A normal branching process $Z\_n$ initialized with $Z\_0=1$ and offspring generated from $Pois(p),p<1,$ has a total progeny / total off spring distribution


$$X=\sum\_{n=0}^\infty Z\_n$$
$X\in \mathbb{N}$ almost surely, because $Z\_n$ goes extinct eventually when $p<1$. $X$ follows the [Borel Distribution](https://en.wikipedia.org/wiki/Borel_distribution). In particular $\mathbb{E}\exp(cX)<\infty$ for small $c$.


**Question:** does the total offspring $X$ also satisfy $\mathbb{E}\exp(cX)<\infty$ for a multitype branching process with poisson offspring?


**Elaboration:**
The relevant multitype branching process can be described as a Markov chain $Z\_n\in \mathbb{N}^m,$ with a $m\times m$ matrix $A$ as parameter. The distribution $Z^{i}\_n\vert Z\_n=v$ is equal in law to


$\sum\_{j=1}^m\sum\_{k=1}^{v\_k} Pois(a\_{ij})$ 


where all poisson summands are independent.


Again, this process goes extinct when $A$ has spectral radius strictly smaller than 1.


I am interested whether the variable


$$X=\sum\_{n=0}^\infty\sum\_{j=1}^m Z^{j}\_n$$


satisfies $\mathbb{E}\exp(cX)<\infty$ for small $c$.



Total offspring of Poisson multitype branching process
Yes, with essentially the same proof. In the scalar case, just notice that $e^{cX\_n+bZ\_n}$ is a supermartingale as long as $p(e^{c+b}-1)\le b$. In the vector case you will need to find positive vectors $c,b\in(0,\infty)^m$ with the property $(e^{c+b}-1)A\le b$ (I use the notation $f(c)=(f(c\_1),\dots,f(c\_m))$ and compare the vectors entry-wise), which is possible if and only if the spectral radius of the matrix $A$ with non-negative entries is strictly less than $1$. 


