Q:

Convergence of the maximum likelihood estimator of a uniform parameter

Suppose there is an sample: $X_{1}, ....., X_{n}$ where each $X_{i}$ is indepedendet and identically distributed as a Uniform $(0,\theta)$.
I know that $X_{max}=max[X_{1},.....,X_{n}]$ is a biased estimator of $\theta$ but a consistent estimator of $\theta$.
If i want to proove that $X_{max}$ converges in probability to $\theta$, can i write this as a (informal) proof: 
Prob($ |X_{max} - \theta| > \epsilon ) = \int_{0}^{\theta} x f(x) dx$ for any $\epsilon > 0$ and then, after the integration y take the limit?
Is it ok as an argument? I am not so worry about the rigorous proof, rather i want to know if the argument is correct.      

A:

The argument is incorrect, and in fact doesn't make much sense. The statement will look more like
$$\lim_{n\to \infty}\Pr(|X_{max} - \theta| > \epsilon) = \lim_{n \to \infty}\int_{\epsilon}^{\theta} |X_{max} - \theta|\, f(x) dx\, $$
where $f(x)$ is the pdf of a Uniform$(0, \theta)$ (So $f(x) = 1/\theta$).
You could work around with the integration to see where you get. An easier way would be to use the properties of maximum. First note that $X_{max} - \theta$ is always negative.
\begin{align*}
\Pr(|X_{max} - \theta| > \epsilon) & = \Pr(\theta - X_{\max} > \epsilon)\\
& = \Pr(X_{max} < \theta - \epsilon)\\
& = \prod_{i = 1}^{n} \Pr(X_i < \theta - \epsilon) \\
& = \left( \dfrac{\theta - \epsilon}{\theta} \right)^n\\
& \to 0  \text{ as $n \to \infty$}
\end{align*}

