Markov Chain stationarity?

From: Bill Taylor (w.taylor_at_math.canterbury.ac.nz)
Date: 01/26/05 

Date: 25 Jan 2005 23:02:35 -0800

Hey kiki - what with you and "lose-mind Lucy" writing very similar queries
at almost exactly the same time, I expect someone somewhere is teaching
a course on this right now. ;-)

"kiki" <lunaliu3@yahoo.com> wrote in message

> A MC has a unique stationary distribution if it is irreducible, positive
> recurrent, aperiodic.

It doesn't have to be aperiodic - there will still be a unique stationary
distribution, it just won't have all the information in it that you
might conceivably be interested in.

> Even when the MC is periodic, its P^n oscillates,

Correct.

> however, the balance equation pi=pi*P still has some solution,
> this solution can be viewed as long-term portion of time spent in the states.

Correct; and it is also the unique stationary distribution!

Though you might prefer to think of it as the *average over the cycle length*
of what you charmingly call the costationary distributions.

It is the stationary distribution of the unique MC that is started
at a *random time* chosen equally among the possible epochs along the period.
(Recall: a MC is only fully specified by its matrix AND starting distribution.)

So you can see that this stationary approach "smooths out" the inherent
cyclicity of the process. So it "loses information". In effect, it's like
looking at the process but having forgotten exactly what the time is!

> If we let the MC starts at this solution to pi=pi*P, can we say the MC is
> stationary, or cyclostationary in this case?

Yes, it is staionary process in this case (and only in this case).

> I guess I am confused about the rows of the limiting matrix P^n and the
> stationary distribution satisfying pi=pi*P...

I hope you are somewhat unconfused now?

> How are they related?

In general, for period k, the average of P^(n+1), ... , P^(n+k)
is the thing that has "pi" in all its rows, (in the limit as n --> inf).
When k = 1, aperiodic, it is just P^n itself, therefore.

All OK now?

-------------------------------------------------------------------------------
          Bill Taylor W.Taylor@math.canterbury.ac.nz
-------------------------------------------------------------------------------
The worst form of inequality is to try to make unequal things equal - Aristotle
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