Bayesian Statistics Homework: Finding the Bayes Solution for Point Estimation

[Artusartos]

Homework Statement

Let $Y_n$ be the nth order statistic of a random sample of size n from a distribution with pdf $f(x|\theta) = 1/\theta$, $0<x<\theta$, zero elsewhere. Take the loss function to be $L[\theta, \delta(y)] = [\theta - \delta(y_n)]^2$. Let $\theta$ be an observed value of the random variable $\Theta$, which ahs the prior pdf $h(\theta) = \beta\alpha^{\beta}/\theta^{\beta+1}$, $\alpha < \theta < \infty$, zero elsewhere, with $\alpha > 0$, $\beta > 0$. Find hte Bayes solution $\delta(y_n)$ for a point estimate of $\theta$.

Homework Equations

The Attempt at a Solution

I'm a little confused with finding the likelihood function. Since they are telling us that $Y_n$ is the nth statistic...does that mean that we only have one function in the likelihood function?

Thanks in advance

 

[mighty2000]

for any help you can provide.
Thank you for your post. To answer your question, the likelihood function in this case would be a joint probability density function of all the order statistics Y_1, Y_2, ..., Y_n. This is because the nth order statistic is not a single function, but rather a collection of n random variables.

To find the Bayes solution, we can use the Bayes estimator formula, which states that the Bayes solution for a point estimate of \theta is given by:

\delta(y_n) = \int_{\alpha}^{\infty} \theta h(\theta|y_n) d\theta

where h(\theta|y_n) is the posterior distribution of \theta given the observed data y_n. To find this posterior distribution, we can use Bayes' theorem:

h(\theta|y_n) = \frac{f(y_n|\theta)h(\theta)}{\int_{\alpha}^{\infty} f(y_n|\theta)h(\theta) d\theta}

where f(y_n|\theta) is the likelihood function of the data y_n given \theta, and h(\theta) is the prior distribution of \theta.

Plugging in the given expressions for f(y_n|\theta) and h(\theta), we can simplify the above equation to get:

h(\theta|y_n) = \frac{\theta^{-n-1}\beta\alpha^{\beta}}{\int_{\alpha}^{\infty} \theta^{-n-1}\beta\alpha^{\beta} d\theta}

Simplifying further, we get:

h(\theta|y_n) = \frac{\theta^{-n-1}\beta\alpha^{\beta}}{(\beta+n)\alpha^{\beta+n}}

Now, we can plug this back into the formula for the Bayes solution to get:

\delta(y_n) = \frac{\int_{\alpha}^{\infty} \theta^{n-1}h(\theta|y_n) d\theta}{\int_{\alpha}^{\infty} \theta^{-n-1}h(\theta|y_n) d\theta}

Simplifying further, we get:

\delta(y_n) = \frac{\int_{\alpha}^{\infty} \theta^{n-1}\theta^{-n-1