Consider a branching Markov chain $(X_n)_{n \geq 0}$, where $X_0 = 1$ and the offspring distribution is given by $$p(0) = \frac 1 6, p(1) = \frac 1 2, p(2) = \frac 1 3.$$ The number of offspring produced by each individual is independent of the number of offspring produced by other individuals. Find the variance of the $n^{th}$ generation.
My working
Let $\mu$ and $\mu_n$ be the mean of the offspring distribution and the $n^{th}$ generation respectively. Similarly, let $\sigma^2$ and $\sigma^2_n$ be the variance of the offspring distribution and the $n^{th}$ generation respectively. From the definition of a random sum, we have
$$\begin{aligned} \mu_n & = \mu_{n - 1}\mu\\[1 mm] & = \mu_0\mu^n\\[1 mm] & = \mu^n \end{aligned}$$
and
$$\begin{aligned} \sigma^2_n & = \mu_{n - 1}\sigma^2 + \sigma^2_{n - 1}\mu^2\\[1 mm] & = \mu^{n - 1}\sigma^2 + \sigma^2_{n - 1}\mu^2\\[1 mm] & = \mu^{n - 1}\sigma^2 + \dots + \mu^{2n - 2}\sigma^2 + \sigma^2_0\mu^{2n}\\[1 mm] & = \sigma^2\mu^{n - 1}(1 + \dots + \mu^{n - 1})\\[1 mm] & = \sigma^2\mu^{n - 1}\left[\frac {\mu^n - 1} {\mu - 1}\right] \end{aligned}$$
I believe my mean of the $n^{th}$ generation is correct, but I am unsure if I have solved the recurrence relation for the variance of the $n^{th}$ generation correctly. If I have gone wrong anywhere, any intuitive explanations will be greatly appreciated :)
You are correct. By law of total variance, \begin{align*} \sigma_n^2:=\mathrm{Var}(X_n)&=\mathrm{Var}(\mathbb E[X_n\mid X_{n-1}])+\mathbb E[\mathrm{Var}(X_n\mid X_{n-1})]\\[.4em] &=\mu^2\,\sigma_{n-1}^2+\mathbb E[X_{n-1}]\,\sigma^2\\[.4em] &=\mu^2\,\sigma_{n–1}^2+\mu^{n-1}\,\sigma^2\\[.4em] &=\mu^2\left(\mu^2\,\sigma_{n-2}^2+\mu^{n-2}\,\sigma^2\right)+\mu^{n-1}\,\sigma^2\\ &=\ldots\\ &=\!\mu^{2n}\,\sigma^2_0+\sigma^2\sum_{i=n-1}^{2n-2}\mu^k\\[.4em] &=\mu^{2n}+\frac{\mu^{2n-1}-\mu^{n-1}}{\mu-1}\sigma^2. \end{align*}