0
$\begingroup$

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.

$\endgroup$
2

2 Answers 2

0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.