Derive Variance of MLE [duplicate]

Question

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?

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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 $\hat{\lambda}$.

Answer 1

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!