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.


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.

  • $\begingroup$ In my problem the invariant measure $\pi$ is a vector $\pi=(\alpha_{0},\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4})$ where $\alpha_{i}\neq 0$ for all $i=0,1,2,3,4$. I don't understand when you say "the measure $\pi$ charges 0". Are you referring to there exists $\alpha_{i}=0$? $\endgroup$ – matematicaActiva Apr 29 '17 at 15:55
  • $\begingroup$ It means that $\pi(0)$ is non-zero. $\endgroup$ – R W Apr 29 '17 at 16:50
  • $\begingroup$ Yeah, that's to what you do you mean? $\endgroup$ – matematicaActiva Apr 29 '17 at 16:59
  • $\begingroup$ But I do not see as clearly as $\pi(0)\neq 0$ implies that the sequence of products $Y_{1}\ldots Y_{n}$ eventually vanishes. $\endgroup$ – matematicaActiva Apr 29 '17 at 17:02
  • $\begingroup$ Because, if $\pi(0)\neq 0$, then $Y_n=0$ for infinitely many $n$ (assuming your chain is irreducible). $\endgroup$ – R W Apr 29 '17 at 17:21

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