Solved – Computing the Variance of an MLE

Suppose we have i.i.d. $n$ observations $(X_1,X_2,…X_n)$ from a
population with density $$f_theta(x)=begin{cases}theta x^{theta-1}&text{ if }0leq xleq 1\0&text{otherwise.}end{cases}$$ Note that $theta>0$. If $T_n$ denotes the
maximum likelihood estimator of $theta$ given that $n$ is the sample
size, then show that $$text{var}(T_n)stackrel{ntoinfty}{longrightarrow} 0$$

This is the problem I am trying to solve: I computed $$T_n=dfrac{-n}{sum_{i=1}^{n}ln(X_i)}$$ But I am stuck in finding $text{var}(T_n)$ and showing that $text{var}(T_n)to0$ as $ntoinfty$.

Combining @Xi'an comments and @wolfies' answer, we have that

$$1/T_n = frac 1n sum_{i=1}^n (-ln X_i)$$

But $-ln X_i sim {rm Exp}(1/theta)$ (where $1/theta$ is the scale parameter), which essentially is a Gamma distribution with shape parameter $1$, and so by the summation properties of the Gamma distribution,

$$sum_{i=1}^n (-ln X_i) sim {rm Gamma}(n, 1/theta)$$.

Using the scaling properties of the Gamma distribution we obtain

$$1/T_n sim {rm Gamma}(n, 1/(ntheta)) $$

Then the inverse of $1/T_n$ (i.e. $T_n$) follows an Inverse Gamma distribution, with same shape paramater and reciprocal scale parameter

$$T_n sim {rm InvGamma}(n, ntheta)$$

that has variance

$$ {rm Var} (T_n) = frac {n^2theta^2}{(n-1)^2(n-2)}$$

as the Mathematica software gave.

The leading term in the numerator is $n^2$ while the leading term in the denominator is $n^3$ so the limit of the variance of $T_n$ with respect to $n$ goes to zero.

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