Maximum Likelihood Estimator for Uniform Distribution

Question

Can somebody please explain this example to me. I am struggling to see why the likelihood is $\frac{1}{\theta^n}$ only if theta is greater than the maximum x. Furthermore, why is it the case that the mle is $x_{(n)}$?

Answer 1

Why the likelihood is $\dfrac {1}{\theta^n}$ only if $\theta$ is greater than the maximum $x$?

There is zero probability that any realization of $x$ will be ever be above the true value of $\theta$. Remember $X$ takes value in $[0,\theta]$. Thus, the likelihood in this case must be zero.

Furthermore, why is it the case that the mle is $x_{(n)}$?

Why don't you plot a graph of the likelihood as suggested? Notice that $L=\dfrac {1}{\theta^n} \nearrow$ when $\theta\searrow$ but $L$ is discontinuous! You can not take derivatives to find the maximum. When $\theta$ goes below $x_{(n)}$ it becomes zero.

So to maximize $L$, you want to pick the lowest $\theta$ such that is possible that your sample realization has positive probability of occurring. That is, the lowest $\theta$ that is above all the $x_{\mathrm{s}}$.