How Does Bayesian Estimation Determine Point Estimates with Prior Distributions?

[rayge]

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 from 0 to \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 has the prior pdf h(\theta)=\frac{\beta \alpha^\beta} {\theta^{\beta + 1}}, \alpha < \theta < \infty, zero elsewhere, with \alpha > 0, \beta > 0. Find the Bayes solution \delta(y_n) for a point estimate of \theta.
The attempt at a solution
I've found that the conditional pdf of Y_n given \theta is:
\frac{n y_n^{n-1}}{\theta^n}
which allows us to find the posterior k(\theta|y_n) by finding what it's proportional to:
k(\theta|y_n) \propto \frac{n y_n^{n-1}}{\theta^n}\frac{\beta \alpha^\beta}{\theta^{\beta + 1}}
Where I'm sketchy is that apparently we can just remove all terms not having to do with theta, come up with a fudge factor to make the distribution integrate to 1 over its support, and call it good. I end up with:
\frac{1}{\theta^{n+\beta}}
When I integrate from \alpha to \infty, and solve for the fudge factor, I get (n+\beta)\alpha^{n+\beta} as the scaling factor, so for my posterior I get:
(n+\beta)\alpha^{n+\beta}\frac{1}{\theta^{n+\beta}}
Which doesn't even have a y_n term in it. Weird.

When I find the expected value of \theta with this distribution, I get 1. Which isn't a very compelling point estimate. So I think I missed a y_n somewhere but I don't know where. Any thoughts? Thanks in advance.

 

[haruspex]

Not an area I'm familiar with, so can't help with your specific question, but one thing does look wrong to me: if you substitute n=0 in your answer, shouldn't you get h(θ)? The power of theta seems to be one off.

 

[rayge]

I wrote a whole reply here while totally missing what you were saying. Thanks for the response! I'll check it out.

 

[Ray Vickson]

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 from 0 to \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 has the prior pdf h(\theta)=\frac{\beta \alpha^\beta} {\theta^{\beta + 1}}, \alpha < \theta < \infty, zero elsewhere, with \alpha > 0, \beta > 0. Find the Bayes solution \delta(y_n) for a point estimate of \theta.
The attempt at a solution
I've found that the conditional pdf of Y_n given \theta is:
\frac{n y_n^{n-1}}{\theta^n}
which allows us to find the posterior k(\theta|y_n) by finding what it's proportional to:
k(\theta|y_n) \propto \frac{n y_n^{n-1}}{\theta^n}\frac{\beta \alpha^\beta}{\theta^{\beta + 1}}
Where I'm sketchy is that apparently we can just remove all terms not having to do with theta, come up with a fudge factor to make the distribution integrate to 1 over its support, and call it good. I end up with:
\frac{1}{\theta^{n+\beta}}
When I integrate from \alpha to \infty, and solve for the fudge factor, I get (n+\beta)\alpha^{n+\beta} as the scaling factor, so for my posterior I get:
(n+\beta)\alpha^{n+\beta}\frac{1}{\theta^{n+\beta}}
Which doesn't even have a y_n term in it. Weird.

When I find the expected value of \theta with this distribution, I get 1. Which isn't a very compelling point estimate. So I think I missed a y_n somewhere but I don't know where. Any thoughts? Thanks in advance.

I'm not sure how the loss-function business enters into the calculation, but you seem to be trying to compute the Bayesian posterior density of $\theta$, given $Y_n = y_n$. You have made an error in that. Below, I will use $Y,y$ instead of $Y_n,y_n$, $C,c$ instead of $\Theta, \theta$ and $a,b$ instead of $\alpha, \beta$---just to make typing easier.

Using the given prior density, the joint density of $(y,c)$ is
f_{Y,C}(y,c) = \frac{b a^b}{c^{b+1}} \frac{n y^{n-1}}{c^n}, 0 < y < c, a < c < \infty
The (prior) density of $Y$ is $f_Y(y) = \int f_{Y,C}(y,c) \, dc$, but you need to be careful about integration limits. For $0 < y < a$ we have
f_Y(y) = \int_{c=a}^{\infty} f_{Y,C}(y,c) \, dc <br /> = \frac{n b y^{n-1}}{a^n (b+n)}, \; 0 &lt; y &lt; a For $y > a$ we have
f_Y(y) = \int_{c=y}^{\infty} f_{Y,C}(y,c) \, dc <br /> = \frac{n b a^b}{y^{b+1}(b+n)}, \: y &gt; a Thus, the posterior density of $C$ will depend on $y$, since the denominator in $f(c|y) = f_{Y,C}(y,c)/f_Y(y)$ has two different forms for $y < a$ and $y > a$.