3
$\begingroup$

My question is highlighted in bold at the end.

$\mathrm{\underline{Background}}$

Consider a product of i.i.d. $d\times d$ random matrices $A_{i}$ (with $\mathbb{E}\log\left\Vert A_{i}\right\Vert <\infty$) acting on a non-zero vecor $X$, i.e. $$ A_{n}\cdots A_{1}X. $$ The Lyapunov exponents are used to describe the exponential growth properties of $$ \left\Vert A_{n}\cdots A_{1}X\right\Vert . $$ Thus we define the the Lyapunov exponent as $$ \lambda\left(X\right):=\lim_{n\to\infty}\frac{1}{n}\log\left\Vert A_{n}\cdots A_{1}X\right\Vert \;\;(1). $$ According to the Muliplicative Ergodic Theorem (or Furstenberg-Kesten Theorem) the number of distinct values $p$ that ($1$) can take is at most $d$, i.e. we have $$ \lambda_{d}\leq\cdots\leq\lambda_{1}. $$ Now, let $\sigma_{i,n}$ be the $i$th singular value of the matrix product $A_{n}\cdots A_{1}$ such that $$ \sigma_{d,n}\leq\cdots\leq\sigma_{1,n}. $$ Furstenberg-Kesten Theorem states that $$ \lim_{n\to\infty}\frac{1}{n}\log\sigma_{i,n}=\lambda_{i}.\;\;(2) $$ The crux of my question (to follow) revolves around ($2$).

$\underline{\mathrm{Question\;Setup: Independent\;nonidentical\;matrices}}$

Consider random scalars $a_{i},b_{i},c_{i}$ where $b_{i}s$ are i.i.d. with $\mathbb{E}\log\left|b_{1}\right|<\infty$, $c_{i}$s are i.i.d. $\mathbb{E}\log\left|c_{1}\right|<\infty$ and $a_{i}s$ are independent but for any $i$ there exists a finite non-zero not-necessarily unitary constant $\alpha_{i}$ such that $a_{1}\overset{d}{=}\alpha_{i}a_{i}$ ($\overset{d}{=}$ denotes equality in distribution) with $\mathbb{E}\log\left|a_{1}\right|<\infty$ . It is easy to show that with $B_{i}$ defined as $$ B_{i}:=\left(\begin{array}{cc} a_{i} & b_{i}\\ 0 & c_{i} \end{array}\right) $$ $$ \mathbb{E}\log\left\Vert B_{i}\right\Vert <\infty\Longleftrightarrow\mathbb{E}\log\left|a_{1}\right|<\infty,\mathbb{E}\log\left|b_{1}\right|<\infty,\mathbb{E}\log\left|c_{1}\right|<\infty. $$ The Lyapunov index of $a_{i}$ is given by $$ \lim_{n\to\infty}\frac{1}{n}\log\left|a_{n}\cdots a_{1}\right|=\lim_{n\to\infty}\frac{1}{n}\log\left|\alpha_{n}\cdots\alpha_{1}\right|+\mathbb{E}\log\left|a_{1}\right|, $$ that of $b_{i}$ is given by $$ \mathbb{E}\log\left|b_{1}\right| $$ and that of $c_{i}$ is given by $$ \mathbb{E}\log\left|c_{1}\right|. $$

With $\sigma_{i,n}$ $i=1,2$ defined as the singular values of $A_{n}\cdots A_{1}$, it can be shown that $$ \frac{1}{n}\log\sigma_{1,n}\to\max\left\{ \lim_{n\to\infty}\frac{1}{n}\log\left|\alpha_{n}\cdots\alpha_{1}\right|+\mathbb{E}\log\left|a_{1}\right|,\mathbb{E}\log\left|c_{1}\right|\right\} \;\; (3) $$ and $$ \frac{1}{n}\log\sigma_{2,n}\to\min\left\{ \lim_{n\to\infty}\frac{1}{n}\log\left|\alpha_{n}\cdots\alpha_{1}\right|+\mathbb{E}\log\left|a_{1}\right|,\mathbb{E}\log\left|c_{1}\right|\right\} \;\; (4). $$ Here is my question: As stated previously, according to Furstenberg for i.i.d. matrices $A_{i}$, the limiting behavior of the singular values coincide with the Lyapunov exponents of $A_{i}$, see ($2$). Is this true if one considers my non-i.i.d. setup? In other words, is it right to say that ($3$) and ($4$) are the Lyapunov exponents of $B_i$?

$\endgroup$

1 Answer 1

4
$\begingroup$

Yes, it is true. However, in your question you mix up a number of things (to begin with, it is Kesten, not Keston). The Multiplicative Ergodic Theorem is not the same as the Furstenberg-Kesten theorem. You can find answers to all your questions in this article.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.