# The Probability Generating Function of the Total Population

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:

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

• Your calculation is correct. – Kavi Rama Murthy Nov 13 '18 at 5:15
• @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 $H_k(z) = E[Z^{S_k}]= E[E[Z^{S_k}|S_{k-1}]] = E[z^{H_{k-1}(s)}] = G[H_{k-1}(s)]$ – Note Nov 14 '18 at 4:28
• 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 $G[H_{k-1} (s)]$? – Kavi Rama Murthy Nov 14 '18 at 5:30
• $G[H_{k-1}(s)] = E[(H_{k-1}(s))^X]$ Since $G$ is pgf of offspring produced by one member, then $X$ ought to equal $1$ so $G[H_{k-1}(s)] = E[(H_{k-1}(s))]$ – Note Nov 14 '18 at 5:44
• 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?. – Kavi Rama Murthy Nov 14 '18 at 5:48