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Thread: Branching process with martingale

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    Branching process with martingale

    Let $\displaystyle \{ X_k^n : n,k \geq 1 \} $ be i.i.d. positive interger-value random variables with $\displaystyle EX_k^n = \mu < \infty $ and $\displaystyle Var(X_k^n) = \sigma ^2 > 0 $. Define $\displaystyle Y_0 =1 $ and recursively define $\displaystyle Y_{n+1}=X^{n+1}_1+ . . . +X_{Y_n}^{n+1} \ \ \ n \geq 0 $

    a) Show that $\displaystyle M_n = \frac {Y_n}{ \mu ^n } $ is a martingale with respect to the filtration $\displaystyle \sigma (Y_0, Y_1, . . . , Y_n) $

    b) Find $\displaystyle E(Y_{n+1}^2 | Y_0,...,Y_n) $ and deduce that M_n has uniformly bounded variance if and only if $\displaystyle \mu > 1 $

    c) For $\displaystyle \mu > 1 $, find $\displaystyle Var(M_ \infty ) $

    My proof so far.

    a) This one is easy, $\displaystyle E(M_{n+1}|Y_1,...,Y_n) = \frac {1}{ \mu ^{n+1} } E(Y_{n+1} | Y_1,...,Y_n) $

    $\displaystyle = \frac {1}{ \mu ^{n+1} }E(X_1^{n+1}+...+X_{Y_n}^{n+1}) $

    $\displaystyle = \frac {1}{ \mu ^{n+1} }(E(X_1^{n+1})+...+E(X_{Y_n}^{n+1})) $

    $\displaystyle = \frac {1}{ \mu ^{n+1} } \frac {Y_n}{ \mu ^n } = M_n$. So that proves that M_n is a martingale.

    b) I'm having problem trying to break down this thing...

    $\displaystyle E(Y_{n+1}^2 | Y_0,...,Y_n) = E[(X_1^{n+1}+...+X_{Y_n}^{n+1})^2|Y_0,...,Y_n]$

    But how would I proceed from here? Thanks.
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    Moo
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    Re: Branching process with martingale

    Hello,

    You know the conditional variance : Var[Y|X]=E[Y^2|X]-E[Y|X]^2
    This will give you E[Y^2|X], and the conditional variance is the sum of the variances, because the variables are independent. Sorry I don't use the latex, nor did I use the correct names for the variables, but it takes too much time to write them down
    Thanks from tttcomrader
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