Let the number of samples n be a random variable following the distribution given by n = N + 1 where N ∼ Poi(λ) with λ > 0. Conditionally on N, let X$_1$, . . . , X$_N$$_+$$_1$ be i.i.d. following an N(μ,σ$^2$) distribution with σ$^2$ < ∞. Compute the expectation and variance of S$^2$ = $\frac{1}{N+1}$$\sum_{i=1}^n $(X$_i$ − μ)$^2$.
I'm not sure how to proceed with this question. My immediate thoughts are to use the formula for iterated conditional expectation, but given I do not have a function f$_($$_X$$_,$$_N$$_)$$_|$$_N$[(x,n)|n] I'm not sure whether this would work.
Any guidance on how to proceed would be much appreciated.
First of all, $$\frac{X_i - \mu}{\sigma} = Z_i \sim \operatorname{Normal}(0,1).$$ This much should be obvious. Consequently, $$\frac{S^2}{\sigma^2} \mid N = \frac{1}{N+1} \sum_{i=1}^n Z_i^2,$$ and we know that the square of a standard normal variable is chi-square with one degree of freedom, and the sum of $N+1$ IID such variables is chi-square with $N+1$ degrees of freedom. It follows that $$\frac{N+1}{\sigma^2} S^2 \mid N \sim \chi^2_n.$$ We then have $$\operatorname{E}[S^2 \mid N] = \frac{\sigma^2}{N+1} (N+1) = \sigma^2,$$ and since this is independent of $N$, the unconditional expectation is $$\operatorname{E}[S^2] = \operatorname{E}[\sigma^2] = \sigma^2.$$ But the variance is different: we calculate $$\operatorname{Var}[S^2 \mid N] = \left(\frac{\sigma^2}{N+1}\right)^2 (2(N+1)),$$ and by the law of total variance, the unconditional expectation is $$\begin{align*}\operatorname{Var}[S^2] &= \operatorname{Var}[\operatorname{E}[S^2 \mid N]] + \operatorname{E}[\operatorname{Var}[S^2 \mid N]] \\ &= \operatorname{Var}[\sigma^2] + \operatorname{E}[2 \sigma^4/(N+1)] \\ &= 0 + 2\sigma^4 \operatorname{E}[(N+1)^{-1}]. \end{align*}$$ The only step remaining, then, is to compute the expectation of $1/(N+1)$ where $N \sim \operatorname{Poisson}(\lambda)$. I leave this as an exercise.