# maximum estimator method more known as MLE of a uniform distribution

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

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If you want to find the maximum likelihood estimate, you first need to derive the likelihood. Did you get that far? Here is a primer: en.wikipedia.org/wiki/Maximum_likelihood_estimator – Emre Jul 5 '11 at 4:57
You asked this question for the method of moments, but you wanted the MLE. I am assuming in that time you've come up with something... surely... what have you tried? What is your effort? I'll write something that will guide you, but I don't want to just write the solution. – mathmath8128 Jul 5 '11 at 4:59
The following video really helped me: youtube.com/watch?v=XaAtkCzdjLE – Dor Aug 31 '15 at 18:06

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 $$\hat{\theta}=x_{\left(n\right)}.$$

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I think you forgot the d theta in the denominator. but good answer! :) – mathmath8128 Jul 5 '11 at 5:41
Thanks aengle...its fixed...:) – Nana Jul 5 '11 at 5:50
@Nana Very old question, but still. Isn't there a problem with endpoints of the given interval? If they were included you solution would be perfectly fine, but the are not. How do deal with it? – Caran-d'Ache Jun 4 '13 at 17:19
I have another queston, I got an unbiased $\frac{n+1}{n}X_(n)$, if given that $\theta$ is greater than 1, will the estimator be changed? – nanopotato Oct 12 '14 at 17:05

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

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your link is broken (at least for me...) :p – mathmath8128 Jul 5 '11 at 5:04
@aengle: Thanks, now it works. – Shai Covo Jul 5 '11 at 5:10