Question #131892
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.
1
Expert's answer
2020-09-08T15:08:20-0400

f(x,θ)=2(θx)θ2f(x,\theta)=\frac{2(\theta-x)}{\theta^2} , 0<x<θ0<x<\theta

L(x,θ)=2(θx)θ2L(x,\theta)=\frac{2(\theta-x)}{\theta^2}

L(x,θ)θ=2θ2(12xθ)\frac{\partial L(x,\theta)}{\partial \theta}=\frac{-2}{\theta^2}(1-\frac{2x}{\theta})

2θ2(12xθ)=0\frac{-2}{\theta^2}(1-\frac{2x}{\theta})=0

1=2xθ1=\frac{2x}{\theta}

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

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

E(2x)=0θ2x(2(θx))θ2dxE(2x)=\smallint_0^{\theta}\frac{2x(2(\theta-x))}{\theta^2}dx

=[2x2(3θ2x)3θ2]0θ=[\frac{2x^2(3\theta-2x)}{3\theta^2}]_0^\theta

=2θ2(3θ2θ)3θ20=\frac{2\theta^2(3\theta-2\theta)}{3\theta^2}-0

=2θ3θ=\frac{2\theta}{3}\ne\theta , Thus, the MLE (2x) is a biased estimator of theta.



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