_{N} \mid \mathcal{F}_M) \geq X_M$.

The hint given is to use that, for any $A \in \mathcal{F}_M$, $$K_A := \begin{cases}M&\text{on }A\\ N&\text{on }A^c \end{cases},$$ is a stopping time.

I was able to show that $K_A$ is indeed a stopping time, but I'm not sure how to proceed, or how to use this stopping time

Answer 1

Hints:

1. Recall (or show) that $\mathbb{E}(X_{\tau}) \geq \mathbb{E}(X_{\sigma})$ for any bounded stopping times $\sigma \leq \tau$.
2. Apply Step 1 to the stopping times $\sigma = K_A$ (defined in your question) and $\tau=N$. Conclude that $$\mathbb{E}(X_N 1_A) \geq \mathbb{E}(X_M 1_A), \qquad A \in \mathcal{F}_M.$$
3. Using the definition of conditional expectation, show that $$\mathbb{E}(X_N 1_A) = \mathbb{E} \big[ \mathbb{E}(X_N \mid \mathcal{F}_M) 1_A \big], \qquad A \in \mathcal{F}_M.$$
4. By Step 2 and 3, it holds that $$\mathbb{E} \big[ \mathbb{E}(X_N \mid \mathcal{F}_M) 1_A \big] \geq \mathbb{E}(X_M 1_A)$$ for all $A \in \mathcal{F}_M$. Conclude that $\mathbb{E}(X_N \mid \mathcal{F}_M) \geq X_M$. (Hint: Consider $A=\{\mathbb{E}(X_N \mid \mathcal{F}_M) < X_M\}$.)