Let $\left(X_n\right)_{n \geq 0}$ be a branching process. We are interested in the probability that the process gets extinct, that is that the process reaches state 0 . Let $H_0$ be the first hitting time of state 0 , that is $ H_0=\inf \left\{n \geq 0: X_n=0\right\} $ We are interested in probability that $H_0$ is finite. Observe that, due to independence, $ \mathbb{P}\left(H_0<\infty \mid X_0=k\right)=\left(\mathbb{P}\left(H_0<\infty \mid X_0=1\right)\right)^k, \quad k \in \mathbb{N} . $ Thus, we will consider a branching process $\left(X_n\right)_{n \geq 0}$ that starts from one individual ( $X_0=1$ ). Let $Y$ be its offspring distribution and $G(s)$ and $F_n(s), n \in \mathbb{N}$, as defined in Definition 5.1.2 Let $\alpha$ denote its extinction probability, that is $ \alpha=\mathbb{P}\left(H_0<\infty \mid X_0=1\right) $ Observe that $ \left\{H_0<\infty\right\}=\bigcup_{n=1}^{\infty}\left\{X_n=0\right\} $ and that the sequence of events $\left\{\left\{X_n=0\right\}, n \in \mathbb{N}\right\}$ is increasing, thus by the continuity of a probability measure, $ \mathbb{P}\left(H_0<\infty \mid X_0=1\right)=\mathbb{P}\left(\bigcup_{n=1}^{\infty}\left\{X_n=0\right\} \mid X_0=1\right)=\lim _{n \rightarrow \infty} \mathbb{P}\left(X_n=0 \mid X_0=1\right) $ Let $ \alpha_n=\mathbb{P}\left(X_n=0 \mid X_0=1\right), \quad n \in \mathbb{N} $ We make the following three observations about the sequence $\left\{\alpha_n, n \in N\right\}$ : (1) the sequence $\left\{\alpha_n, n \in \mathbb{N}\right\}$ is increasing: since $ \left\{X_{\mathrm{n}}=0\right\} \subseteq\left\{X_{\mathrm{n}+1}=0\right\}, \quad n \in \mathbb{N} $ it follows that $ \alpha_n=\mathrm{P}\left(X_n=0 \mid X_0=1\right) \leq \mathbb{P}\left(X_{n+1}=0 \mid X_0=1\right)=\alpha_{n+1}, \quad n \in \mathbb{N} $ (2) the sequence $\left\{\alpha_n, n \in \mathbb{N}\right\}$ is increasing and bounded above by 1 (since for a fixed $n \in \mathbb{N}, \alpha_n$ is a probability), hence it is convergent and $ \alpha=\lim _{n \rightarrow \infty} \alpha_n $ (3) $\alpha_{n+1}=G\left(\alpha_n\right), \quad n \in N:$ since (3) $\alpha_{n+1}=G\left(\alpha_n\right), n \in \mathbb{N}$ : since $ \alpha_{n+1}=\mathrm{P}\left(X_{n+1}=0 \mid X_0=1\right)=F_{n+1}(0), \quad n \in \mathbb{N} $ by Proposition 5.1.1, $ \alpha_{n+1}=F_{n+1}(0)=G\left(F_n(0)\right)=G\left(\alpha_n\right), \quad n \in \mathbb{N} $ Thus, $\left\{\alpha_n, n \in \mathbb{N}\right\}$ is a convergent sequence and $\alpha_{n+1}=G\left(\alpha_n\right), n \in \mathbb{N}$. Since $G(s)$ is a power series which converges for $|s|<1$, it is continuous for $|s|<1$, hence $ \alpha=\lim _{n \rightarrow \infty} \alpha_{n+1}=\lim _{n \rightarrow \infty} G\left(\alpha_n\right)=G\left(\lim _{n \rightarrow \infty} \alpha_n\right)=G(\alpha) $ Therefore, the extinction probability $\alpha$ satisfies $ \alpha=G(\alpha) $ and also $\alpha \in[0,1]$ as it is a probability. The equation $s=G(s)$ might have more than one solution in $[0,1]$. For instance, 1 is always a solution since $G(1)=1$. The following proposition clarifies which solution of $s=G(s)$ on $[0,1]$ is the extinction probability: > [!NOTE] Proposition > Let $\left(X_n\right)_{n \geq 0}$ be a branching process with $X_0=1$ and the offspring distribution $Y$. Let $G(s)$ be the probability generating function of $Y$. The extinction probability > > $ > \alpha=\mathbb{P}\left(H_0<\infty \mid X_0=1\right) > $ > > is the smallest non-negative solution of the equation > > $ > s=G(s) \quad \text { for } s \in[0,1] > $ > **Proof**: We've seen that the extinction probability $\alpha$ is a non-negative solution of the equation $s=G(s)$. It remains to show that it is the smallest non-negative solution on $[0,1]$ of the given equation. We start by showing that $G(s)$ is a non-decreasing function on $[0,1]$. We have that $G^{(1)}(s)=\sum_{k=1}^{\infty} k s^{k-1} \mathbb{P}(Y=k)=\mathbb{P}(Y=1)+2 s \mathbb{P}(Y=2)+3 s^2 \mathbb{P}(Y=3)+\ldots$thus $G^{(1)}(s) \geq 0, \quad s \in[0,1]$that is $G(s)$ is a non-decreasing function on $[0,1]$. Let $\beta \in[0,1]$ be a solution of the equation $s=G(s)$. Then, since $G(s)$ is a non-decreasing function on $[0,1]$, $ 0 \leq \beta \Rightarrow G(0) \leq G(\beta)=\beta $ Recall that $ \begin{gathered} \alpha_n=\mathbb{P}\left(X_n=0 \mid X_0=1\right), \quad n \in \mathbb{N} \\ \alpha_1=\mathbb{P}\left(X_1=0 \mid X_0=1\right)=\mathbb{P}(Y=0)=G(0) \\ \alpha_{n+1}=G\left(\alpha_n\right), \quad n \in \mathbb{N} \end{gathered} $ and that $ \alpha=\lim _{n \rightarrow \infty} \alpha_n $ Hence, $ \begin{aligned} \alpha_1=G(0) & \leq G(\beta)=\beta \\ \alpha_2=G\left(\alpha_1\right) & \leq G(\beta)=\beta \\ \alpha_3=G\left(\alpha_2\right) & \leq G(\beta)=\beta \\ \cdots & \\ \alpha_{n+1}=G\left(\alpha_n\right) & \leq G(\beta)=\beta, \quad n \in \mathbb{N}, \\ \alpha=\lim _{n \rightarrow \infty} \alpha_n & \leq \beta . \end{aligned} $ Therefore, the extinction probability is the smallest non-negative solution of the equation $ G(s)=s $