Variance recursion formula in Galton-Watson process

Question

Consider a Galton-Watson process with expected offspring $\mathbb{E}[\xi]=\mu<\infty$ and variance $\text{Var}(\xi)=\sigma^2<\infty$ where the offspring in generation $t\in\mathbb{N}$ is given by $Z_t$. Suppose I introduce a new random variable $W_t$ given by $W_t=\mu^{-t}Z_t$.

Noting this setup, in the proof of Proposition $1.4$ on page $7$ of this source they say:

"a straightforward calculation yields $$\text{Var}(W_t)=\frac{\sigma^2}{\mu^{t+1}}+\text{Var}(W_{t-1})"$$ but I am unable to see what straightforward calculation this follows from.

Any help would be appreciated.

Answer 1

I've found a solution that uses the law of total variance conditioning on $Z_{t-1}$:

$$ \begin{align} \text{Var}(Z_t)&=\mathbb{E}[\text{Var}(Z_t|Z_{t-1})]+\text{Var}(\mathbb{E}[Z_t|Z_{t-1}])\\ &=\mathbb{E}[Z_{t-1}\cdot\text{Var}(\xi)]+\text{Var}(Z_{t-1}\cdot\mathbb{E}[\xi])\\ &=\sigma^2\mu^{t-1}+\mu^2\text{Var}(Z_{t-1})\\ \end{align} $$

from which we multiply through by $\mu^{-2t}$ to get

$$ \mu^{-2t}\text{Var}(Z_t)=\frac{\sigma^2}{\mu^{t+1}}+\mu^{-2(t-1)}\text{Var}(Z_{t-1}). $$

Taking the exponents of $\mu$ inside the variances yields

$$ \text{Var}(\mu^{-t}Z_t)=\frac{\sigma^2}{\mu^{t+1}}+\text{Var}(\mu^{-(t-1)}Z_{t-1}) $$

which can be written in terms of $W_t=\mu^{-t}Z_t$ as

$$ \text{Var}(W_t)=\frac{\sigma^2}{\mu^{t+1}}+\text{Var}(W_{t-1}). $$