Finding $\mathbb{E}\left[\hat{\sigma}^2\right] = \frac{\sum_{i=1}^{n}{\left(\ln(x)-\mu\right)^2}}{n}$ where $X \sim \mathrm{LogNorm}$

Question

I'm tasked with finding $\mathbb{E}\left(\hat{\sigma}^2\right)$ where $\hat{\sigma}^2 = \frac{\sum_{i=1}^{n}{\left(\ln(x)-\mu\right)^2}}{n}$, $x \sim \mathrm{Lognorm}(\mu, \sigma^{2}$) . So far I get,$$\frac{1}{n}\sum_{i=1}^{n}{ \mathbb{E}\left((\ln(x)-\mu)^2\right)}$$ and can simplify to $$\frac{1}{n}\sum_{i=1}^{n}{\sigma^2} \\ \frac{n\sigma^2}{n} = \sigma^2$$ but I know this isn't right because the MLE estimate of $\sigma^2$ shouldn't be unbiased. Can anyone assist on what I did wrong?

Answer 1

You did no mistake: $X \sim \textrm{Lognorm}(\mu,\sigma^2) \implies \ln(X)\sim\textrm{Normal}(\mu,\sigma^2)$ then set $Y=\ln(X)$ we end up with $$\hat{\sigma}^2=\frac{1}{n}\sum_{k=1}^n(Y-\mu)^2 \implies \mathbb{E}[\hat{\sigma}^2]=\frac{1}{n}n\sigma^2=\sigma^2$$ This is unbiased. However: assume $\mu$ not known. Then $$s^2=\frac{1}{n}\sum_{k=1}^n(Y-\bar{Y})^2 \implies \mathbb{E}[s^2]=\frac{n-1}{n}\sigma^2$$ which is biased. You probabily expected this one. But when $\mu$ is known, the naive variance estimator is unbiased.

Answer 2

Note that $\mathbb{E}[(\ln(X)-\mu)^{2}]=\sigma^{2}$ because $\ln(X)\sim \mathcal{N}(\mu,\sigma^{2})$, so $\mathbb{E}[(\ln(X)-\mu)^{2}]=\text{Var}(\ln(X))=\sigma^{2}$.