Expected value of a series of random variables in a markov chain I have a Markov Chain such that $X_n = max(X_{n-1}+\xi _n,0)$ where the $\xi_n$ series is independent and identically distributed. I want to show that if $\mathbb E(\xi_n) > 0$ (where  $\mathbb E(\xi_n)$ is the expected value of $\xi_n$) then$\frac{X_n}{n}$ tends to $\mathbb E(\xi_n)$ as n approaches infinity for any choice of $X_0$.
And... I have no idea how to show this. I know that repeated application of the above will yield $X_n = max(X_0+\sum\limits_{i=1}^n\xi_i, \sum\limits_{i=2}^n\xi_i, \sum\limits_{i=3}^n\xi_i, ..., 0)$, but I'm stuck here.
 A: $$X_n = \max(X_0+\sum\limits_{i=1}^n\xi_i, \sum\limits_{i=2}^n\xi_i, \sum\limits_{i=3}^n\xi_i, ..., 0)$$
$$\Rightarrow \frac 1nX_n = \max\left(\frac 1n(X_0+\sum\limits_{i=1}^n\xi_i), \frac 1n\sum\limits_{i=2}^n\xi_i, \frac 1n\sum\limits_{i=3}^n\xi_i, ..., \frac 1n\cdot0\right)$$
$$\Rightarrow \operatorname{plim} \frac 1nX_n = \operatorname{plim}\max\left(\frac 1n(X_0+\sum\limits_{i=1}^n\xi_i), \frac 1n\sum\limits_{i=2}^n\xi_i, \frac 1n\sum\limits_{i=3}^n\xi_i, ..., \frac 1n\cdot0\right)$$
The $\max$ is a continuous function. Also, since the $\xi$-series is i.i.d, the Law of Large Numbers holds. Then 
$$\operatorname{plim} \frac 1nX_n = \max\left(\operatorname{plim}\frac 1n(X_0+\sum\limits_{i=1}^n\xi_i), \operatorname{plim}\frac 1n\sum\limits_{i=2}^n\xi_i, \operatorname{plim}\frac 1n\sum\limits_{i=3}^n\xi_i, ..., \operatorname{plim}\frac 1n\cdot0\right)$$
$$\rightarrow_{p} \max\left(\frac 1n\sum\limits_{i=1}^nE(\xi_i), \frac 1n\sum\limits_{i=2}^nE(\xi_i), \frac 1n\sum\limits_{i=3}^nE(\xi_i), ..., 0\right)$$
$$= \max\left(\frac 1nnE(\xi), \frac 1n(n-1)E(\xi), \frac 1n(n-2)E(\xi),..., 0\right) = E(\xi)$$
under the assumption that $E(\xi)>0$. QED
