Finding the Probability of a branching process given $A(z)$

Question

I have a past exam question that shows the following:

Consider the branching process $\{X_n, n = 0, 1, 2, \ldots\}$ where $X_n$ is the population size of the $n$th generation. Assume $P(X_0 = 1) = 1$ and that the probability generating function of the offspring distribution is $A(z) = (1 − p) + pz^2$ for some $0 < p < 1$.

Question: What is $P(X_{100} = 3)$?

The solution states that $X_n$ is always even for $n\ge1$, so $P(X_{100} = 3) = 0$.

Why is this so, and what is $A(z)$ trying to show?

Thank you.

Answer 1

That $A(z)$ is the generating function for the next generation means that if you write $A(z)$ in the form

$$ A(z)=\sum_{n=0}^\infty a_nz^n\;, $$

then $a_n$ is the probability that an individual will have $n$ offspring in the next generation. Since in your case only $a_0$ and $a_2$ are non-zero, the number of offspring is always guaranteed to be even.