Question

3. Suppose that Xi,.... Xn is a random sample from a uniform distribution over [0,0) That is, 0 elsewhere Also suppose that the prior distribution of θ is a Pareto distribution with density 0 elsewhere where θ0 > 0 and α > 1. (a) Determine (b) Show , θ > max(T1 , . . . ,Zn,%) and hence deduce the posterior density of θ given x, . . . ,Zn is (c) Compute the mean of the posterior distribution and use this to estimate . [3] (d) What is the conjugate prior distribution for the uniform distribution?[1]

Answer 1

Given $X_1,X_2,..,X_n$ are random sample from the uniform distribution $\left [ 0,\theta \right ]$ . That is

$f\left ( x|\theta \right )=\frac{1}{\theta };0\leqslant x\leqslant \theta$

And the prior distribution of $\theta$ is Pareto distributed as ,

$h\left ( \theta \right )=\frac{\alpha \theta _0^\alpha }{\theta ^{\alpha +1}};\theta >\theta _0$ Where $\theta _0>0,\alpha >1$

a) The joint PDF of $X_1,X_2,..,X_n$ given $\theta$ is

$f\left ( x_1,x_2,..,x_n|\theta \right )=\prod_{i=1}^{n}f\left ( x|\theta \right )\\ {\color{Blue} f\left ( x_1,x_2,..,x_n|\theta \right )=\frac{1}{\theta^n };0\leqslant x\leqslant \theta}$

b) The posterior distribution of $\theta$ is

$k\left ( \theta |x_1,x_2,..,x_n\right ) \propto f\left ( x_1,x_2,..,x_n|\theta \right )h\left ( \theta \right )\\ k\left ( \theta |x_1,x_2,..,x_n\right ) \propto \frac{1}{\theta^n }\frac{\alpha \theta _0^\alpha }{\theta ^{\alpha +1}}\\$

$f\left ( \theta |x_1,x_2,..,x_n \right )\propto\frac{\alpha \theta _0^\alpha }{\theta ^{n+\alpha +1}};0\leqslant x\leqslant \theta\\ f\left ( \theta |x_1,x_2,..,x_n \right ) \propto \frac{\alpha \theta _0^\alpha }{\theta ^{n+\alpha +1}};0\leqslant x\leqslant \theta\\ {\color{Blue} f\left ( \theta |x_1,x_2,..,x_n \right ) \propto\frac{1 }{\theta ^{n+\alpha +1}};0\leqslant x\leqslant \theta;\theta \geqslant \textup{max}\left ( x_1,x_2,..,x_n,\theta _0 \right )}$

Thus the posterior PDF of $\theta$ is

$f\left ( \theta |x_1,x_2,..,x_n \right ) =\frac{\frac{1 }{\theta ^{n+\alpha +1}}}{\int_{x_0}^{\infty } \frac{d\theta }{\theta ^{n+\alpha +1}}};\\ f\left ( \theta |x_1,x_2,..,x_n \right ) =\frac{\frac{1 }{\theta ^{n+\alpha +1}}}{ \frac{1 }{\left ( n+\alpha \right )x_0 ^{n+\alpha }}};\\ f\left ( \theta |x_1,x_2,..,x_n \right ) =\frac{\left ( n+\alpha \right )x_0 ^{n+\alpha } }{\theta ^{n+\alpha +1}};\theta \geqslant x_0$

The proof is complete. This is a Pareto distribution.

c) The mean of the posterior( Bayes estimator) distribution is

$\mu _B=\int_{x_0}^{\infty }\theta f\left ( \theta |x_1,x_2,..,x_n \right )d\theta \\ \mu _B=\int_{x_0}^{\infty }\frac{\left ( n+\alpha \right )x_0 ^{n+\alpha } }{\theta ^{n+\alpha }}d\theta \\ \mu _B=-\left [ \frac{\left ( n+\alpha \right )\left ( n+\alpha -1 \right )x_0 ^{n+\alpha } }{\theta ^{n+\alpha-1 }} \right ]_{x_0}^\infty \\ {\color{Blue} \mu _B=\left ( n+\alpha \right )\left ( n+\alpha -1 \right )x_0}$

The posterior mean or Bayes estimator is found as above

We are required to solve only 4 parts. Please post the remaining questions as another post.