**Proof plan.** Solve the master equation via the generating [function](/page/Function) method ([method of characteristics](/page/Method%20of%20Characteristics)), extract the probability of extinction $P_0(t) = \mathcal{P}(0, t)$, and take $t \to \infty$. **Step 1 (Generating function PDE).** Let $N(t)$ denote the population size with $N(0) = N_0$. The master equation for $P_n(t) := \mathbb{P}(N(t) = n)$ is \begin{align*} \frac{dP_0}{dt} &= \mu P_1, \\ \frac{dP_n}{dt} &= \lambda(n-1)P_{n-1} - (\lambda + \mu)nP_n + \mu(n+1)P_{n+1}, \quad n \geq 1. \end{align*} Defining the generating function $\mathcal{P}(z, t) = \sum_{n=0}^\infty P_n(t)\,z^n$, multiplying the master equation by $z^n$ and summing yields \begin{align*} \frac{\partial \mathcal{P}}{\partial t} = (z - 1)(\lambda z - \mu)\frac{\partial \mathcal{P}}{\partial z}. \end{align*} The initial condition $N(0) = N_0$ gives $\mathcal{P}(z, 0) = z^{N_0}$. **Step 2 (Method of characteristics).** The characteristic equations are \begin{align*} \frac{dt}{d\tau} = 1, \quad \frac{dz}{d\tau} = (1 - z)(\lambda z - \mu), \quad \frac{d\mathcal{P}}{d\tau} = 0. \end{align*} The third equation gives $\mathcal{P} = \text{const.}$ along characteristics. From the first, $t = \tau$. The second is a [separable](/page/Separable) ODE whose solution (for $\lambda \neq \mu$) gives \begin{align*} \frac{z - \mu/\lambda}{1 - z} = c_0\, e^{(\lambda - \mu)t} \end{align*} for a constant $c_0$ determined by the initial data. Setting $t = 0$ and $z = s$ (the Cauchy data parameter): $c_0 = (s - \mu/\lambda)/(1 - s)$. Solving for $s$ in terms of $(z, t)$ and substituting into $\mathcal{P} = s^{N_0}$: \begin{align*} \mathcal{P}(z, t) = \left(\frac{e^{(\mu - \lambda)t}(\lambda z - \mu) - \mu(z - 1)}{e^{(\mu - \lambda)t}(\lambda z - \mu) - \lambda(z - 1)}\right)^{N_0}. \end{align*} **Step 3 (Extinction probability).** Setting $z = 0$: \begin{align*} P_0(t) = \mathcal{P}(0, t) = \left(\frac{\mu - \mu e^{(\mu - \lambda)t}}{\lambda - \mu e^{(\mu - \lambda)t}}\right)^{N_0} = \left(\frac{\mu(1 - e^{(\mu - \lambda)t})}{\lambda - \mu e^{(\mu - \lambda)t}}\right)^{N_0}. \end{align*} **Step 4 (Long-time [limit](/page/Limit)).** If $\lambda > \mu$, then $(\mu - \lambda) < 0$ and $e^{(\mu - \lambda)t} \to 0$ as $t \to \infty$. Hence \begin{align*} P_0(\infty) = \left(\frac{\mu}{\lambda}\right)^{N_0}. \end{align*} If $\lambda \leq \mu$, then $e^{(\mu - \lambda)t} \to \infty$ (when $\lambda < \mu$) or remains $1$ (when $\lambda = \mu$). In both cases, $P_0(t) \to 1$: extinction is certain. **Step 5 (Mean and variance).** Differentiating $\mathcal{P}$ with respect to $z$ and evaluating at $z = 1$: \begin{align*} m(t) &= \frac{\partial \mathcal{P}}{\partial z}\bigg|_{z=1} = N_0\, e^{(\lambda - \mu)t}, \\ \sigma^2(t) &= N_0\,\frac{\lambda + \mu}{\lambda - \mu}\,e^{(\lambda - \mu)t}\left(e^{(\lambda - \mu)t} - 1\right). \end{align*} The mean satisfies the deterministic ODE $dm/dt = (\lambda - \mu)m$, confirming that the ODE model captures the expected behaviour. The variance grows exponentially, reflecting the increasing spread of possible population sizes.