Using CBS-principle to show submartingales

Question

I need to give a proof for some lemma which states the following:

If $(X_n,\mathbb{F}_n)_{n\geq 1}$ is a submartingale and $\tau$ is a stopping time, then $(X_{\tau\wedge n},\mathbb{F}_{n})_{n\geq 1}$ is also a submartingale.

Now the CBS-principle ("Can't Beat the System") is a result. I've shown that if $(X_n,\mathbb{F}_n)_{n\geq 1}$ is a submartingale, and $(C_n)_{n\geq 1}$ is predictable and satisfies that each $C_n$ is non-negative and boundend, then $\big((C\cdot X)_n,\mathbb{F}_n\big)_{n\geq 1}$ is a submartingale.

I know that I can write $(C\cdot X)_n$ as: $$(C\cdot X)_n = \sum_{i=2}^n C_i\Delta X_i$$ Since $(C\cdot X)_1 = 0$ by assumption and $\Delta$ is just the standard difference operator.

Now how do I proceed from this? I guess I have to find some relation between $(C\cdot X)_n$ and $X_{\tau\wedge n}$.

Answer 1

Hint:

You need to pick the right $C$:

-Keep adding the increments for $i$ satisfying $\{i\le\tau\}$. Note that this event is in ${\cal F}_{i-1}$.

-Don't add any increments for larger $i$.

In other words, take $C_i = {\bf 1}_{\{i\le \tau\}}$, and then $(C\cdot X)_n = \sum_{i=2}^n {\bf 1}_{\{i\le \tau\}} (X_i -X_{i-1}) = \sum_{i=2}^{\tau \wedge n} (X_i -X_{i-1}) = X_{n \wedge \tau} -X_1$. It therefore follows that $X_{n \wedge \tau} = X_1 + (C\cdot X)_n$ is a martingale ($X_1$ is measurable with respect to ${\cal F}_n$ for all $n\ge 1$).