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I have a process $X_{n+1} = X_n\xi_n$ where $\xi_n\sim\mathcal N(1,1)$ and $\xi_n$ is independent of $X_n$. I need to prove that if $X_0\neq0$ then $ \mathsf P\{|X_n|>1\text{ for some }n\geq0\} = 1. $ From this I construct a random walk: $Y_n = \log|X_n|$ so $ Y_{n+1} = Y_n+\eta_n $ where $\eta_n = \log|\xi_n|$. I guess that from here I should apply the Law of Large Numbers - but I'm stacked with it. Could you help me? For now I should prove that $Y_n$ will eventually be positive a.s. starting from any point.

On the other hand, $X_n$ is a martingale which maybe also useful for deriving the desired result. If it helps, one can take $\xi_n\sim\mathcal N(m,1)$ for some $m\geq1$.

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    @Didier: Does the way of taking this integral matter for this question?2011-07-04

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Didier Piau perfectly showed an equivalence of this problem and unboundness of a random walk and also gave a solution for the latter problem in this question: When random walk is upper unbounded

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Hint:

$E(x_{n} | x_{n-1}) = x_{n-1} E(\xi_{n}) = x_{n-1}$

$E(x_n) = E( E(x_{n} | x_{n-1})) = E(x_{n-1})$

Hence $E(x_n)=x_0$

Doing the same for $E(x_n^2)$ we can show that the variance tends to infinity with $n$. Then, you are done (we are not done, we need more than this).

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    MMmm no, come to think of it, we need something stronger than that. For example, a Cauchy variable has infinite variance but the probability that it's modulus is above 1 is certainly not 1. So my "you're done" sentence is false.2011-07-04
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What is the relationship between the $\xi_n$ sequence and the $X_n$ sequence? The statement is clearly false when $\xi_n = 1/X_n$.

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    $\xi_n\perp \mathcal F_n = \sigma\{X_k,k\leq n\}$ and $\xi_n$ are iid r.v.2011-07-05