# Thread: uniform order stats question

1. ## uniform order stats question

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 kth 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.

2. Originally Posted by 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 kth 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.
In both cases you should make the substitution $\displaystyle u = \frac{y}{\theta} \Rightarrow y = u \theta$ and $\displaystyle dy = \theta \, du$.

Then you have:

$\displaystyle E(Y_{(k)}) = \frac{\theta \, n!}{(k-1)! \, (n-k)!} \int_0^1 u^k \, (1 - u)^{n-k} \, du = \frac{\theta \, n!}{(k-1)! \, (n-k)!} B(k+1, \, n - k + 1)$

and

$\displaystyle E(Y_{(k)}^2) = \frac{\theta^2 \, n!}{(k-1)! \, (n-k)!} \int_0^1 u^{k+1} \, (1 - u)^{n-k} \, du = \frac{\theta^2 \, n!}{(k-1)! \, (n-k)!} B(k+2, \, n - k + 1)$.

(I haven't checked your pdf calculation but it integrates to 1 so looks OK).

3. would it be possible to find the Variance, since you would get:

$\displaystyle \frac{\theta^2 \, n!}{(k-1)! \, (n-k)!} B(k+2, \, n - k + 1) -\left( \frac{\theta \, n!}{(k-1)! \, (n-k)!} B(k+1, \, n - k + 1) \right)^2$

this might sound like a stupid question, but is it even possible to square a binomial representation or a Beta function?

4. Originally Posted by lllll
would it be possible to find the Variance, since you would get:

$\displaystyle \frac{\theta^2 \, n!}{(k-1)! \, (n-k)!} B(k+2, \, n - k + 1) -\left( \frac{\theta \, n!}{(k-1)! \, (n-k)!} B(k+1, \, n - k + 1) \right)^2$

[snip]
Yes.

Note that if x and y are positive integers: $\displaystyle B(x, y) = \frac{(x-1)! \, (y-1)!}{(x + y - 1)!}$.