Branching process: Prove that $g_Z(t) = g(tg_Z(t))$ (generative function)

27 Views Asked by At

Let $(Z_t)_{t \in \mathbb{N}_0}$ be a subcritical branching process such that $Z_0 = 1$ and $\mu = E[Z_1] \le 1.$ Furthermore, let $Z := \sum_{k = 0}^{\infty} Z_t$ be the number of overall successors. Show that $$g_Z(t) = g(tg_Z(t))$$ with $g$ being the generative function of $Z_1.$

The generative function in general is defined as

$$g_X(z) = E(z^X) = \sum_{k = 0}^{\infty} P(X = k)z^k.$$

We were given the hint to condition on the value of $Z_1,$ but I don't see how that is supposed to help me yet. I started out with: $$g(tg_Z(t)) = g_{Z_1}(tg_Z(t)) = \sum_{l = 0}^{\infty} P(Z_1 = l) (tg_Z(t))^l = \sum_{l = 0}^{\infty} P(Z_1 = l) t^l (g_Z(t))^l.$$

It might be useful to note that

$$(g_Z(t))^l = g_{lZ}(t).$$

Besides that, conditioning on the value of $Z_1$ might yield something like this:

$$P(Z = k | Z_1 = l) = \frac{P(Z = k, Z_1 = l)}{P(Z_1 = l)}.$$

So overall, we could rewrite the power series from above as

$$\sum_{l=0}^{\infty} \frac{P(Z = k | Z_1 = l)}{P(Z = k, Z_1 = l)} t^l g_{lZ}(t).$$