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**lllll** I'm having some difficulty finding the mean and the variance of the following question:

Let $\displaystyle Y_1, Y_2,... \ ,Y_n$ be independent uniformly dist. r.v. on the interval $\displaystyle [0, \theta] $

I found my density function for *k*th order stat between 1 and n, which is:

$\displaystyle \frac{n!}{(k-1)!(n-k)!} \times \left( \frac{1}{\theta} y \right)^{k-1} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \times \frac{1}{\theta} $ where $\displaystyle 0 \leq y \leq \theta$

now using what I found if I try to find the $\displaystyle E(Y_{(k)})$ I get:

$\displaystyle \int_{0}^{\theta} \frac{n!}{(k-1)!(n-k)!} \times \left( \frac{1}{\theta} y \right)^{k-1} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \times \frac{1}{\theta} \times y \ dy$ where $\displaystyle 0 \leq y \leq \theta$

= $\displaystyle \frac{n!}{(k-1)!(n-k)!} \int_{0}^{\theta} \left( \frac{1}{\theta} y \right)^{k} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \ dy$ where $\displaystyle 0 \leq y \leq \theta$

now the integral looks a lot like a *Beta Function*, but what puzzles me is that the beta function goes from 0 to 1, and I have 0 to $\displaystyle \theta$. If I were to make the substitution. I would get:

$\displaystyle \frac{n!}{(k-1)!(n-k)!} B[k+1, \ n-k+1]$ where $\displaystyle 0 \leq y \leq \theta$

And for the $\displaystyle E(Y_{(k)}^2)$ I get:

$\displaystyle \int_{0}^{\theta} \frac{n!}{(k-1)!(n-k)!} \times \left( \frac{1}{\theta} y \right)^{k-1} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \times \frac{1}{\theta} \times y^2 \ dy$ where $\displaystyle 0 \leq y \leq \theta$

= $\displaystyle \frac{n!}{(k-1)!(n-k)!} \int_{0}^{\theta} \left( \frac{1}{\theta} y \right)^{k} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \times y \ dy$ where $\displaystyle 0 \leq y \leq \theta$

Again the first part of the integral looks like the *Beta Function*, but since there's an additional y within it, I'm a little confused on how to integrate this out, since I don't think that a closed form of this function exists.