Let $(\Omega,\mathcal{F},\mathbb{P}):=(M^{\mathbb{N}\_{0}},\mathcal{M}^{\mathbb{N}\_{0}},\mathbb{P})$ be a probability space where $M=\left\{0,1,2,3,4\right\}$, $\mathcal{M}^{\mathbb{N}\_{0}}$ is product $\sigma$-algebra. Note that if we consider the cylinder $I=\left[x\_{0},x\_{1},\ldots,x\_{k}\right]$ defined by 
\begin{equation}I=\left[x\_{0},x\_{1},\ldots,x\_{n}\right]:=\left\{\mathbf{y}=(y\_{i})\_{i\in\mathbb{N}}\in M^{\mathbb{N}\_{0}} \left|y\_{i}=x\_{i} \mbox{ para }i=0,1,\ldots,n \right. \right\}.\end{equation}
Then **the collection of all cylinders form a semi-algebra that we called $\mathcal{S}$**, futhermore, by measure theory we know that \begin{equation}\mathcal{M}^{\mathbb{N}\_{0}}=\sigma(\mathcal{S}).\end{equation}
In this sense, given a fixed transition matrix $\Pi$ of size $5\times 5$ and your invariant measure $\pi$ we define $\mathbb{P}$ in $S$ as
$$\mathbb{P}(I):=\pi(x\_{0})\Pi(x\_{0},x\_{1})\Pi(x\_{1},x\_{2})\cdots \Pi(x\_{k-1},x\_{k}).$$ 


**The question:** In this context, let the stationary sequence $(Y\_{n})\_{n\in\mathbb{N}}$ defined by $Y\_{n}:=Y\_{0}\circ \varphi^{n} $ where $\varphi$ is the 1-left shift (i.e. $\varphi(x\_{0},x\_{1},x\_{2},x\_{3},\ldots)= (x\_{1},x\_{2},x\_{3},x\_{4},\ldots)$) and $Y\_{0}:=\mathrm{Proy}\_{0}$ (i.e. $\mathrm{Proy}\_{0}(x\_{0},x\_{1},x\_{2},x\_{3},\ldots)=x\_{0}$). We define the random matrix
$$B\_{n}:=\left(\begin{array}{cc} Y\_{n} & 0 \\ 0& 2\end{array}\right).$$
and $$A\_{n}:=B\_{n}B\_{n-1}\cdots B\_{1}B\_{0}.$$
Determine the Lyapunov exponents $\lambda\_{1}$ and $\lambda\_{2}$ of $(A\_{n})\_{n\in\mathbb{N}}$. [Note: Consider $Y\_{0}\sim \pi $]


**My attempt:** Considering the **Theorem 6.15** (Furstenberg-Kesten) in this [link](https://www.mathematik.hu-berlin.de/~fiebig/paper_Imkeller/stochastic_dynamics.pdf) we should calculate $\widehat{\Omega}$, I think that $$\widehat{\Omega}:=\left\{\mathbf{y}=(y\_{n})\_{n\in\mathbb{N}}\in M^{\mathbb{N}} | y\_{i}=0 \mbox{ for some }i\right\}.\tag{1}$$
In this sense, we know that the singular values of $A\_{n}$ are $$\delta\_{1}(A\_{n})=2^{n+1} \qquad \mbox{ and }\qquad \delta\_{2}(A\_{n})=Y\_{n}Y\_{n-1}\cdots Y\_{1}Y\_{0}.$$
Then, for any $\omega\in \widehat{\Omega}$ the Lyapunov exponents are
$$\Lambda\_{1}(\omega)=\lim\_{n\rightarrow \infty} \frac{1}{n} \log\delta\_{1}(A\_{n})=\log(2) $$
and for $n$ large we have


$$\Lambda\_{2}(\omega)=\lim\_{n\rightarrow \infty} \frac{1}{n} \log\delta\_{2}(A\_{n})=\lim\_{n\rightarrow \infty} \frac{1}{n} \log\left(Y\_{n}(\omega)Y\_{n-1}(\omega)\cdots Y\_{1}(\omega)Y\_{0}(\omega)\right)=\lim\_{n\rightarrow \infty} \frac{1}{n} \log\left(0\right)=-\infty. $$


**The problem is show that $\widehat{\Omega}$ in (1) is the given in Theorem 6.15.**



Calculate of Lyapunov Exponents of a sequence of random matrices
Strictly speaking, Lyapunov exponents are only defined for products of non-degenerate matrices. Since you are dealing with diagonal matrices, you are just asking about the exponential growth rates of the diagonal entries, which can be considered separately, so that you don't really need to talk about matrix products at all. The second diagonal term being just powers of 2, the first one is the product $Y\_1\dots Y\_n$. As far as I understand $Y\_i$ is precisely your Markov chain on $\{0,1,2,3,4\}$ with stationary measure $\pi$. If the measure $\pi$ charges 0, then it is obvious that the sequence of products $Y\_1\dots Y\_n$ eventually vanishes.


