I am trying to solve following problem.
Problem. Suppose that $X_n = \rho X_{n-1} + \epsilon_n$ with $|\rho| < 1$ and $X_0 = 0$, where $\epsilon_n$ are iid r.v.'s with mean $0$ and variance $1$. Show that $\max_{1\le k \le n} |X_k|/\sqrt{n} \to 0~$ a.s.
My idea is using Borel-Canteli lemma to show the a.s. convergence.
Since
$$\max_{1\le k \le n} |X_k|/\sqrt{n} \to 0 ~\text{ a.s. } \iff
\forall\epsilon > 0: P(\max_{1\le k \le n} |X_k|/\sqrt{n} > \epsilon ~\text{ i.o.}) = 0,$$
and $$\{\max_{1\le k \le n} |X_k|/\sqrt{n} > \epsilon ~\text{ i.o.}\} = \{ |X_n|/\sqrt{n} > \epsilon ~\text{ i.o.}\},$$
(is it true?)
I think that it is enough to show that
$|X_n|/\sqrt{n} \to 0 ~\text{ a.s. }$
To apply BC lemma, I am trying to bound $\sum_n P(|X_n|/\sqrt n > \epsilon)$. But Markov inequality only shows that
$$
P(|X_n|/\sqrt n > \epsilon) \le \frac{var(X_n)}{n\epsilon^2} = \frac{1-\rho^{2n}}{(1-\rho^2)n\epsilon^2}.
$$
But this cannot bound $\sum_n P(|X_n|/\sqrt n > \epsilon)$. How can I precede? Shall we need a 4th finite moment for $X_n$?
Best Answer
Hints: $X_n=\rho^{n-1}\epsilon_1+\rho^{n-2} \epsilon_2+...+\epsilon_n$ by iteration. This shows that $X_n$ converges a.s.. [ You can use Kolmogorov's Three Series Theorem, for example]. Note that if a sequence $(x_n)$ of real numbers is bounded then $\max \{|x_1|,|x_2|,...|x_n|\} / \sqrt n \to 0$.