# Why can't we use calculus in finding the M.L.E. of Uniform(-theta, theta)?

I've understood that we use maximum/minimum of x's as MLE of theta. But no one so far has explained the reason why differentiation won't work. Please explain.

• For the same reason we cannot use calculus for $U(0,\theta)$ and in general, when support of the distribution depends on the parameter of interest. See also math.stackexchange.com/questions/649678/…. – StubbornAtom Oct 6 '18 at 6:54
• MLE derived here. – StubbornAtom Oct 6 '18 at 6:56
• We can use calculus. In elementary math one learns to look at critical points which include the zeros of the first derivative, if they exist, but not restricted to them. – Math-fun Oct 7 '18 at 7:46

Differentiation won't work because the likelihood function, as a function of the parameter $$\theta$$, doesn't achieve its maximum where its derivative is zero.
The density function of the Uniform$$[-\theta,\theta]$$ distribution is $$f_\theta(x) =\frac1{2\theta}I(|x|\le\theta)$$ so the likelihood function for a sample $$x_1,x_2,\ldots,x_n$$ is $$L(\theta)=\prod_{i=1}^nf_\theta(x_i)=\frac1{(2\theta)^n}\prod_{i=1}^n I(|x_i|\le\theta) =\frac1{(2\theta)^n}I(\theta \ge\max_i |x_i|).$$ The likelihood is a function of $$\theta$$ which is nonzero on the interval $$I:=[\max_i|x_i|,\infty)$$. Over this interval the function is strictly decreasing, so its derivative is never zero there. Outside of $$I$$ the function is zero so its maximum doesn't occur there. Thus calculus is no help in finding the maximum of $$L$$. On the other hand, since $$L$$ is nonzero and decreasing over $$I$$, it achieves its maximum at the left endpoint of $$I$$: $$\hat\theta:=\max_i |x_i|$$ and this is the MLE of $$\theta$$.