(a) What does it mean to say that $\left(M_{n}, \mathcal{F}_{n}\right)_{n \geqslant 0}$ is a martingale? (b) Let $\left(X_{n}\right)_{n \geqslant 0}$ be a Markov chain defined by $X_{0}=0$ and

$$\begin{aligned} & \mathbb{P}\left[X_{n}=0 \mid X_{n-1}=0\right]=1-\frac{1}{n} \end{aligned}$$

$$\begin{aligned} &\mathbb{P}\left[X_{n}=1 \mid X_{n-1}=0\right]=\mathbb{P}\left[X_{n}=-1 \mid X_{n-1}=0\right]=\frac{1}{2 n} \\ &\mathbb{P}\left[X_{n}=0 \mid X_{n-1}=0\right]=1-\frac{1}{n} \end{aligned}$$

and

$$\mathbb{P}\left[X_{n}=n X_{n-1} \mid X_{n-1} \neq 0\right]=\frac{1}{n}, \quad \mathbb{P}\left[X_{n}=0 \mid X_{n-1} \neq 0\right]=1-\frac{1}{n}$$

for $n \geqslant 1$. Show that $\left(X_{n}\right)_{n \geqslant 0}$ is a martingale with respect to the filtration $\left(\mathcal{F}_{n}\right)_{n} \geqslant 0$ where $\mathcal{F}_{0}$ is trivial and $\mathcal{F}_{n}=\sigma\left(X_{1}, \ldots, X_{n}\right)$ for $n \geqslant 1$.

(c) Let $M=\left(M_{n}\right)_{n \geqslant 0}$ be adapted with respect to a filtration $\left(\mathcal{F}_{n}\right)_{n \geqslant 0}$ with $\mathbb{E}\left[\left|M_{n}\right|\right]<\infty$ for all $n$. Show that the following are equivalent:

(i) $M$ is a martingale.

(ii) For every stopping time $\tau$, the stopped process $M^{\tau}$ defined by $M_{n}^{\tau}:=M_{n \wedge \tau}$, $n \geqslant 0$, is a martingale.

(iii) $\mathbb{E}\left[M_{n \wedge \tau}\right]=\mathbb{E}\left[M_{0}\right]$ for all $n \geqslant 0$ and every stopping time $\tau$.

[Hint: To show that (iii) implies (i) you might find it useful to consider the stopping time

$$T(\omega):= \begin{cases}n & \text { if } \omega \in A, \\ n+1 & \text { if } \omega \notin A,\end{cases}$$

for any $\left.A \in \mathcal{F}_{n .} .\right]$