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
maximum estimator method more known as MLE of a uniform distribution
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0The following video really helped me: https://www.youtube.com/watch?v=XaAtkCzdjLE – 2015-08-31
2 Answers
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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1https://math.stackexchange.com/questions/649678/how-do-you-differentiate-the-likelihood-function-for-the-uniform-distribution-in?rq=1 – 2018-10-05