How to find the MLE for uniform distributed random variables

Question

I am trying to understand the maximum likelihood estimation on an example.

Given random variables Y,X that are independent and uniformly distributed on $[0,\theta]$. Find the MLE

What I know:

For $Z=(X,Y)$ the likelihood function is $L(\theta)=f(\mathbf x;\theta)$. And the MLE is defined as $\Theta=\operatorname {arg\,max}_{\theta \in \Theta}L(\theta)$

The Solution only states that $\Theta=\max(x_1,x_2)$ I hope someone could explain to me how to deal with this problem. Thanks in advance

Answer 1

The likelihood is $\frac{1}{\theta^2}$ thus it is strictly decreasing in all its support.

But you know that

$$0\leq x\leq \theta$$

$$0\leq y\leq \theta$$

That is

$$\theta\geq \max(x,y)$$

So, as the likelihood is strictly decreasing its argmax is attained at the frontier...

Ps: the MLE is not the argmax but the argsup...try the same exercise with support $(0;\theta)$ and you will realize that the argmax does not exist

Answer 2

If $X, Y\sim \operatorname{Unit}([0,\theta])$, and $X$ and $Y$ are independent, then $f(x,y;\theta)=\frac{1}{\theta^2}\mathbb{1}_{[0,\theta]}(x)\mathbb{1}_{[0,\theta]}(y)=\mathbb{1}_{[0,\theta]}(\max(x,y))$

Notice that $f$ attains values $0$ or $\frac{1}{\theta^2}$. So, when $(x,y)$ (your data) is given, then any value of $\theta$ that is larger or equal than $\max(x,y)$ will give you $f(x,y;\theta)>0$. In particular, you can take $\hat{\theta}(x,y)=\max(x,y)$ (an estimator should be a function of your data). Other statistics such statistics will give you $f>0$ but $\hat{\theta}$ given above at least can be classified a minimal.