Answer to Question #131892 in Statistics and Probability for abhi

Question

Show that the maximum likelihood estimator of the parameter theta of a population having the density function (2/ theta^2) (theta - x), 0 < x < theta, for a sample of unit size is 2x, where x is the sample value. Also, show that the estimate obtained is biased.

Answer 1

$f(x,\theta)=\frac{2(\theta-x)}{\theta^2}$ , $0<x<\theta$

$L(x,\theta)=\frac{2(\theta-x)}{\theta^2}$

$\frac{\partial L(x,\theta)}{\partial \theta}=\frac{-2}{\theta^2}(1-\frac{2x}{\theta})$

$\frac{-2}{\theta^2}(1-\frac{2x}{\theta})=0$

$1=\frac{2x}{\theta}$

$2x=\hat\theta$ Thus, the MLE of $\theta$ is 2x.

To check for biasedness, expectation of 2x shoud not be equal to theta.

$E(2x)=\smallint_0^{\theta}\frac{2x(2(\theta-x))}{\theta^2}dx$

$=[\frac{2x^2(3\theta-2x)}{3\theta^2}]_0^\theta$

$=\frac{2\theta^2(3\theta-2\theta)}{3\theta^2}-0$

$=\frac{2\theta}{3}\ne\theta$ , Thus, the MLE (2x) is a biased estimator of theta.