If we have that $\tau \leq \nu$ are stopping times and that $X_n$ is a submartingale, how to show $E(X_{\tau \wedge n}) \leq E(X_{\nu \wedge n})$

Question

If we have that $\tau, \nu$ are stopping times and that $\tau \leq \nu$, if $X_n$ is a submartingale, how can I show that $E(X_{\tau \wedge n}) \leq E(X_{\nu \wedge n})$? Is there a way to characterize it by a Martingale transform or is there a simpler idea? thanks

Answer 1

Define $C_n=1\{\tau<n\le\nu\}$ ($C_n$ is a predictable process $\because\{n\le \nu\},\{n\le \tau\}\in \mathcal{F}_{n-1}$). Then

$$ (C\cdot M)_n=\sum_{k=1}^n1\{\tau<k\le \nu\}(M_k-M_{k-1})=M_{\nu\wedge n}-M_{\tau\wedge n} $$

and since $(C\cdot M)_n$ is a submartingale we have

$$ \mathbb{E}(C\cdot M)_n\ge\mathbb{E}(C\cdot M)_0=0. $$

Referring to your previous question, assume $\tau\le \nu\le N$ (are bounded). Then

$$ \mathbb{E}[M_{\nu}-M_{\tau}]=\mathbb{E}(C\cdot M)_N\ge 0. $$

Answer 2

You actually can reduce this problem to a martingale transform. What you need to prove is that $$X_{\tau \wedge n} \le E(X_{\nu \wedge n} | \mathcal{F}_{\tau \wedge n})$$

Then the result will follow by inegrating. To prove this consider $A \in \mathcal{F}_{\tau \wedge n}.$

Now consider $H_k = 1_A 1_{\{\tau \wedge n < k \le \nu \wedge n\}}$, this is a predicible process. Then $(H \cdot X)_k$ is a submartingale by martingale transform, and $$(H \cdot X)_n = X_{\nu \wedge n}1_A - X_{\tau \wedge n}1_A$$ By integrating you get what you need.