0
$\begingroup$

The question reads : Find the maximum likelihood estimator for theta based on the sample of size n from a distribution with density $$f(x|\theta)=\frac{2\theta^2}{x^3};x>\theta.$$ According to my calculations it is observed that the maximum is not attained at a critical point, however I thought it fit to assume theta to be more than zero but not very much sure whether my conclusion will be nice that way.

$$L(\theta)=2n\ln(\theta)+n\ln2-3(\sum \ln(x))$$

$L'(\theta)=\frac{2}{\theta}$ which maximizes the the pdf but I am finding it difficult to conclude from the derivative with respect to theta.

Thank you in advance.

$\endgroup$
  • $\begingroup$ What is your question tho? To validate your result? $\endgroup$ – FisherDisinformation Jul 14 '15 at 13:15
  • $\begingroup$ What is the best way of concluding on the properties of estimators that lead to such result,. $\endgroup$ – DOCTOR NGILAZI BANDA JOSHUA Jul 14 '15 at 13:26
  • 2
    $\begingroup$ The loglikelihood L is increasing hence one wants to choose theta as large as possible. But "as large as possible" here is severely restricted by the condition that x>theta for every x in the sample, otherwise L=-oo. Thus the MLE is hat-theta=min x. $\endgroup$ – Did Jul 14 '15 at 13:43
  • 1
    $\begingroup$ @Did somewhat of a pedantic point here, but if $\hat\theta = \min_i x_i$ then $\hat\theta=x_j$ for some $j$, and so we don't have $x_i>\theta$ for all $i$. So I amended my answer to use $\geqslant$ instead - no harm since this is an absolutely continuous distribution. $\endgroup$ – Math1000 Jul 14 '15 at 13:51
  • 1
    $\begingroup$ @Math1000 Indeed, there might not be any MLE but a "supremum likelihood estimate", so to speak. $\endgroup$ – Did Jul 14 '15 at 13:55
3
$\begingroup$

Elaborating on on @Did's comment, we find that $$\frac\partial{\partial \theta}\ell(\theta)=\frac{2n}\theta>0$$ for all $\theta$. Hence the MLE is $$\operatorname{argmax}_\theta \ell(\theta) = \max\{\theta : x_i \geqslant \theta\; \forall i\}=\min_i x_i. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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