**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 $B_i$?