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

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
Commented Jul 5, 2011 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. Commented Jul 5, 2011 at 4:59
• The following video really helped me: youtube.com/watch?v=XaAtkCzdjLE
– Dor
Commented Aug 31, 2015 at 18:06
• I see no reason why this question is off-topic. Commented Mar 30, 2021 at 19:53

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! :) Commented Jul 5, 2011 at 5:41
• Thanks aengle...its fixed...:)
– Nana
Commented Jul 5, 2011 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? Commented Jun 4, 2013 at 17:19
• How is differentiating valid here?? Commented May 25, 2018 at 21:22
• math.stackexchange.com/questions/649678/… Commented Oct 5, 2018 at 18:07

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

• your link is broken (at least for me...) :p Commented Jul 5, 2011 at 5:04
• @aengle: Thanks, now it works. Commented Jul 5, 2011 at 5:10