Let $X_1,...,X_n\sim \text{Uniform}(0,\theta)$. Show that the maximum likelihood estimation (MLE) is consistent.

Setting $Y=\text{max}\{X_1,...,X_n\}$ I know that for any constant $c\in\mathbb{R}$,$$ \mathbb{P}(Y<c)=\mathbb{P}(X_1<c)\mathbb{P}(X_2<c)\cdots \mathbb{P}(X_n<c) $$ but I haven't been able to show consistency yet. Any ideas?

  • $\begingroup$ The MLE of $\theta$? Are they iid? $\endgroup$
    – Therkel
    Oct 22, 2017 at 17:02
  • $\begingroup$ Yes, and yes they are. $\endgroup$
    – sam wolfe
    Oct 22, 2017 at 17:04
  • $\begingroup$ Have you shown that the MLE of $\theta$ is $Y$? Or do you know that? $\endgroup$
    – Therkel
    Oct 22, 2017 at 17:07

1 Answer 1


Note that $P(\max_{1\leq i\leq n} X_i \leq t)=\begin{cases} 0 &if &t\leq 0 \\ \left(\frac t \theta \right)^n &if &t \in [0,\theta]\\ 1 &if &t\geq \theta \end{cases}$

For $\epsilon>0$ , $$P(|\max_{1\leq i\leq n} X_i - \theta|>\epsilon)=P(\max_{1\leq i\leq n} X_i \geq \theta + \epsilon )+P(\max_{1\leq i\leq n} X_i \leq \theta - \epsilon )= \begin{cases} \left(\frac{\theta - \epsilon }{\theta}\right)^n &if &\epsilon <\theta \\ 0 &if &\epsilon \geq \theta \end{cases}$$

which goes to $0$ as $n\to \infty$.

  • 2
    $\begingroup$ Note that since the events $\{\max_i X_i \leq \theta - \epsilon\}$ are monotonely increasing this convergence in probability implies a.s. convergence. $\endgroup$
    – Therkel
    Oct 23, 2017 at 6:52

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