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Bayesian point estimate

  1. May 9, 2014 #1
    The problem statement, all variables and given/known data
    Let [itex]Y_n[/itex] be the nth order statistic of a random sample of size n from a distribution with pdf [itex]f(x|\theta)=1/\theta[/itex] from [itex]0[/itex] to [itex]\theta[/itex], zero elsewhere. Take the loss function to be [itex]L(\theta, \delta(y))=[\theta-\delta(y_n)]^2[/itex]. Let [itex]\theta[/itex] be an observed value of the random variable [itex]\Theta[/itex], which has the prior pdf [itex]h(\theta)=\frac{\beta \alpha^\beta} {\theta^{\beta + 1}}, \alpha < \theta < \infty[/itex], zero elsewhere, with [itex]\alpha > 0, \beta > 0[/itex]. Find the Bayes solution [itex]\delta(y_n)[/itex] for a point estimate of [itex]\theta[/itex].
    The attempt at a solution
    I've found that the conditional pdf of [itex]Y_n[/itex] given [itex]\theta[/itex] is:
    [tex]\frac{n y_n^{n-1}}{\theta^n}[/tex]
    which allows us to find the posterior [itex]k(\theta|y_n)[/itex] by finding what it's proportional to:
    [tex]k(\theta|y_n) \propto \frac{n y_n^{n-1}}{\theta^n}\frac{\beta \alpha^\beta}{\theta^{\beta + 1}}[/tex]
    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:
    [tex]\frac{1}{\theta^{n+\beta}}[/tex]
    When I integrate from [itex]\alpha[/itex] to [itex]\infty[/itex], and solve for the fudge factor, I get [itex](n+\beta)\alpha^{n+\beta}[/itex] as the scaling factor, so for my posterior I get:
    [tex](n+\beta)\alpha^{n+\beta}\frac{1}{\theta^{n+\beta}}[/tex]
    Which doesn't even have a [itex]y_n[/itex] term in it. Weird.

    When I find the expected value of [itex]\theta[/itex] with this distribution, I get 1. Which isn't a very compelling point estimate. So I think I missed a [itex]y_n[/itex] somewhere but I don't know where. Any thoughts? Thanks in advance.
     
  2. jcsd
  3. May 9, 2014 #2

    haruspex

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    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.
     
  4. May 10, 2014 #3
    I wrote a whole reply here while totally missing what you were saying. Thanks for the response! I'll check it out.
     
  5. May 10, 2014 #4

    Ray Vickson

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    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
    [tex] 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 [/tex]
    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
    [tex] f_Y(y) = \int_{c=a}^{\infty} f_{Y,C}(y,c) \, dc
    = \frac{n b y^{n-1}}{a^n (b+n)}, \; 0 < y < a [/tex] For ##y > a## we have
    [tex] f_Y(y) = \int_{c=y}^{\infty} f_{Y,C}(y,c) \, dc
    = \frac{n b a^b}{y^{b+1}(b+n)}, \: y > a [/tex] 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##.
     
    Last edited: May 10, 2014
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