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

  1. #1
    fex
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    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 $\{Z_n\}_{n \ge 1} $ be a sequence of i.i.d. random variables with values in a denumerable space $E$.
    Let $\{X_{n+1} = f(X_n,Z_{n+1})\}_{n \ge 0}$ a new sequence of random variable with values in a denumerable space $F$, with $f:F \times E \rightarrow E$ a generic function ( $X_0$ a random variable on a denumerable set $F$).

    If for all $k,k_1,\dots,k_n \in E ,i,i_0,\dots i_{n-1} \in F $:
    $P(Z_{n+1} = K | X_n = i, X_{n-1} = i_{n-1},\dots , Z_n = k_n , \dots , Z_1 = k_1)$ $=$  P(Z_{n+1} = k | X_n = i)$

    then $\{X_n\}_{n \ge 0}$ is a Markov chain.

    In other words, the previous hypothesis implies that: $P(X_{n+1} = j | X_n = i, X_{n-1} = i_{n-1}, \dots , X_0 = i_0) = P(X_{n+1} = j | X_{n} = i)$

    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!
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  2. #2
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    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).
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  3. #3
    fex
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    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  $\{Z_n\}$ are not i.i.d. (the way the theorem is proposed by the author fooled me ).
    You're right, when the random variables \{$Z_n\}$ 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 $k, k_1, \dots ,k_n \in E, i, i_0, \dots , i_{n-1} \in F $ ...").
    The point is, what happens if I can't state $P(Z_{n+1}=j | Z_{n}=i_n, \dots , Z_{0}) = P(Z_{n+1} = j)$ ?

    fex
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