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

-
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. – platinumtucan Jul 5 '11 at 4:59

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

-
 I think you forgot the d theta in the denominator. but good answer! :) – platinumtucan Jul 5 '11 at 5:41 Thanks aengle...its fixed...:) – Nana Jul 5 '11 at 5:50

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

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