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Let $X_1, ...X_n$ be IID random variables with uniform$[ -\theta , \theta ]$ . I need to find the Maximum Likelihood estimator (MLE) of $\theta$.

My work is as follows,

The likelihood function is , $L(\theta)$ = $1/2\theta^n$ $I(|X_{(n)}| \leq \theta )$.

Since this is an decreasing function of $\theta$ , MLE of $\theta$ is $ |X_{(n)}|$.

is this correct?

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    $\begingroup$ Yes, this is correct, only the indicator function in the likelihood should be $I(|X_{(n)}| \leq \theta)$ since the support of the distribution is closed (I edited the question). $\endgroup$ – dsaxton Aug 4 '18 at 16:01
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    $\begingroup$ MLE would be the sufficient statistic found here using the Factorisation theorem, namely $\max |X_i|=\max(-X_{(1)},X_{(n)})=\max(|X_{(1)}|,|X_{(n)}|)$. Also see this post on Math.SE : math.stackexchange.com/questions/2795320/…. $\endgroup$ – StubbornAtom Aug 4 '18 at 17:19
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You must be more careful in deducing the likelihood function, otherwise your thinking seems correct. The density function for one $X_i$ is $$ f(x_i)=\frac1{2\theta}\cdot I\{ -\theta<X_i < \theta \} $$ So the likelihood function can be written as \begin{align*} \mathcal{L}(\theta)&=\prod_{i=1}^n \frac1{2\theta}\cdot I\{ -\theta<X_i < \theta \} \\ &=2^{-n}\theta^{-n} \cdot \prod_i I\{-\theta<X_i < \theta \} \\ &= 2^{-n}\theta^{-n} \cdot I\{-\theta < \min_i x_i \le \max_i x_i <\theta \} \\ &= 2^{-n}\theta^{-n} \cdot I\{0\le \max[|\min_i x_i|, |\max_i x_i|]\} < \theta \end{align*} and this is zero if $\theta$ is too small, that is, lesser than $\max[|\min_i x_i|, |\max_i x_i|]$. At that point it becomes positive, and then decreasing from there. That gives the MLE as $$ \hat{\theta}_\text{MLE}=\max[|\min_i x_i|, |\max_i x_i|] $$ Another way to the solution is by noting that $|X_i| \sim \mathcal{U}(0,\theta)$ and use the solution for that case (which leads to an equivalent likelihood function), se for instance here

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