# 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\$.

Contents

\$\$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.

Rate this post