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Consider the normal distribution when $\theta=\sigma^2$: $N(\theta,\theta)$ with probability density function:

  • $f(x;\theta)=\frac{1}{\sqrt{2\pi\theta}}e^{-\frac{1}{2\theta}(x-\theta)^2}$

I was asked to find the maximum likelihood estimator of the law. In other words, how to find the maximum value of:

  • $L(\theta)=\frac{1}{(2\pi\theta)^{n/2}}e^{-\frac{1}{2\theta}\sum(X_i-\theta)^2}$

where $X_1,...,X_n$ are iid following $N(\theta,\theta)$.

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  • $\begingroup$ It would be nice to know what have you tried. The maximum value of the likelihood occurs when the first derivative is equal to zero, given some mild conditions. $\endgroup$ – Tony Hellmuth May 29 '18 at 12:58
  • $\begingroup$ @TonyHellmuth I did not achieve to solve $L'(\theta)=0$, maybe there is a trick or am I just bad at algebra? $\endgroup$ – Jean Dupuis May 29 '18 at 13:01
  • $\begingroup$ How about the natural log of $L(\theta)$. Same holds at least here. $\endgroup$ – Tony Hellmuth May 29 '18 at 13:03
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HINT:

It would be much easier if you just used the log-likelihood $l(\theta)$ giving $$l(\theta)=-\frac n2\log2\pi\theta-\frac1{2\theta}\sum_i(X_i-\theta)^2$$ Then solve $$\frac{\partial l}{\partial \hat{\theta}}=0.$$

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  • $\begingroup$ Should come out nicely :) $\endgroup$ – Tony Hellmuth May 29 '18 at 13:06

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