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Let $\left(\theta_{n}\right)_{n\in\mathbb{N}}$ be a sequence of random variables iid with uniform distribution in $\left\{-1,1\right\}$. We consider $X_{n}$ recurrent random sequence given by $$X_{n}=a X_{n-1}+\theta_{n}$$ where $a\in (1,\infty)$.

Show that $$\mathrm{P}\left( \lim_{n\rightarrow \infty} X_{n} \in \left\{-\infty,\infty \right\} \right)=1.$$

Remark: I need a suggestion to prove this fact.

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    Are you sure that the recurrence goes backwards like that, giving $X_n$ in terms of $X_{n+1}$? – 2017-02-13
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    I haven't worked with these before, but you get easily that $\mathbb E[X_{n+1}] = a\mathbb E[X_n]$, so if $\mathbb E[X_0] = c$, then $\mathbb E[X_n] = ca^n$, so the expectation approaches infinity/negative infinity depending on whether $E[X_0]$ is positive/negative. The only way this can happen is if $X_n$ is either increasing unboundedly or decreasing unboundedly (as otherwise the expectation would be finite). This is assuming @ByronSchmuland 's comment is correct, as otherwise $\lim_{n\to\infty}\mathbb E[X_n] = 0$. – 2017-02-13
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    "The only way this can happen is if Xn is either increasing unboundedly or decreasing unboundedly (as otherwise the expectation would be finite)" This is really not true. The pathwise behaviour can be oscillatory and yet the expectations, go to infinity. – 2017-02-13

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Define a random variable $Y = \sum_{k=0}^{\infty} a^{-k}\theta_k$. Note that this sum converges almost surely since $|\theta_k|<1$ for all $k$, and because $\sum_k a^{-k}=a/(a-1)<\infty$.

Next, notice by induction that $X_n = a^n \theta_0+ a^{n-1} \theta_1 + ... + a \theta_{n-1} + \theta_n$. Therefore $a^{-n}X_n = \sum_{k=0}^n a^{-k}\theta_k$.

So we see that $|a^{-n}X_n-Y| \leq \sum_{k=n+1}^{\infty} a^{-k}|\theta_k| \leq \sum_{k=n+1}^{\infty} a^{-k}= a^{-n}/(1-a)$. Multiplying both sides by $a^n$, we see that $|X_n-a^nY| \leq 1/(1-a)=:C$ almost surely.

Hence, on the event $\{Y>0\}$, we see that $X_n>a^nY - C$ which diverges to $+\infty$ as $n \to \infty$ (since $Y>0$ and $a^n \to \infty$). Similarly, on the event $\{Y<0\}$, we see that $X_n < C - a^n(-Y)\to -\infty$. Now, it is not difficult to see that $Y$ has no atoms, so that $P(Y=0)=0$. So we are finished.

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    You say that $|\theta_k|<1$, but $\theta_k \in \left\{-1,1\right\}$. – 2017-02-13
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    @diegofonseca I meant that $|\theta_k|\leq 1$. Note that it makes absolutely no difference in the proof. – 2017-02-14