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I am trying to understand the derivation of following PGF Formula in branching process

$$H_n(s)=H_{n-1}(H(s))$$

I presume that the initial steps for this derivation are standard...so am avoiding entering them.

The derivation is clear up to the following step

$$H_n(s) = \sum_{i}\Pr(Z_{n-1}=i)\left(\sum_k \Pr(Y_1+Y_2+ \cdots + Y_k=k)s^k\right)$$

The next statement is

Since $\{Y_1, Y_2, \ldots\}$ are iid random variables and PGF of $Y_i$ is $H(S)$

PGF of $Y_1+Y_2+\cdots+Y_i$ is $(H(s))^i$ and hence $H_n(s) = \sum_{i} \Pr(Z_{n-1} = i)(H(s))^i$

I am just unable to fathom this last step..may I request a simple explanation for this last step please?

heropup
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SAK
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  • treat $ (H(s))^i$ as s, so that result becomes PGF. – Lella Nov 01 '16 at 05:47
  • There was a typo in your expression. it should read as $H_{n}(s)=\sum_{i} p(Z_{n-1} = i) [H(s)]^{i}$ – Lella Nov 01 '16 at 05:54
  • Yes, thanks ..it was a typo. I have edited it. Could you please elaborate on your explanation.. I still dont seem to understand it. ..( I am studying maths as a hobby... and dont have the advantage of asking doubts to professors/fellow students) – SAK Nov 01 '16 at 06:05
  • The square bracketed term is the pmf of sum of iid rv's. $Y_{1}+Y_{2}+\cdots+Y_{k}$. This together $s^{k}$ defines a PGF of sum of iid variables. – Lella Nov 01 '16 at 06:09
  • $\sum_{i}P(Z_{n-1}=i)\left(M_{Y_{1}+Y_{2}+\cdots+Y_{k}}(s)\right)$ – Lella Nov 01 '16 at 06:11
  • OK... thanks got it. – SAK Nov 01 '16 at 06:16

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