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Let the density of $X$ be $\displaystyle\frac{1}{2\theta}\mathbb{I}_{(-\theta,\theta)}(x)$.

I did some calculations and since the derivative is always negative, I thought the MLE would be $\hat \theta=\max(\{x_i,-x_i\})$. However, at that point the density is zero...

The density doesn't determine completely the distribution, so we could instead suppose that the density is $\displaystyle\frac{1}{2\theta}\mathbb{I}_{[-\theta,\theta]}(x)$, and since $[x,\infty)\cap [-x,\infty)=[\max\{x,-x\},\infty)$, we could now have $\hat \theta=\max(\{x_i,-x_i\})$ without the density being zero.

Am I correct?

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  • $\begingroup$ Nope. You can't do that, since as you said the density is actually $0$. MLE is just not the way to go here. $\endgroup$ – Stan Tendijck Feb 18 at 9:57
  • $\begingroup$ Mle, which definitely exists here, cannot be found using differentiation. Presumably, you have a sample of $n$ observations. And it makes no difference whether you write $[-\theta,\theta]$ or $(-\theta,\theta)$. $\endgroup$ – StubbornAtom Feb 18 at 10:05
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    $\begingroup$ math.stackexchange.com/q/2795320/321264. $\endgroup$ – StubbornAtom Feb 18 at 10:26
  • $\begingroup$ @StubbornAtom $\max_{i\in I} |x_i|=\min_{i\in I}\{x_i,-x_i\}$. But my question is still not answered. $\endgroup$ – An old man in the sea. Feb 18 at 10:52
  • $\begingroup$ $\min \{x_i,-x_i\}$ is just $-|x_i|$, and it is not an MLE here. I don't know how you reached this answer. $\endgroup$ – StubbornAtom Feb 18 at 13:04
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Two functions $f(x)=\displaystyle\frac{1}{2\theta}\mathbb{I}_{(-\theta,\theta)}(x)$ and $f(x)=\displaystyle\frac{1}{2\theta}\mathbb{I}_{[-\theta,\theta]}(x)$ are PDFs of the same uniform distribution. No matter what kind of brackets is written here.

Please note also, that it is not "boundary of parameter space". Here parameter space is $\theta>0$, and a.s. $\hat\theta=\max_{i\in I}|x_i|>0$ too. It lies inside the space of all possible values of parameter.

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