the book of Ross & Pekoz have the next statement:
A branching process follows the size of a population over succeeding generations. It supposes that, independent of what occurred in prior generations, each individual in generation $n$ independently has $j$ offspring with probability $p_j$, $j \geq 0$.
The offspring of individuals of generation $n$ then make up generation $n + 1$. Let $X_n$ denote the number of individuals in generation $n$.
Assuming that m = $\sum_{j}jp_j$, the mean number of offspring of an individual, is finite it is east to verify that $Z_n = X_n/m^n, n ≥ 0$, is a martingale.
and I'm trying to verify that $Z_n$ is indeed a martingale, so I check that:
$Z_n$ is measurable respect $\sigma(X_1,...,X_n)$
$Z_n$ is adapted to $\sigma(X_1,...,X_n)$
But in the next part I don't see how to continue:
- $E[Z_n+1|\sigma(X-1,...,X_n)]=Z_n$
Because I only come with $E[Z_{n+1}|\sigma(X_1,...,X_n)]=\frac{E[X_{n+1}|\sigma(X_1,...,X_n)]}{m^{n+1}}.$
Any help is appreciated, thanks.
Note that since this is in discrete time, and $X$ and $Z$ generate the same $\sigma$-Algebra ($Z$ is just some constant times $X$) we only have to show that $$\mathbb{E}[Z_{n+1}|X_{n}] = Z_{n}.$$
Now, given that we know that in generation $n$ we have $X_{n}$ members of the population, which each generate $j$ offspring with probability $p_j$ independently, the expected value for the $n+1$-th generation, divided by $m^{n+1}$ is
$$\mathbb{E}[Z_{n+1}|X_{n}] = \frac{1}{m^{n+1}} \sum_{i=1}^{X_{n}} \sum_j p_j = \frac{1}{m^{n+1}}\sum_{i=1}^{X_{n}} m = X_{n}\frac{m}{m^{n+1}} = \frac{X_{n}}{m^{n}} = Z_n$$