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

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 $$Var(X)$$?

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 $$\frac{ \hat \theta^2}{12} = \frac{ (\max X_i) ^2}{12} = \frac{ \max X_i^2}{12} \quad (\text{since} \ X_i \geq 0)$$