$X \sim U(0, \theta^*]$, how to find the MLE?

Question

Given $X \sim U(0, \theta ^*]$. How can I show that $\frac{1}{12}max_{1 \leq i \leq n}X_i^2$ is an MLE of $Var(X)$?

Answer 1

If $X \sim \mathcal{U}]0,\theta_0]$ then $\text{Var}(X)=\frac{\theta_0^2}{12}$.

So you what you need is the MLE for $\theta_0$.

So let $X_1,\dots,X_n \sim \mathcal{U}[0,\theta_0]$ iid.
The likelihood of this sample is

\begin{align*} L(\theta) &= \prod_{i=1}^n f(X_i ; \theta) \\ &= \prod_{i=1}^n \frac{1}{\theta} I( X_i \leq \theta) \\ &= \frac{1}{\theta^n} \prod_{i=1}^n I( X_i \leq \theta) \end{align*}

The MLE is the value $\hat \theta$ that maximizes $L$.

If $\theta < \max X_i$, then there is at least one $i$ such that $I(X_i \leq \theta) =0$ and thus $L(\theta) = 0$.

Now if $\theta \geq \max X_i$, $\prod_{i=1}^n I( X_i \leq \theta) =1$ and $L(\theta) = \frac{1}{\theta^n}$ which is decreasing in $\theta$.
Thus $L$ reaches its maximum at the value $\hat \theta = \max X_i$.

So the MLE for the variance of $X$ is $$\frac{ \hat \theta^2}{12} = \frac{ (\max X_i) ^2}{12} = \frac{ \max X_i^2}{12} \quad (\text{since} \ X_i \geq 0)$$