Markov Chain Homework: Expressing Transition Matrix P in Terms of G

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SUMMARY

The discussion focuses on expressing the transition matrix P of a Markov chain defined by the function G: I x [0, 1] -> I. The sequence (Xn)n≥0 is established as a Markov chain where Xn+1 = G(Xn, Yn+1), with Yn+1 being independent random variables uniformly distributed on [0, 1]. The key conclusion is that the transition probability can be expressed as P[Xn+1 = xn+1 | Xn = xn] = P[G(xn, y) = xn+1], where y is the uniformly distributed random variable.

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hsong9
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Please Help! Markov Chain

Homework Statement


Let X0 be a random variable with values in a countable set I. Let Y1, Y2, ... be a
sequence of independent random variables, uniformly distributed on [0, 1].
Suppose we are given a function G : I x [0, 1] -> I
and define inductively for n >= 0, Xn+1 = G(Xn, Yn+1).
Show that (Xn)n>=0 is a Markov chain and
express its transition matrix P in terms of G.



Homework Equations





The Attempt at a Solution


I know that I need to show that Xn+1 depends on Xn by checking the condition in the definition of Markov chain, and then
try to find some formula for P(Xn+1 = j | Xn=i) in terms of G.

Actually, my background for Markov chain lacks a little, so I have no how I find some formula for P in terms of G.. How do I handle terms of G?
Anybody give me some hints or answer?
 
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I guess you should only need to show P[Xn+1 = xn+1 | X1 = x1, X2 = x2, ..., Xn = xn] = P[Xn+1 = xn+1 | Xn = xn]. (Markov property).

In your case it holds, since G is only a function of Xn; and Yi are independent random variables. P[Xn+1 = xn+1 | Xn = xn] = P[G(xn, y)=xn+1] where y is the uniformly distributed random variable.
 

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