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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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  • $\begingroup$ Are you sure that the normal is $\mathcal N(1,1)$ (mean 1)? $\endgroup$
    – leonbloy
    Commented Jul 4, 2011 at 14:54
  • $\begingroup$ Yes, it's not a misprint. $\endgroup$
    – SBF
    Commented Jul 4, 2011 at 14:54
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    $\begingroup$ The random variable $\log|\xi|$ is integrable. $\endgroup$
    – Did
    Commented Jul 4, 2011 at 15:09
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    $\begingroup$ So your question reduces to determining whether $E\log|\xi|$ is $\ge0$ (you win) or $<0$ (you lose). $\endgroup$
    – Did
    Commented Jul 4, 2011 at 15:26
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    $\begingroup$ WolframAlpha? Yes you lose for $m=1$, but you win for $m\ge1.3$ or something, though. $\endgroup$
    – Did
    Commented Jul 4, 2011 at 15:46

3 Answers 3

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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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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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  • $\begingroup$ I assume $\xi_n$ is independent of $x_n$ $\endgroup$
    – leonbloy
    Commented Jul 4, 2011 at 15:26
  • $\begingroup$ Yes, they are. Edited. $\endgroup$
    – SBF
    Commented Jul 4, 2011 at 15:39
  • $\begingroup$ Is the r.v. $\xi_n$ independent of r.v. $X_n$ or the sequence $\{\xi_n\}$ independent of the sequence $\{X_n\}$? These are different conditions. Also, are the $\xi_n$ assumed to be i.i.d? Again, the statement can be false if the $\xi_n$ have some sort of relationship among themselves. $\endgroup$
    – user765195
    Commented Jul 4, 2011 at 20:11
  • $\begingroup$ $\xi_n\perp \mathcal F_n = \sigma\{X_k,k\leq n\}$ and $\xi_n$ are iid r.v. $\endgroup$
    – SBF
    Commented Jul 5, 2011 at 6:48
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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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  • $\begingroup$ Do you mean I should apply Chebyshev inequality? $\endgroup$
    – SBF
    Commented Jul 4, 2011 at 15:06
  • $\begingroup$ 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. $\endgroup$
    – leonbloy
    Commented Jul 4, 2011 at 15:18

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