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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 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.

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