TITLE: Zeroes of the random Fibonacci sequence
QUESTION [23 upvotes]: Let $X_n$ be the "random Fibonacci sequence," defined as follows:
$X_0 = 0, X_1 = 1$;
$X_n = \pm X_{n-1} \pm X_{n-2}$, where the signs are chosen by independent 50/50 coinflips.
It is known that $|X_n|$ almost surely grows exponentially by a (much more general) theorem of Furstenberg and Kesten about random matrix products:  the base of the exponent was determined explicitly by Viswanath to be $1.13\ldots$
I am not too proud to say that I learned all this from Wikipedia:
http://en.wikipedia.org/wiki/Random_Fibonacci_sequence
What I did not learn from Wikipedia, or any of the references I gathered therefrom, was:
Question: what, if anything, do we know about the probability that $X_n = 0$, as a function of $n$?

REPLY [12 votes]: First of all, it is somewhat misleading to attribute the claim concerning exponential growth to Furstenberg and Kesten. What they proved (in 1960) is that for partial products of a random stationary sequence of matrices there exists almost surely an exponential rate of growth. However, it does not exclude possibility that this growth rate is zero. Its positivity for i.i.d. sequences was proved (under reasonable irreducubility conditions) by Furstenberg in 1963.
Now to your question. The reduction to matrix products consists in observing that
$$
\left( X_{n+1} \atop X_n \right) = \left( 1 \;\pm 1 \atop1\;\;\;\;\;\ 0 \right) \left( X_n \atop X_{n-1} \right) \;,
$$
whence 
$$
\left( X_{n+1} \atop X_n \right) = B_n \left( 1 \atop 0 \right) \;,
$$
where
$$
B_n = A_n A_{n-1} \dots A_1
$$
is the product of i.i.d. matrices $A_i$ which are either $M=\left( 1 \;1 \atop1\; 0 \right)$ or
$M'=\left( 1 \;-1 \atop1\;\;\; 0 \right)$ with equal probabilities 1/2 (as explained by JSE, one can omit the first $\pm$ in the recurrence relation). Therefore, $X_n=0$ iff $B_n$ is an upper triangular matrix with $\pm 1$ on the diagonal. The Schreier graph of the quotient of $GL(2,\mathbb Z)$ by the group of such triangular matrices is non-amenable, whence one can deduce exponential decay of the return probabilities.
PS The answer to Mark's question is NO. The reason is that there exist amenable groups (e.g., the higher dimensional lamplighter groups), for which the linear rate of escape of the simple random walk is strictly positive (which is analogous to positivity of the exponent for matrix products), but nonetheless the probability of returning to zero decays subexponentially (because of amenability).