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I am trying to calculate the maximun likelihood estimator, but there is something which is wrong and I can't find the error.

$f(x;\theta)=\theta x^{-2} e^{-\theta/x}$ for x>0.

The likelihood is $\displaystyle\prod_{i=1}^n f(x;\theta) = \displaystyle\prod_{i=1}^n \theta x^{-2} e^{-\theta/x}$ = $\theta^{n} (\displaystyle\sum_{i=1}^n x_{i}^{-2}) e^{-\theta \displaystyle\sum_{i=1}^n x_{i}^{-1}} $

and the log-likelihood is

$l(\theta;x) = log L(\theta;x) = nlog\theta (\displaystyle\sum_{i=1}^n x_{i}^{-2}) - \theta \displaystyle\sum_{i=1}^n x_{i}^{-1}$

To maximize this we differentiate to obtain

$l'(\theta) = \frac{n}{\theta} \displaystyle\sum_{i=1}^n x_{i}^{-2} - \displaystyle\sum_{i=1}^n x_{i}^{-1} $

and then solve $l'(\hat{\theta})=0$ to obtain

$\frac{n}{\theta} \displaystyle\sum_{i=1}^n x_{i}^{-2} = \displaystyle\sum_{i=1}^n x_{i}^{-1} $ $\implies$ $\frac{n}{\hat{\theta}} \frac{1}{\bar{x^2}} = \frac{1}{\bar{x}}$ $\implies$ $\hat{\theta} = \frac{n\bar{x}}{\bar{x^2}} $

I would appreciate if someone could tell me what I am doing wrong.

Thank you.

  • The error comes at the end of the first sentence. You do not have two $\sum$. Start directly taking the sum of the logarithms. – Claude Leibovici Mar 28 '17 at 12:19
  • when loging the product, the sum with $x^{-2}$ comes inside the log. so when differentiating and setting equal = 0, you should get log(sum) - sum =0 instead of sum-sum=0. – mathreadler Mar 28 '17 at 12:43
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    First line should read $$\theta^n\left(\prod_{i=1}^n x_i^{-2}\right)\mathrm{e}^{-\theta\sum_{i=1}^nx_i^{-1}}$$ – Chinny84 Mar 28 '17 at 12:58

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