Random products of Jordan blocks — what distribution are they converging to?

Question

Let $$ M_1=\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right), \qquad M_2=\left( \begin{array}{cc} 1 & 0 \\ 1 & 1 \\ \end{array} \right),\qquad v = \left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right) $$

and consider the $n$-fold random product, for $n \in \mathbb N$. More precisely, let

$$\Psi_n := \log \| M_{i_1}M_{i_2}\cdots M_{i_n}v \|_{L^1}:$$ where each $i_k$ are i.i.d., equal to 1 or 2 with equal probability.

Scaling by $2^{-n}$, in this question we consider $\Phi_n$ as a compactly supported probability measure on $\mathbb R$ (i.e. a weighted sum of Dirac deltas).

Central Limit Theorem

For probability measures $\mu_n, \mu$, we here say

$$\mu_n\stackrel{\text{dist.}}\longrightarrow \mu$$

if, for each $-∞ < a \leq b < ∞$,

$$\mu_n\big([a,b]\big) \to \mu([a,b]) .$$

Under certain hypotheses on the $M_i$, one would see, for generic $v$, that $\Psi_n$ satisfies a kind of Central Limit Theorem:

I.e., that there exist constants $\lambda_\ast > 0$ and $\sigma >0$ such that

$$ \frac{\Psi_n - n \lambda_\ast}{\sigma^2\sqrt n} \stackrel{\text{dist.}}\longrightarrow N(0,1) \qquad\text{as }n \to ∞,$$

where $N(0,1)$ is the standard normal distribution.

Reality

Calculating some histograms in Mathematica, the shape of $\Psi_n$ seems to not be converging to a Gaussian, but to some kind of skewed distribution: see pictures below.

Questions:

• What can $\Phi_n$ be converging to, if anything?
• How would one prove that $\Phi_n$ is not converging to a Gaussian?

Answer 1

This is a partial answer, but given the lack of other answers hopefully it is helpful.

1. When you say that $\Psi_{n}$ should converge to normal, be careful: the CLT says only that $$ Z_{n} = \frac{\frac{\Psi_{n}}{n} - \lambda^{\ast}}{\sigma/\sqrt{n}} $$ converges to anything normal, and really that is only true if $\frac{1}{n}\Psi_{n}$ behaves sufficiently like a sample mean ($\Psi_{n}$ is not a sum, so this is not obvious).
2. The CLT theorem depends on the original sampling distribution. For instance, it does not apply if the sampling distribution is Cauchy. More generally, if the sampling distribution comes from a family of stable distributions, the CLT does not necessarily apply (I assume there are other ways for the CLT to fail, but I find this a particularly illuminating case).
3. Lastly, as pointed out by Peter Morfe in the comments, the paper "Central Limit Theorems for Linear Groups" (Benoist and Quint 2016) seems to answer this problem.