# Derive Variance of MLE [duplicate]

I want to calculate the variance of the MLE of an iid sample $$X_1,\dots,X_n$$ if $$f(x)=\alpha x^{\alpha-1}, 0 \leq x \leq 1, \alpha >0$$ The MLE is fairly easy calculated as: $$\hat{\alpha}=-\frac{n}{\sum_{i=1}^n\ln(x_i)}$$ I think about deriving the density of $$\hat{\alpha}$$, and then calculating $$E(\hat{\alpha}^2)$$ and $$E(\hat{\alpha})$$ to end up with $$V(\hat{\alpha})=E(\hat{\alpha}^2)-E(\hat{\alpha})^2$$, however, this approach seems to be very cumbersome. Is there an easier way to derive the variance?

To give an example of how an easier way may look like. Recently, I wanted to derive the variance of the MLE if $$X_i \overset{iid}{\sim}\textrm{Exp}(\lambda)$$. In this case, the MLE is $$\hat{\lambda}=\frac{n}{\bar{x}}$$. However, if $$X_i\sim \textrm{Exp}(\lambda)$$, then we could also say that $$X_i\sim \textrm{Gamma}(1,\lambda)$$ and $$\sum_{i=1}^nX_i\sim \textrm{Gamma}(n,\lambda)$$. Therefore, $$\frac{1}{\bar{x}}\sim \textrm{InvGamma}(n,\lambda)$$, which we can then use to calculate the moments of $$\hat{\lambda}$$.

First, let's note that your $$x \sim \text{Beta}(\alpha,1)$$. This makes life easier for you, given your previous work, as $$-\ln x \sim \text{Exp}(\alpha)$$ (we can save ourselves the derivation and look this up on the Beta distribution's Wikipedia page.) We observe that $$\hat{\alpha} = 1 / \overline{-\ln(x)}$$, the inverse of the sample mean of the $$-ln\,x_i$$, i.e., the inverse of the sample mean of an Exponential variate.
Your derivation of the distribution of $$1/\bar{x}$$ where $$x \sim \text{Exp}(\lambda)$$ shows us, with next to no work, that the distribution of $$\hat{\alpha}$$ will be $$\text{Inverse Gamma}(n, \alpha)$$. You can get all the desired moments from that!