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I'm trying to solve this problem

Let X be a random absolutely continuous variable with probability density function $$f_{\lambda\mu}(x) = \sqrt{\frac{\lambda}{2\pi x^3}}\exp{\left\{-\frac{\lambda}{2\mu^2x}(x-\mu)^2\right\}} \quad x>0$$ with $\mu,\lambda>0$. Find the MLE of $\mu$ and $\lambda$ for a sample of size $n$.

I have no problem for finding the MLE for $\lambda$. My question is about the parameter $\mu$. I proceed like this:

Likelihood function: $$f_{\lambda\mu}(x_1,\ldots,x_n) = \left( \frac{\lambda}{2\pi}\right)^{n/2}\prod x_i^{-3/2}\exp{\left\{ -\frac{\lambda}{2\mu^2}\sum\frac{(x_i-\mu)^2}{x_i} \right\}}\quad x_1,\ldots,x_n >0$$ Finding the maximum: $$\log f_{\lambda\mu}(x_1,\ldots,x_n) = \frac{n}{2}\log \left( \frac{\lambda}{2\pi}\right) -\frac{3}{2}\sum \log x_i -\frac{\lambda}{2\mu^2}\sum\frac{(x_i-\mu)^2}{x_i} \quad x_1,\ldots,x_n >0$$

$$ \frac{\partial}{\partial \mu} \log f_{\lambda\mu}(x_1,\ldots,x_n) = \frac{\lambda}{\mu^3}\sum\frac{(x_i-\mu)^2}{x_i} + \frac{\lambda}{\mu^2}\sum \frac{(x_i-\mu)}{x_i} = 0 \quad x_1,\ldots,x_n >0 $$ But I cannot solve this for $\mu$: $$ \frac{1}{\mu}\sum\frac{(x_i-\mu)^2}{x_i} + \sum \frac{(x_i-\mu)}{x_i} = 0 $$ Should I just stop here or there is any form to solve it for $\mu$?

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I would have thought you could expand your final line to

$$\displaystyle \frac{1}{\mu}\sum\frac{x_i^2 }{x_i}+\frac{1}{\mu}\sum\frac{-2 x_i \mu }{x_i}+\frac{1}{\mu}\sum\frac{\mu^2}{x_i} + \sum \frac{x_i}{x_i} + \sum \frac{-\mu}{x_i} =0 $$

and with some tidying up this would lead to the natural estimator

$$\displaystyle \mu =\frac{1}{n}\sum{x_i}.$$

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  • $\begingroup$ Thank you. I didn't expect it would be so easy just expanding it! $\endgroup$
    – Danowsky
    Commented Apr 29, 2015 at 7:16

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