# Total offspring of Poisson multitype branching process

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. 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 $$c$$.

## 1 Answer

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 $$1$$.