$\mathrm{Var}\left(\frac{1}{n}\sum_{i=1}^{n}{(\ln(x)-\mu)^2}\right)$ where $x \sim \mathrm{LogNorm}(\mu, \sigma^2)$ I apologize for asking so many questions.  My last question, I am stuck with evaluating $\mathrm{Var}\left(\frac{1}{n}\sum_{i=1}^{n}{(\ln(x)-\mu)^2}\right)$, where $x \sim \mathrm{LogNorm}(\mu, \sigma^2)$. I know this simplifies to $\frac{1}{n^2}\mathrm{Var}\left(n\sigma^2\right)$, but I'm not confident this is equal to 0. Can anyone offer assistance?

For a normal distribution with mean $\mu$ and variance $\sigma^2$, the moments are $\mathbb E(X)=\mu, \mathbb E(X^2)=\mu^2+\sigma^2, \mathbb E(X^3)=\mu^3+3\sigma^2\mu, \mathbb E(X^4)=\mu^4+6\sigma^2\mu^2+3\sigma^4$
So let $Y=\log X\sim \text{Normal}$ $$\begin{split}\operatorname{Var}\left(\frac 1 n \sum_i \left(Y-\mu\right)^2\right)
&=\frac 1 {n^2}\sum_iVar(Y-\mu)^2\\
&=\frac 1 {n^2}\sum_i\left[E(Y-\mu)^4-[E(Y-\mu)^2]^2\right]\\
&=\frac 1 {n^2}\sum_i\left[E(Y^4-4Y^3\mu+6Y^2\mu^2-4Y\mu^3+\mu^4)-[E(Y^2-2\mu Y+\mu^2)]^2\right]\\
&=\frac 1 {n^2}\sum_i 2\sigma^4\\
&=\frac{2\sigma^4}{n}\end{split}$$
