Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
2  
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. –  ae0709 Jul 5 '11 at 4:59

2 Answers 2

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

share|improve this answer
    
I think you forgot the d theta in the denominator. but good answer! :) –  ae0709 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

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

share|improve this answer
    
your link is broken (at least for me...) :p –  ae0709 Jul 5 '11 at 5:04
    
@aengle: Thanks, now it works. –  Shai Covo Jul 5 '11 at 5:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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