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.
Best Answer
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 $$\varphi(t)= \psi \bigl(\varphi(t/\mu)\bigr) \,.$$