$E[Z_{n}^2]$ in terms of $E[Z_{n-1}^2]$

Question

After calculating part i),

I am getting $E[Z_{n}^2]$ = $(\sigma^2 + \mu^2)$ $E[Z_{n-1}]$ which I think is incorrect.

$E[Z_{n}^2]$ = $\sum_{k=0}^{inf}$ $E(Z_{n}^2/ Z_{n-1}=k) * P(Z_{n-1}=k)$ = $\sum_{k=0}^{inf}$$k(\sigma^2 + \mu^2$)$P(Z_{n-1}=k)$

which gives the above result. Any help is appreciated.

Answer 1

Note that, $\text{E}(Z_n | Z_{n-1} = k) = k\mu$, and $\text{Var}(Z_n | Z_{n-1} = k) = \text{Var}\left(\sum_{i=1}^k \xi_{n-1, i} \right) = k \text{Var}(\xi_{1,1}) = k\sigma^2.$ So, $\text{E}(Z_n^2 | Z_{n-1} = k) = k\sigma^2 + k^2 \mu^2.$

Therefore, $$\text{E}(Z_n^2) = \sum_{k} \text{E}(Z_n^2 | Z_{n-1} = k) \Pr(Z_{n-1} = k) = \sigma^2 \text{E}(Z_{n-1}) + \mu^2 \text{E}(Z_{n-1}^2).$$