Expected value of a series of random variables in a markov chain

Question

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 $X_n = \max\left(X_0+\sum\limits_{i=1}^n\xi_i, \sum\limits_{i=2}^n\xi_i, \sum\limits_{i=3}^n\xi_i, \dots, 0\right)$, but I'm stuck here.

Answer 1

If I say $S_n = \sum\limits_{i=1}^n\xi_i$, I have that $X_n = \max(X_0+S_n, S_n-S_1, S_n-S_2, \dots, 0)$ and that

$$\frac{X_n}n = \max\left(\frac{X_0+S_n}n, \frac{S_n-S_1}n, \frac{S_n-S_2}n, \dots, 0\right)$$

So maybe if I take the limit for $n\rightarrow +\infty$, the above has that

$$\lim\limits_{n\rightarrow +\infty}\frac{X_n}n = \lim\limits_{n\rightarrow +\infty}\frac{S_n}n = \mathbb E(\xi_n)$$

where the last equality holds because of the law of large numbers, and since the expected value is positive, it is the maximum of that list. Is that reasoning correct?

Answer 2

Repeating the relation $X_n \geq X_{n-1} + \xi_n$ gives $X_n \geq X_0 + S_n$, where $S_n = \sum_{i=1}^{n} \xi_i$. So $\frac{X_n}{n} \geq \frac{X_0}{n} + \frac{S_n}{n}$, then the law of large number gives $\liminf_{n \rightarrow +\infty} \frac{X_n}{n} \geq E(\xi_1)$ almost surely. Since $E(\xi_1) > 0$, it ensures that almost surely for $\omega \in \Omega$, $X_n$ is strictly positive when $n$ is bigger than a certain $N(\omega)$. So after $N(\omega)$, the given iteration actually becomes $X_n = X_{n-1} + \xi_n$, then the result follows obviously