Equivalent condition for $ (X_n, \mathcal{F}_n) $ to be a martingale

Question

I've encountered an interesting problem and am not quite able to solve it. It is to prove the following statement ($ X_n $ denotes a sequence adapted to a filtration $\mathcal{F}_n) $:

$$ (X_n,\mathcal{F}_n) \text{ is a martingale} \iff \text{ for any bounded stopping moment } \tau ~ \mathbb{E} X_\tau = \mathbb{E}X_0$$

I'm not quite sure where to begin with - it's possible to prove that for a martingale it holds that

$$ \mathbb{E}\left( X_n - X_{n-1}\right) = 0 $$

but $ \tau $ is a random variable, so I can't think of a way to skip that;

I would appreciate some help

Answer 1

Hints:

• "$\Rightarrow$": Use Doob's optional stopping theorem and the fact that $\tau$ is bounded.
• "$\Leftarrow$": It holds that $$\mathbb{E}(X_{\tau}) = \mathbb{E}(X_{\sigma})$$ for any two bounded stopping times $\tau$, $\sigma$. Show that $$\tau :=n \cdot 1_{B} \qquad \quad \sigma := k \cdot 1_{B}$$ are bounded stopping times for $k \leq n$ and $B \in \mathcal{F}_k$. Conclude from $$\mathbb{E}(X_n 1_B) = \mathbb{E}(X_k 1_B), \qquad \text{for all} \, \, k \leq n$$ that $(X_n,\mathcal{F}_n)_{n \geq 0}$ is a martingale.