maximum estimator method more known as MLE of a uniform distribution

Question

Let $ X_1, ... X_n $ a sample of independent random variables with uniform distribution $(0,$$ \theta $$ ) $ Find a $ $$ \widehat\theta $$ $ estimator for theta using the maximun estimator method more known as MLE

Answer 1

First note that $f\left({\bf x}|\theta\right)=\frac{1}{\theta}$ , for $0\leq x\leq\theta$ and $0$ elsewhere.

Let $x_{\left(1\right)}\leq x_{\left(2\right)}\leq\cdots\leq x_{\left(n\right)}$ be the order statistics. Then it is easy to see that the likelihood function is given by $$L\left(\theta|{\bf x}\right) = \prod^n_{i=1}\frac{1}{\theta}=\theta^{-n}\,\,\,\,\,(*)$$ for $0\leq x_{(1)}$ and $\theta \geq x_{(n)}$ and $0$ elsewhere.
Now taking the derivative of the log Likelihood wrt $\theta$ gives:

$$\frac{\text{d}\ln L\left(\theta|{\bf x}\right)}{\text{d}\theta}=-\frac{n}{\theta}<0.$$ So we can say that $L\left(\theta|{\bf x}\right)=\theta^{-n}$ is a decreasing function for $\theta\geq x_{\left(n\right)}.$ Using this information and (*) we see that $L\left(\theta|{\bf x}\right)$ is maximized at $\theta=x_{\left(n\right)}.$ Hence the maximum likelihood estimator for $\theta$ is given by $$ \hat{\theta}=x_{\left(n\right)}.$$

Answer 2

This example is worked out in detail here (pages 13-14).