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Math Help - Markov chain

  1. #1
    Moo
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    Markov chain

    Hi guys !

    I'm pretty desperate : it's a simple question, but I don't manage to write the solution properly...

    We have (a Galton-Watson process) Z_0=a and Z_n=\sum_{i=1}^{Z_{n-1}} X_{i,n}, where the sequence (X_{i,n})_{i,n} is an iid sequence.

    How can I show that Z_n is a homogeneous Markov Chain ? I know all the basic methods :
    - prove that P(Z(n+1)=z(n+1)|Z0=a,Z1=z1,...,Zn=zn)=P(Z(n+1)=z(n +1)|Zn=zn) <--- I've tried it, but it wasn't rigourous enough to me
    - prove that the process can be expressed as a function of Zn and an random variable U(n+1) independent with Zn <--- the problem here is that either Zn is in the index of the independent rv, either I have an infinite number of random variables. So I don't really know how to extend it...

    Thanks
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  2. #2
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    Quote Originally Posted by Moo View Post
    Hi guys !

    I'm pretty desperate : it's a simple question, but I don't manage to write the solution properly...

    We have (a Galton-Watson process) Z_0=a and Z_n=\sum_{i=1}^{Z_{n-1}} X_{i,n}, where the sequence (X_{i,n})_{i,n} is an iid sequence.

    How can I show that Z_n is a homogeneous Markov Chain ? I know all the basic methods :
    - prove that P(Z(n+1)=z(n+1)|Z0=a,Z1=z1,...,Zn=zn)=P(Z(n+1)=z(n +1)|Zn=zn) <--- I've tried it, but it wasn't rigourous enough to me
    I think you are trying too hard. On the event {Z_{n-1}=N} the sum is from 1 to N. On the event {Z_1=C_1,...,Z_{n-1}=N}, the sum is again from one to N.

    - prove that the process can be expressed as a function of Zn and an random variable U(n+1) independent with Zn <--- the problem here is that either Zn is in the index of the independent rv, either I have an infinite number of random variables. So I don't really know how to extend it...
    It might help to prove this using the c.f. :

    <br />
\mathbb{E}[e^{i \theta Z_n}|Z_{n-1}=N]=\prod_{i=1}^N \mathbb{E}[e^{i \theta X_{i,n}}]=\mathbb{E}[e^{i \theta X_{1,n}}]^N=:\phi(\theta)^N<br />
    By the tower law,
    <br />
\mathbb{E}[e^{i \theta Z_n}]=\mathbb{E}[\phi(\theta)^{Z_{n-1}}]<br />

    Characteristic functions uniquely determine distributions and the above is a function of phi and Z_n.
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  3. #3
    MHF Contributor

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    I too think you're trying too hard Here are my suggestions...

    Quote Originally Posted by Moo View Post
    - prove that P(Z(n+1)=z(n+1)|Z0=a,Z1=z1,...,Zn=zn)=P(Z(n+1)=z(n +1)|Zn=zn) <--- I've tried it, but it wasn't rigourous enough to me
    We have P(Z_{n+1}=z_{n+1}|Z_0,\ldots,Z_n=z_n)=P\left(\sum_  {i=1}^{z_n}X_{i,n+1}=z_{n+1}\middle|Z_0,\ldots,Z_n  =z_n\right) (z_n,z_{n+1}))" alt="=P\left(\sum_{i=1}^{z_n}X_{i,n+1}=z_{n+1}\right)(=(z_n,z_{n+1}))" /> using the fact that the random variable in the probability is independent of those in the condition.


    - prove that the process can be expressed as a function of Zn and an random variable U(n+1) independent with Zn <--- the problem here is that either Zn is in the index of the independent rv, either I have an infinite number of random variables. So I don't really know how to extend it...
    Why not say that Z_{n+1} is a (time-independent) function of Z_n and U_{n+1}=(X_{i,n+1})_{i\geq 1}, which is independent of \mathcal{F}_n (the past before time n) and whose law \mu does not depend on n? Then U_{n+1} is not real, but this doesn't matter in order to deduce that (Z_n)_n is an homogeneous Markov chain:

    E[\varphi(Z_{n+1})|\mathcal{F}_n]=E[\varphi(f(Z_n,U_{n+1}))|\mathcal{F}_n] =\int \varphi(f(Z_n,u)) d\mu(u) is a time-independent function of Z_n (choose \varphi(z)={\bf 1}_{(z=z_{n+1})} and take expectation to get the above definition).
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  4. #4
    Moo
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    One thing I didn't need much time to figure out is that I agree that I was overcomplicating it But I really worry for the way I write it...

    I sort of combined your replies (though I didn't see Laurent's yet), or so I think...

    (a_n)_{n\in N} a sequence of integers.
    P(Z_{n+1}=a_{n+1}\mid Z_0=a_0,\dots,Z_n=a_n) =P\big(\sum_{i=1}^{a_n} X_{i,n+1}=a_{n+1}\mid Z_0=a_0,\dots,Z_n=a_n\big)

    Whatever values a_0,\dots,a_{n-1} take, the probability won't be changed as none of them intervene in \sum_{i=1}^{a_n} X_{i,n+1}.

    So P(Z_{n+1}=a_{n+1}\mid Z_0,\dots,Z_n)=P(Z_{n+1}=a_{n+1}\mid Z_n)

    Hence (Zn) is a Markov chain.
    As to show that it's homogeneous, that's no big deal.

    I like the way of the characteristic functions, but I already show something similar later on and can't be removed, sorry

    As for defining U_{n+1}, I'm happy, so the formula can be generalized if it's defined this way... I wasn't sure at all about that.

    If the above text is not correct, I'll go to this stuff with U_{n+1} But I'd be happy if one could point out what's wrong in it

    Thanks guys !
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