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

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 $n\to \infty$.

  • 1
    $\begingroup$ 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 '17 at 6:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.