Markov chains as function of a sequence of i.i.d. random variables

Hi!

I'm a new user and first of all congratulations, this forum is very useful, I'm finding out a lot of interesting answers, I hope I can contribute soon.

I've got a question about Markov chains.

Studying a manual I've found out the following theorem, related to one of the canonical representations of Markov chains:

Let be a sequence of i.i.d. random variables with values in a denumerable space .

Let a new sequence of random variable with values in a denumerable space , with a generic function ( a random variable on a denumerable set ).

If for all :

then is a Markov chain.

In other words, the previous hypothesis implies that:

The manual doesn't show the proof and I tried to prove it but I didn't succeed.

Could anybody help me obtaing the proof or links to other lecture notes with the solution?

Thanks in advance!

Re: Markov chains as function of a sequence of i.i.d. random variables

Hey fex.

Since we know Z_n is I.I.D then this implies P(Z_n+1|Z_n,Z_n-1,...) = P(Z_n+1|anything) = P(Z_n+1) (complete independence).

So now we have a distribution involving X_n and Z_n but since Z_n is I.I.D then the probability for f(Xn.Zn) will be P(Xn)*P(Zn).

With this factorization you should be able to do some re-arranging to get P(Xn+1|Xn,Xn-1,Xn-2,...) = P(Xn+1|Xn).

Re: Markov chains as function of a sequence of i.i.d. random variables

Hi chiro,

thanks for your reply. I'm so sorry, there was an error in my previous post, the random variables are not i.i.d. (the way the theorem is proposed by the author fooled me (Worried)).

You're right, when the random variables are i.i.d., following you're line of reasoning we can find out a way to prove it (indeed we don't need even the statement "for all ...").

The point is, what happens if I can't state ?

fex