Verify that $X_n$ is a martingale

Solution 1:

Hope this helps.

For part (a), to check that $E[X_{n+1}|\mathscr{F}_n]=X_n$, we have that the process is Markov with discrete state space, so we can express the conditional expectation of $X_{n+1}$ wrt $\mathscr{F}_n$ as $$E[X_{n+1}|\mathscr{F}_{n}]=\sum_{k\in \mathbb{N}\cup \{0\}}E[X_{n+1}|X_{n}=k]\mathbf{1}_{\{X_n=k\}}$$ and (see Poisson distribution) $$E[X_{n+1}|X_{n}=k]=\sum_{j\in \mathbb{N}}j\frac{e^{-k}k^j}{j!}=k$$ So $$E[X_{n+1}|\mathscr{F}_{n}]=\sum_{k\in \mathbb{N}}k\mathbf{1}_{\{X_n=k\}}=X_n$$

For part (b) we use Doob's maximal inequality. Suppose $X_0=j$. Recall that a martingale is also a supermartingale so $$aP\bigg(\max_{k\leq n}X_k\geq a\bigg)\stackrel{\textrm{Doob}}{\leq} E[X_n]=j$$ and (consider that the state space is discrete) $$\bigg\{\sup_{n \in \mathbb{N}}X_n\geq a\bigg\}=\bigg\{\sup_{n \in \mathbb{N}}\max_{k\leq n}X_k\geq a\bigg\}=\bigcup_{n \in \mathbb{N}}\bigg\{\max_{k\leq n}X_k\geq a\bigg\}$$ and since $\{\max_{k\leq n}X_k \geq a\}\subseteq \{\max_{k\leq n+1}X_k\geq a\},\,\forall n$, we get by continuity of measures $$aP\bigg(\sup_{n \in \mathbb{N}}X_n\geq a\bigg)\leq j$$

For part (c) I will just present an informal argument. I will assume that the stationary distribution $\pi$ exists and we find it: $$\pi_j=\sum_{k \in \mathbb{N}\cup\{0\}}P_{kj}\pi_k\in [0,1],\,\forall j$$ consider that $$\pi_0=\sum_{k \in \mathbb{N}\cup\{0\}}e^{-k}\pi_k=\pi_0+\pi_1e^{-1}+(...)$$ By subtracting $\pi_0$ from both sides, we obtain that $\pi_k=0,\,\forall k >0$, so $\pi_0=1$ necessarily.