# Finding maximum likelihood estimator, symmetric uniform distribution

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?

• 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). – dsaxton Aug 4 '18 at 16:01
• 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/…. – StubbornAtom Aug 4 '18 at 17:19

You must be more careful in deducing the likelihood function, otherwise your thinking seems correct. The density function for one $$X_i$$ is $$f(x_i)=\frac1{2\theta}\cdot I\{ -\theta So the likelihood function can be written as \begin{align*} \mathcal{L}(\theta)&=\prod_{i=1}^n \frac1{2\theta}\cdot I\{ -\theta and this is zero if $$\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