$E(M_{n\wedge \tau}) = E(M_0)$ implies $M$ is a martingale

Question

$(M_n)_{n\geq 0}$ is an adapted family such that $E|M_n|<\infty$, then, I need to show that $E(M_{n\wedge \tau}) = E(M_0)$ implies $M$ is a martingale, for every stopping $\tau$.

Honestly, I do not know how to start. Any hint? Thanks

Answer 1

Consider $A \in \mathcal{F}_k$, we need to show that $\mathbb{E}[\chi_A \cdot M_k] = \mathbb{E}[\chi_A \cdot M_{k+1}]$.

Consider $\tau = k + \chi_{A}$, this is clearly a stopping time. Therefore $\mathbb{E}[M_\tau] = \mathbb{E}[M_0] = \mathbb{E}[M_k]$, i.e. $$ \mathbb{E}[M_k] = \mathbb{E}[M_k \chi_{A^c} + M_{k+1}\chi_A]$$ which implies our thesis.

Answer 2

This is some kind of variation of no-arbitrage principle.
So, for any $k<n$, we define the following time: $$\tau_k:= \begin{cases} k, & \text{if}\ M_{k} \ge \mathbb{E}(M_{k+1}|F_k) \\ k+1, & \text{other wise} \end{cases} $$ Clearly, $\tau_k$ is a stopping time.
On the other hand we have:
$$ \mathbb{E}(M_{k}) = \mathbb{E}(M_0)= \mathbb{E}( M_{n \wedge \tau_k})= \mathbb{E}( M_k \mathbb{1}_{\ M_{k} \ge \mathbb{E}(M_{k+1}|F_k)} +M_{k+1} \mathbb{1}_{\ M_{k} < \mathbb{E}(M_{k+1}|F_k)} ) $$ Which implies $\mathbb{P}( M_k < \mathbb{E}(M_{k+1}|F_k) )=0$ Vice versa, we have: $$\mathbb{P}( M_k > \mathbb{E}(M_{k+1}|F_k) )=0$$ Hence, $$ \mathbb{E}(M_{k+1}|F_k)= M_k$$ Thus, the conclusion.