# Functional equation related to the supercritical Galton-Watson process

Consider the Galton-Watson process $$(Z_n)_n$$ and the martingale $$M_n = \mu^{-n}Z_n$$. We assume that $$\mu > 1$$, i.e. our Galton-Watson process is supercritical and $$\mathbb{E}(M_\infty) < \infty$$. Prove that the Laplace transformation $$\varphi(t) := \mathbb{E}(e^{-tM_\infty})$$ of the limit random RV $$M_\infty$$ and the probability generating function $$\psi$$ of the offspring distribution, i.e. $$\psi(s) = \mathbb{E}(s^X) = \sum_{k \ge 0} p_ks^k$$ for $$\lvert s \rvert \le 1$$, satisfy the functional equation

$$\varphi(t) = \psi(\varphi(t/\mu)), \qquad \varphi(0) = 1, \qquad \partial^+ \varphi(0) = -\mathbb{E}(M_\infty)$$

The second and third points are clear to me, but I do not understand how to prove the first one. I tried to just plug in the Laplacian in $$\psi$$, but the resulting term is just messy.

• Is the answer I wrote below clear? Commented Nov 24, 2022 at 16:31
• Yeah, I just forgot to accept. I am sorry. Thanks for your answer. Commented Nov 24, 2022 at 18:05

Let $$Z_{n-1,j}$$ denote the number of descendants in generation $$n$$ of the $$j$$'th member of the first generation. Then $$Z_n=\sum_{j=1}^{Z_1} Z_{n-1,j} \,,$$ where the summands are i.i.d. (with the law of $$Z_{n-1}$$) and independent of $$Z_1$$. Thus $$M_{n-1,j}:=Z_{n-1,j}/\mu^{n-1}$$ satisfy $$M_n=\frac1{\mu}\sum_{j=1}^{Z_1}M_{n-1,j} \,.$$ Let $$\varphi_n(t):=E(e^{-tM_n})$$. Then $$E(e^{-tM_n} | Z_1)= \varphi_{n-1}(t/\mu)^{Z_1} \tag{*}\,,$$ so taking expectation of both sides (thinking of the RHS of $$(*)$$ as $$s^{Z_1}$$), gives $$\varphi_n(t)=E(e^{-tM_n})= E\Bigl(\varphi_{n-1}(t/\mu)^{Z_1}\Bigr)= \psi \bigl(\varphi_{n-1}(t/\mu)\bigr) \tag{**}\,.$$ Since $$\varphi_n(t) \to \varphi(t)$$ by bounded convergence and $$\psi$$ is continuous, we infer from $$(**)$$ that $$\varphi(t)= \psi \bigl(\varphi(t/\mu)\bigr) \,.$$