I have recently been toying (very superficially) with the random Fibonacci sequence, i.e., defined by $F_0=1=F_1=1$ and $$ F_{n} = F_{n-1} + \varepsilon_n F_{n-2} $$ where $(\varepsilon_n)_{n\geq 2}$ is a sequence of i.i.d. Rademacher random variables.

It is known that this sequence is such that $$ |F_n|^{1/n} \xrightarrow[n\to\infty]{\rm a.s.} \nu $$ where $\nu \approx 1.13$ is the Viswanath constant (D. Viswanath, 1999).

I am interested in the case where the $(\varepsilon_n)_{n\geq 2}$ are no longer uniform in $\{-1,1\}$, but instead $1$ with probability $p\in[0,1]$ (so the standard Fibonacci sequence corresponds to $p=1$, and the standard random Fibonacci sequence to $p=1/2$). Then my understanding is that the convergence proof goes through, i.e., $$ |F_n|^{1/n} \xrightarrow[n\to\infty]{\rm a.s.} \nu(p) $$ I am interested in the behavior of $\nu(p)$ as a function of $p$. Is there anything known about it?

Through some basic simulations, the behavior seems very regular, but I am not sure what else to infer from it.<span class=$\nu(p)$ as a function of p" />

Note that the paper by D. Viswanath does (briefly) discuss a biased version (see, e.g., Figure 5); however, the recurrence considered there is of the form $F_{n} = \varepsilon_n F_{n-1} + \varepsilon_n' F_{n-2}$ where $\varepsilon'_n,\varepsilon_n$ are i.i.d. The result is the same (because of the absolute values) when $p=1/2$, but it is not the case for $p\neq 1/2$.

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    $\begingroup$ The paper How do random Fibonacci sequences grow? has some results on this problem and a nonlinear generalization. A preprint is also available at arXiv:math/0611860 in case you don't have access to PTRF. $\endgroup$
    – HMPanzo
    Nov 8, 2020 at 2:40
  • $\begingroup$ Thanks -- Based on the abstract it looks very relevant. $\endgroup$
    – Clement C.
    Nov 8, 2020 at 2:43
  • $\begingroup$ Mmmh. That made me realize I made a mistake in my simulations and attempts, so even better... $\endgroup$
    – Clement C.
    Nov 8, 2020 at 2:57
  • $\begingroup$ @HMPanzo If you want to make this comment an answer, I definitely would upvote it. This would be a very good reference to have archived permanently here. $\endgroup$
    – Clement C.
    Nov 10, 2020 at 18:26

1 Answer 1


This example is discussed in my 1990 PhD thesis, where is it is shown that $\nu(p)$ is a real analytic function of $p$. The thesis was written in Hebrew, but this result was generalized in two papers [1], [2]. Note that the existence of the limit you call the Viswanath constant (and more generally, $\nu(p)$) is an immediate consequence of the Furstenberg-Kesten Theorem [3] on the existence of the top Lyapunov exponent for random matrix products. This is because the vector $(F_n,F_{n+1})$ is obtained by multiplying $(F_{n-1},F_n)$ by a random matrix that takes two possible values.

[1] Peres, Yuval Analytic dependence of Lyapunov exponents on transition probabilities. Lyapunov exponents (Oberwolfach, 1990), 64–80, Lecture Notes in Math., 1486, Springer, Berlin, 1991.

[2] Peres, Yuval Domains of analytic continuation for the top Lyapunov exponent. Ann. Inst. H. Poincaré Probab. Statist. 28 (1992), no. 1, 131–148. (Reviewer: Philippe Bougerol) http://www.numdam.org/article/AIHPB_1992__28_1_131_0.pdf

[3] Furstenberg, Harry, and Harry Kesten. "Products of random matrices." The Annals of Mathematical Statistics 31, no. 2 (1960): 457-469.


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