Order Statistics, Unbiasedness, and Expected Values

by Providence88
Tags: expected, order, statistics, unbiasedness, values
Providence88 is offline
Jul26-09, 06:31 PM
P: 7
1. The problem statement, all variables and given/known data

Let Y1, Y2, ..., Yn denote a random sample from the uniform distribution on the interval ([tex]\theta[/tex], [tex]\theta + 1[/tex]). Let [tex]\hat{\theta} = Y_{(n)} - \frac{n}{n+1}[/tex]

Show that [tex]\hat{\theta}[/tex] is an unbiased estimator for [tex]\theta[/tex]

2. Relevant equations

Well, to check for unbiasedness, E([tex]\hat{\theta}[/tex]) should = [tex]\theta[/tex].

The difficulty for me arises when calculating [tex]g_{(n)}(y)[/tex], needed to find E[[tex]\hat{\theta}[/tex]]. The interval ([tex]\theta[/tex], [tex]\theta + 1[/tex]) seems to make this integral very complicated:

[tex]E[\hat{\theta}][/tex] = [tex]\int^{\theta + 1}_{\theta} yg_{(n)}(y)[/tex]

3. The attempt at a solution

I attempted to find [tex]g_{(n)}(y)[/tex], which I thought to be [tex]ny^{n-1}[/tex], but according to our solutions manual, it's actually [tex]n(y-\theta)^{n-1}[/tex], which I have no idea how that is concluded. And even if that is the true value of [tex]g_{(n)}(y)[/tex], the integral is still looking very daunting.

Any help? Thanks!
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Cyosis is offline
Jul27-09, 10:09 AM
HW Helper
P: 1,495
I am not quite sure what [itex]g_n(y)[/itex] is so you will have to explain why you thought it would equal [itex]n y^{n-1}[/itex]. As for the integral it is not so hard. You can solve it by partial integration.

\int_{\theta}^{\theta+1} ny(y-\theta)^{n-1}dy=y (y-\theta)^n ]_\theta^{\theta+1}-\int_{\theta}^{\theta+1} (y-\theta)^n dy
Providence88 is offline
Jul27-09, 10:16 AM
P: 7
Oh, [tex]g_{(n)}(y)[/tex] is the density function for [tex]Y_{(n)}[/tex]=max(Y1, Y2, ..., Yn)

[tex]g_{(n)}(y) = n[F(Y)]^{n-1}*f(y)[/tex], where F(Y) is the distribution function of Y and f(y) is the density function. Since the bounds are theta and theta plus one, I assumed that f(y), by definition, is 1/(theta + one - theta), which equals one. If f(y) = 1, then F(Y) = y + C. I'm starting to think that the plus C would be -(theta).

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