# Order Statistics, Unbiasedness, and Expected Values

1. ### Providence88

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 ($$\theta$$, $$\theta + 1$$). Let $$\hat{\theta} = Y_{(n)} - \frac{n}{n+1}$$

Show that $$\hat{\theta}$$ is an unbiased estimator for $$\theta$$

2. Relevant equations

Well, to check for unbiasedness, E($$\hat{\theta}$$) should = $$\theta$$.

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

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

3. The attempt at a solution

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

Any help? Thanks!

Last edited: Jul 26, 2009
2. ### Cyosis

1,495
I am not quite sure what $g_n(y)$ is so you will have to explain why you thought it would equal $n y^{n-1}$. As for the integral it is not so hard. You can solve it by partial integration.

$$\int_{\theta}^{\theta+1} ny(y-\theta)^{n-1}dy=y (y-\theta)^n ]_\theta^{\theta+1}-\int_{\theta}^{\theta+1} (y-\theta)^n dy$$

3. ### Providence88

7
Oh, $$g_{(n)}(y)$$ is the density function for $$Y_{(n)}$$=max(Y1, Y2, ..., Yn)

$$g_{(n)}(y) = n[F(Y)]^{n-1}*f(y)$$, 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).