I would like to compute the estimators $\alpha , \beta$ (in the Maximum Likelihood sense) of: $f(x) = \frac{x-\alpha}{\beta^2} e^{-\frac{(x-\alpha)^2}{2\beta^2}}$ .

I compute the likelihood function: $L(\alpha, \beta) = \prod_{i=1}^{N}\frac{x_i-\alpha}{\beta^2}e^{-\frac{(x_i-\alpha)^2}{2\beta^2}}$

Then, the log-likelihood function:

$\log(L(\alpha, \beta)) = \frac{-1}{2} \sum_{i=1}^{N} \frac{x_i-\alpha}{2\beta^2}\log(\frac{x_i-\alpha}{\beta^2})$

After computing the derivative of $\log(L(\alpha, \beta))$ with respect to $\alpha$ and $\beta$, I ended up with:

$\sum_{i=1}^{N}(x_i-\alpha) \left [ -2\log \left (\frac{x_i-\alpha}{\beta^2} \right )- 1 \right ] \overset{!}{=} 0$

$\sum_{i=1}^{N}(x_i-\alpha)^2 \left [ \log \left (\frac{x_i-\alpha}{\beta^2} \right )- \frac{1}{x_i-\alpha} \right ] \overset{!}{=} 0$

However, I do not see how to solve such a system.

  • $\begingroup$ I have read that for certain distributions, it can't be possible to analytically find a solution. Is it the case here? $\endgroup$
    – floflo29
    Commented May 27, 2016 at 19:21


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