Find asymptotic efficiency of MLE to UMVUE

Question

Let $\{X_i\}_{i=1}^n $ be a sequence of i.i.d random variables with common pdf: $$ f(x;a,\theta) =\theta a^\theta x^{-(\theta+1)} \boldsymbol 1_{(a,\infty)}(x) \, \,\text{; where } \theta, a > 0$$ I would like to find the asymptotic relative efficiency of the MLE of a with respect to the UMVUE of a. I believe I have already found the correct MLE for $a$ but I'm not sure how to find the efficiency in this case. From the likelihood it seems like the first order statistic is the MLE. $$\mathcal{L}(X,a,\theta) = \theta^n a^{n\theta} \prod_{i=1}^n\left( x_i^{-(\theta+1)} \boldsymbol \cdot 1_{(a,\infty)}(x_i)\right) \implies \min_{1 \leq i \leq n}(x_i) = \hat{a} $$

Typically I would use the score to find the Fisher Information and then take the ratio but I don't think that works here. Thanks for your help.

Answer 1

MLE of $a$ is indeed the first order statistic $X_{(1)}=\min\limits_{1\le i\le n}X_i$ because the likelihood is non-decreasing in $a$ subject to the restriction $a<X_{(1)}$. Because the population distribution is Pareto, you can verify that $X_{(1)}$ also has a Pareto distribution from which you can get its exact variance.

UMVUE of $a$ however depends on whether $\theta$ is known or not. In any case, it is found using the Lehmann-Scheffé theorem.

• If $\theta$ is known, then $X_{(1)}$ is a complete sufficient statistic and UMVUE of $a$ is of the form $c(\theta) X_{(1)}$ for some function $c$.

• If $\theta$ is not known, then $\left(\prod\limits_{i=1}^n X_i,X_{(1)}\right)$ or equivalently $\left(U, X_{(1)}\right)$ is a complete sufficient statistic where $U=\sum\limits_{i=1}^n (\ln X_i-\ln X_{(1)})$. Here $U$ has a certain Gamma distribution, and $U$ and $X_{(1)}$ can be shown to be independent. The resulting UMVUE of $a$ would be of the form $g(U)X_{(1)}$ for some function $g$.

Using the points above, you can find the exact variance of both UMVUE and MLE of $a$. Asymptotic relative efficiency of MLE with respect to UMVUE is then the limit of the ratio $\operatorname{Var}(\hat a)/\operatorname{Var}(X_{(1)})$ as $n\to \infty$ where $\hat a$ is UMVUE of $a$. Note that Fisher information is not usually defined for non-regular distributions like this where support of the distribution depends on the parameter of interest.

Answer 2

Since the MLE is the first order statistic it has the pdf: $n \theta a^{n\theta} x^{-(n\theta+1)} \boldsymbol 1_{(a,\infty)}(x) $.

Naturally then $\displaystyle \mathbb{E}[X_{(1)}] = \frac{a n\theta}{n\theta-1 } $, so that $\displaystyle \frac{(n\theta-1)X_{(1)} }{n\theta } = T(x)$ is the UMVUE by Lehmann-Scheffé.

$$\implies \frac{Var(T(x)) }{Var(X_{1})} = \left(\frac{(n\theta -1) }{n\theta }\right)^2 = 1 - \frac{2}{n \theta} + \frac{1}{n^2\theta^2}$$

So that the asymptotic efficiency is indeed 1.