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Branching Process

if initial population $0$ is equal to $1$, i.e. $X_0 \equiv 1$

and $S_k$ is the total population size up to generation $k$ meaning $S_k = 1 +X_1 +...+X_k$.

If $G(z)$ is the Probability Generating Function of the number of offspring produced by a single member.

and $H_k(z) $ is the The Probability Generating Function of $S_k$.

what is the relationship between $H_k(z)$ and $H_{k-1}(z)$

My calculation gave me the following relationship, but I am not sure:

$$H_k(z) = G(H_{k-1}(z))$$

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    $\begingroup$ Your calculation is correct. $\endgroup$ – Kavi Rama Murthy Nov 13 '18 at 5:15
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    $\begingroup$ @KaviRamaMurthy, why do I need the fact that population zero equals $1$ and where do I use that fact. Also is this a sound proof $H_k(z) = E[Z^{S_k}]= E[E[Z^{S_k}|S_{k-1}]] = E[z^{H_{k-1}(s)}] = G[H_{k-1}(s)]$ $\endgroup$ – Note Nov 14 '18 at 4:28
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    $\begingroup$ If the population at time $0$ is $n$ then $G$ gets replaced by $G^{n}$. Your argument for showing that $H_k=G\circ H_{k-1}$ is not correct. How do you know that $Ez^{S_k}|S_{k-1} =z^{H_{k-1}(s)}$?. Besides the last step is also wrong because $z^{H_{k-1}}$ is just a number not a random variable. How do you write this as $G[H_{k-1} (s)]$? $\endgroup$ – Kavi Rama Murthy Nov 14 '18 at 5:30
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    $\begingroup$ $G[H_{k-1}(s)] = E[(H_{k-1}(s))^X]$ Since $G$ is pgf of offspring produced by one member, then $X$ ought to equal $1$ so $G[H_{k-1}(s)] = E[(H_{k-1}(s))]$ $\endgroup$ – Note Nov 14 '18 at 5:44
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    $\begingroup$ You are getting a bit confused with definition of generating function. $G(t)=Et^{S_1}$. $H_{k-1}(s)$ is just a number, not a random variable. Why are you taking its expectation?. $\endgroup$ – Kavi Rama Murthy Nov 14 '18 at 5:48

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