exercise:5184e6637c

[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

(Suggested by Peter Doyle) In the proof of Theorem, we assumed the existence of a fixed vector [math]\mat{w}[/math]. To avoid this assumption, beef up the coupling argument to show (without assuming the existence of a stationary distribution [math]\mat{w}[/math]) that for appropriate constants [math]C[/math] and [math]r \lt 1[/math], the distance between [math]\alpha P^n[/math] and [math]\beta P^n[/math] is at most [math]C r^n[/math] for any starting distributions [math]\alpha[/math] and [math]\beta[/math]. Apply this in the case where [math]\beta = \alpha P[/math] to conclude that the sequence [math]\alpha P^n[/math] is a Cauchy sequence, and that its limit is a matrix [math]W[/math] whose rows are all equal to a probability vector [math]w[/math] with [math]wP=w[/math]. Note that the distance between [math]\alpha P^n[/math] and [math]w[/math] is at most [math]C r^n[/math], so in freeing ourselves from the assumption about having a fixed vector we've proved that the convergence to equilibrium takes place exponentially fast.