Let $ X = (X_n, n \in \Bbb N_0)$ be a stochastic process with state space $\Bbb N_0$, with unitary mean value for each $n \geqslant 1$, with independent increments and such that $P(X_0 = 0) = 1.$ I wish to prove that $X$ is a martingale.
My attempt. We have that
$$ E(X_n) = 1, \quad \text{ for all } n \geqslant 1 $$
and, on the other hand, we also have that
$$ E(X_0) = P(X_0 = 0) \cdot 0 = 1 \cdot 0 =0 . $$
Hence, we conclude that $E(|X_n|) = E(X_n) < \infty$ for all $n \in \Bbb N_0$. Thus, if we wish to show that $X$ is a martingale, all we have to do is to show that $X$ satisfies the following property:
$$ E(X_{t_{n+1}} \, \, | \, \, X_{t_0} = x_0, \dots,X_{t_n} = x_n) = x_n, \quad \forall n \in \Bbb N, \, \forall t_0 < t_1 < \dots < t_{n+1} \in \Bbb N_0, \, \forall x_0,\dots,x_{n+1} \in \Bbb N_0. $$
With this in mind, I elaborated the following:
$$ E(X_{t_{n+1}} \, \, | \, \, X_{t_0} = x_0, X_{t_1} = x_1, \dots,X_{t_n} = x_n) = E(X_{t_{n+1}} \, \, | \, \, X_{t_0} = x_0, X_{t_1} - X_{t_0} = x_1-x_0, \dots,X_{t_n} - X_{t_{n-1}} = x_n - x_{n-1}) $$
Now, note that $X_{t_{n+1}} = X_{t_{n+1}} - X_{t_n} + X_{t_n}$ and thus we can write
$$ E(X_{t_{n+1}} \, \, | \, \, X_{t_0} = x_0, X_{t_1} - X_{t_0} = x_1-x_0, \dots,X_{t_n} - X_{t_{n-1}} = x_n - x_{n-1}) $$
$$ = E(X_{t_{n+1}} - X_{t_n} \, \, | \, \, X_{t_0} = x_0, X_{t_1} - X_{t_0} = x_1-x_0, \dots,X_{t_n} - X_{t_{n-1}} = x_n - x_{n-1}) $$
$$ + E(X_{t_n} \, \, | \, \, X_{t_0} = x_0, X_{t_1} - X_{t_0} = x_1-x_0, \dots,X_{t_n} - X_{t_{n-1}} = x_n - x_{n-1}).$$
Now, we know that $X$ has independent increments and thus $X_{t_1} - X_{t_0},\dots,X_{t_{n+1}} - X_{t_n}$ are independent variables and so we almost can "forget" the conditionality on the first mean value. Basically, I am having a hard time dealing with $X_{t_0},$ because without it I think I would be able to solve this.
Thanks for any help in advance.
It's much simpler than that. Let $m>n$ and note that \begin{align} E(X_m | \mathcal{F}_n ) &= E(X_n + (X_m-X_n)| \mathcal{F}_n) \\ &= X_n + E( X_m-X_n| \mathcal{F}_n) \\ &= X_n + E( X_m-X_n) \\ &= X_n \end{align} Where the third line is due to the independence of increments.