# How to prove the Martingale's property if we have a special stopping time?

$$(X_n)$$ is a sequence of $$(F_n)$$-adapted integrable random variables, where $$(F_n)$$ is a Filtration and $$X_0=0$$.

I have to prove that

1)$$X_n$$ is a martingale with a respect to $$F_n$$

iff

2)for any finite stopping-time $$t$$ that takes at most two values, we have $$E[X_t]=0$$

what i have:

"1) to 2)" we can prove that $$X_n$$ is uniformly integrable and then use Optional Stopping Theorem to show that $$E[X_t]=E[X_0]=0.$$

"2) to 1)": we have that $$X_n$$ is integrable and adapted.

How can i show the Martingale property of $$(X_n)$$ here, and am i wrong in the first part?

1. Try first stopping times taking just one value; say, $$t=n$$ for a positive integer $$n$$. (Implicit in 2) is the integrability of each random variable $$X_n$$.)
2. Let $$m be non-negative integers and let $$A\in F_m$$. Consider the stopping time $$t(\omega)=\cases{ m,&\omega\in A;\cr n,&\omega\in A^c.\cr}$$
• You get $E[X_0]=0$, which is part of being a martingale (since $X_0=0$). Commented Aug 28, 2022 at 19:55
• Using the stopping time $t$ suggested, from $E[X_t]=0$ you deduce that $$0=E[X_m\cdot1_{A^c}]+E[X_n\cdot 1_A]=-E[X_m\cdot1_{A}]+E[X_n\cdot 1_A],$$ where the second equality uses $E[X_m]=0$. You have deduced that $$E[X_n\cdot 1_A]=E[X_m\cdot1_{A}],$$ for each event $A\in F_m$. Refer now to your definition of conditional expectation to see that this implies (indeed it equivalent to) $$E[X_n\mid F_m]=X_m.$$ Commented Aug 28, 2022 at 20:02