# Thread: Order Statistics with a dash of Covariance.

1. ## Order Statistics with a dash of Covariance.

Here's the question:

Let x1,x2,x3 be a random sample from a distribution with a PDF: f(x) = 2x, 0<x<1.

a) find the probability that the largest of x1,x2,x3 is less than the distribution median.
b) Find Cov(x(2),x(3))

Using quantiles, I found the distribution median to be = sqrt(1/2);
I found the max to be the pdf 6x^5 0<x<1;

In order to solve this, P(6x^5 < sqrt(1/2)) solved for x. This gave me an integral 2x with bounds from 0 to (((sqrt(.5))/6))^(1/5)) yielding 0.42514

Did I go through this correctly?

For the Covariance I am also not confident.

using the definition that cov(x,y) = E[(x - E(x))(Y - E(y))]

I found to be something like E[(12x^3-12x^5-24/35)(6x^5-6/7)] which I calculated to be around 78/245.

Thanks in advance for you help!

2. Originally Posted by hasbren
Here's the question:

Let x1,x2,x3 be a random sample from a distribution with a PDF: f(x) = 2x, 0<x<1.

a) find the probability that the largest of x1,x2,x3 is less than the distribution median.
b) Find Cov(x(2),x(3))

Using quantiles, I found the distribution median to be = sqrt(1/2);
I found the max to be the pdf 6x^5 0<x<1;

In order to solve this, P(6x^5 < sqrt(1/2)) solved for x. This gave me an integral 2x with bounds from 0 to (((sqrt(.5))/6))^(1/5)) yielding 0.42514

Did I go through this correctly?

For the Covariance I am also not confident.

using the definition that cov(x,y) = E[(x - E(x))(Y - E(y))]

I found to be something like E[(12x^3-12x^5-24/35)(6x^5-6/7)] which I calculated to be around 78/245.

Thanks in advance for you help!
I don't have time to work through the question and therefore check the details, but your approach looks fine. (All your pdf's integrate to 1 which is a good sign and your means are between 0 and 1 which is another good sign).

3. Thanks for your approval Mr. Fantastic

4. To obtain the covariance between two order statistics you need the joint density.

$E(X_{(2)}X_{(3)})=\int_0^1\int_0^{x_3}x_2x_3f_{X_{ (2)}X_{(3)}}(x_2,x_3)dx_2dx_3$

you can obtain $E(X_{(2)})$ and $E(X_{(3)})$ via the marginals.

5. So can I just multiply the two together to get the joint? I don't know if they are independent... oh wait- do I since they are order statistics?

6. no, they are clearly dependent x(2)<x(3)

7. So, how would I go about getting the joint then?

8. It's in many books, most likely your textbook.
I found it here with r=2, n=s=3....
Joint Distribution of and

$f_{X_{(2)},X_{(3)}}(x_2,x_3)=3![F(x_2)]f(x_2)f(x_3)$

$f(x)=2x$ and $F(x)=x^2$ on 0<x<1

so your joint density should be $f_{X_{(2)},X_{(3)}}(x_2,x_3)=24x_2^3x_3$

ON the set $0

which I just checked is a valid density, it does integrate to one...

$24\int_0^1\int_0^{x_3}x_2^3x_3dx_2dx_3=1$

9. I appreciate the help my friend-