<p>(This question has been unsuccessfully asked on mathoverflow: <a href="https://mathoverflow.net/questions/161728/convergence-of-iterated-stochastic-matrices">https://mathoverflow.net/questions/161728/convergence-of-iterated-stochastic-matrices</a>)</p>

<p>It is well-known that for a stochastic aperiodic matrix $M$,
the sequence of iterated powers $(M^n)_n$ converges.</p>

<p>Here I would like to a have a more precise analysis. Consider now a sequence of stochastic matrices $(M_n)_n$, converging to $A$. We even assume that there exists $0 &lt; \alpha &lt; 1$ such that for all $n$, we have $||M_n - A|| \le \alpha^n$.
Is it true that $||M_n^n - A^n||$ converges to $0$?</p>

CC BY-SA 3.0
2014-03-29T20:52:02.673Z
Is the following analysis helpful? $$||M_n^n-A^n|| = || \sum_{i=0}^{n-1} M_n^{n-i-1} (M_n - A) A^i || \le \sum_{i=0}^{n-1} ||M_n||^{n-i-1} ||M_n - A|| ||A||^i \leq n \alpha^n \to 0$$ I assume that the norm is submultiplicative and the norm of a stochastic matrix is at most 1. In particular, $||M|| = \max_{||x||_1=1} ||xM||_1$ satisfies these properties.
Thomas Steinke
CC BY-SA 3.0
2014-03-29T23:30:36.133Z
If want to use a different norm, you can go via the 1-norm. This seems to answer the question as stated, unless I'm missing something.
Thomas Steinke
CC BY-SA 3.0
2014-03-30T11:39:04.457Z
It does answer the question! I was using the same equality, but missing a norm which is both submultiplicative and where stochastic matrices are unitary. Thank you very much
Nathanaël Fijalkow
<pr.probability><matrices>
Convergence of iterated stochastic matrices
Nathanaël Fijalkow
# Convergence of iterated stochastic matrices

(Asked by: username_0 on 2014-03-29)

(This question has been unsuccessfully asked on mathoverflow: <https://mathoverflow.net/questions/161728/convergence-of-iterated-stochastic-matrices>)

It is well-known that for a stochastic aperiodic matrix $M$,
the sequence of iterated powers $(M^n)\_n$ converges.

Here I would like to a have a more precise analysis. Consider now a sequence of stochastic matrices $(M\_n)\_n$, converging to $A$. We even assume that there exists $0 < \alpha < 1$ such that for all $n$, we have $||M\_n - A|| \le \alpha^n$.
Is it true that $||M\_n^n - A^n||$ converges to $0$?

**Tags:** pr.probability, matrices
> **Comments:**
> **username_1**: Is the following analysis helpful? $$||M_n^n-A^n|| = || \sum_{i=0}^{n-1} M_n^{n-i-1} (M_n - A) A^i || \le \sum_{i=0}^{n-1} ||M_n||^{n-i-1} ||M_n - A|| ||A||^i \leq n \alpha^n \to 0$$ I assume that the norm is submultiplicative and the norm of a stochastic matrix is at most 1. In particular, $||M|| = \max_{||x||_1=1} ||xM||_1$ satisfies these properties.  
> **username_1**: If want to use a different norm, you can go via the 1-norm. This seems to answer the question as stated, unless I'm missing something.  
> **username_0**: It does answer the question! I was using the same equality, but missing a norm which is both submultiplicative and where stochastic matrices are unitary. Thank you very much  

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