[proofplan] For part (i), we use the tower property: $\mathbb{E}[Z_{n+1}] = \mathbb{E}[\mathbb{E}[Z_{n+1} \mid Z_n]]$, and since $Z_{n+1}$ is the sum of $Z_n$ independent offspring variables each with mean $\mu$, we get $\mathbb{E}[Z_{n+1}] = \mu \,\mathbb{E}[Z_n]$. Iterating from $\mathbb{E}[Z_0] = 1$ gives $\mathbb{E}[Z_n] = \mu^n$. For part (ii), we use the conditional variance formula $\operatorname{Var}(Z_{n+1}) = \mathbb{E}[\operatorname{Var}(Z_{n+1} \mid Z_n)] + \operatorname{Var}(\mathbb{E}[Z_{n+1} \mid Z_n])$ to establish a recurrence for $\operatorname{Var}(Z_n)$, then solve it. [/proofplan] [step:Compute $\mathbb{E}[Z_n] = \mu^n$ by iterating the conditional expectation] Given $Z_n = k$, the next generation is $Z_{n+1} = \sum_{i=1}^{k} X_i$, where $X_1, X_2, \ldots$ are i.i.d. offspring variables with $\mathbb{E}[X_i] = \mu$. By [linearity of expectation](/theorems/1117), \begin{align*} \mathbb{E}[Z_{n+1} \mid Z_n = k] = k\mu, \end{align*} so $\mathbb{E}[Z_{n+1} \mid Z_n] = \mu Z_n$. Taking expectations of both sides ([tower property](/theorems/1121)), \begin{align*} \mathbb{E}[Z_{n+1}] = \mu\, \mathbb{E}[Z_n]. \end{align*} Since $Z_0 = 1$, iterating gives $\mathbb{E}[Z_n] = \mu^n$ for all $n \ge 0$. [/step] [step:Establish a recurrence for $\operatorname{Var}(Z_n)$ using the conditional variance formula] The conditional variance formula states $\operatorname{Var}(Z_{n+1}) = \mathbb{E}[\operatorname{Var}(Z_{n+1} \mid Z_n)] + \operatorname{Var}(\mathbb{E}[Z_{n+1} \mid Z_n])$. We compute each term. **First term.** Given $Z_n = k$, the variable $Z_{n+1} = \sum_{i=1}^{k} X_i$ is a sum of $k$ independent copies with variance $\sigma^2$ each, so $\operatorname{Var}(Z_{n+1} \mid Z_n = k) = k\sigma^2$. Hence $\operatorname{Var}(Z_{n+1} \mid Z_n) = \sigma^2 Z_n$, and \begin{align*} \mathbb{E}[\operatorname{Var}(Z_{n+1} \mid Z_n)] = \sigma^2 \,\mathbb{E}[Z_n] = \sigma^2 \mu^n. \end{align*} **Second term.** From the previous step, $\mathbb{E}[Z_{n+1} \mid Z_n] = \mu Z_n$, so \begin{align*} \operatorname{Var}(\mathbb{E}[Z_{n+1} \mid Z_n]) = \operatorname{Var}(\mu Z_n) = \mu^2 \operatorname{Var}(Z_n). \end{align*} Combining, \begin{align*} \operatorname{Var}(Z_{n+1}) = \sigma^2 \mu^n + \mu^2 \operatorname{Var}(Z_n). \end{align*} [/step] [step:Solve the recurrence to obtain the closed-form variance] Let $v_n = \operatorname{Var}(Z_n)$. The recurrence is $v_{n+1} = \mu^2 v_n + \sigma^2 \mu^n$ with $v_0 = 0$ (since $Z_0 = 1$ is deterministic). **Case $\mu \ne 1$.** We claim $v_n = \sigma^2 \mu^{n-1} \cdot \frac{\mu^n - 1}{\mu - 1}$. Unrolling the recurrence, \begin{align*} v_n &= \mu^2 v_{n-1} + \sigma^2 \mu^{n-1} \\ &= \mu^2(\mu^2 v_{n-2} + \sigma^2 \mu^{n-2}) + \sigma^2 \mu^{n-1} \\ &= \mu^4 v_{n-2} + \sigma^2 \mu^n + \sigma^2 \mu^{n-1}. \end{align*} Continuing to unroll (applying the recurrence $n$ times down to $v_0 = 0$), \begin{align*} v_n = \sigma^2 \sum_{j=0}^{n-1} \mu^{2j} \cdot \mu^{n-1-j} = \sigma^2 \mu^{n-1} \sum_{j=0}^{n-1} \mu^j = \sigma^2 \mu^{n-1} \cdot \frac{\mu^n - 1}{\mu - 1}. \end{align*} **Case $\mu = 1$.** The recurrence becomes $v_{n+1} = v_n + \sigma^2$, giving $v_n = n\sigma^2$. [/step]