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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.

Dr. NGILAZI BANDA
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  • What is your question tho? To validate your result? – FisherDisinformation Jul 14 '15 at 13:15
  • What is the best way of concluding on the properties of estimators that lead to such result,. – Dr. NGILAZI BANDA Jul 14 '15 at 13:26
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    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. – Did Jul 14 '15 at 13:43
  • @Did Thank you very much for your effort,its not that easy to get such a friendly response – Dr. NGILAZI BANDA Jul 14 '15 at 13:46
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    @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. – Math1000 Jul 14 '15 at 13:51
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    @Math1000 Indeed, there might not be any MLE but a "supremum likelihood estimate", so to speak. – Did Jul 14 '15 at 13:55

1 Answers1

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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 $$\operatorname{argmax}_\theta \ell(\theta) = \max\{\theta : x_i \geqslant \theta\; \forall i\}=\min_i x_i. $$

Math1000
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