# Thread: [SOLVED] beta function

1. ## [SOLVED] beta function

Hi,
I'm starting with
$X_1, X_2, ... iid ~ U(0,1)$
and I'm trying to find the density of
$S_N = X_{k:n} - X_{k-2:n}.$

I look at the joint pdf, then do a Jacobian transformation and get very close to the expression of a beta function, but it's not quite it (but should be):
$\frac{n!} {(k-3)!(n-k)!} \ (y-z)^{k-3} * z * (1-y)^{n-k}$

Any suggestions on how I can get it into the form of a beta function or where I may have gone awry?

Thanks!

2. I'm having trouble with the subscripts of $S_N$

Are these two order stats that differ by 2?

3. ## order statistics subscripts

Hi,
Yes, k:n is the kth order statistic out of n, and k-2:n the (k-2)th order statistic out of n. The final answer should depend on 2, not on k. I guess it will integrate out once I get the right expression.
Thanks!

4. The joint PDF should be easy.
Show me the 2-2 transformation and the other rv before you integrate it out.
I get

$\frac{n!} {(k-3)!(n-k)!} \ (z)^{k-3} *(y- z) * (1-y)^{n-k} I(0

for the joint pdf of the 2 order stats.

5. ## transformation

Matheagle,

Thanks. My 2-2 transformation ends up looking very similar. I set one variable (I'll rename them from my original post so we don't run into confusion with the variables that you use) to be

$
s = X_{(k)} - X_{(k-2)}
$

and

$
t= X_{(k)}
$

I then get that

$
X_{(k-2)} = t - s
$

and

$
X_{(k)} = t
$

The Jacobian turns out to be equal to 1 in that transformation and I plug in the new variables and get

$
\frac{n!} {(k-3)!(n-k)!} \ (t-s)^{k-3} *(s) * (1-t)^{n-k}
$

I'm not sure how to manipulate the variables to get it into the form of a beta eventually (before or after integration).

Thanks!

6. ## integration

Hi,

so I'm thinking that I can re-arrange it the following way and then try to integrate it over t:

$
\frac{n! * s} {(k-3)!(n-k)!} \int (t-s)^{k-3} * (1-t)^{n-k} * dt
$

where t goes from s to 1 (?). But I don't seem to have a clue on how to come up with a way to integrate this...

Thanks again!

7. the density is correct
the support is 0<t-s<t<1
you need to draw that and figure out the bounds of integration.

You have s<t, s>0 and t<1
so integrate out t you have s<t<1....

$f_s(s)=\int_s^1 f(s,t)dt$

8. Thanks yet again. Sorry to keep going, but I am not sure how to integrate it since I have "s" in each of the () and don't see being able to use udv = ... or anything along those lines.

9. you owe me
I just did it with s=y-z BUT t=z
There's a basic calculus sub I did to make it a beta.

10. ## basic calculus

matheagle,

Thanks! I do owe you. I didn't see the forest for the trees (or maybe I banged my head against the wall one too many times).

I found another reference as well (Order Statistics, 3rd ed., David & Nagajara) - they use

t = v * (1-s) and integrate over v.

I'll still have to mull it over for a while on how I can get it to be more intuitive. (Still don't quite get the beta with the easier substitution since the s outside the integral comes back into the integral for me when I do this.)

Thanks!

Order statistics - Google Books

11. you can read this as well
SpringerLink - Journal Article
I wanted you to try that other substitution
and see if you obtained the same result I did.
After dinner I may post my work.............

BACK to my variables... we have

$\frac{n!} {(k-3)!(n-k)!} \ (z)^{k-3} *(y- z) * (1-y)^{n-k} I(0

Let s=y-z but now t=z, the smaller of the two order stats. The jacobian should be one again
That doesn't need checking in these cases.
The density of these two is...

$\frac{n!} {(k-3)!(n-k)!} \ (t)^{k-3} *s * (1-s-t)^{n-k} I(0

That region becomes t>0, s>0 and s+t<1

So the density of s becomes

$\frac{n!s} {(k-3)!(n-k)!} \int_0^{1-s} t^{k-3} (1-s-t)^{n-k} dt$

NOW use calc one, to make the bounds go from 0 to 1, let

$w={t\over 1-s}$ which gives you $t=(1-s)w$ and $dt=(1-s)dw$

and all I see are beta constants now...........

It looks like your reference did what I did.
I have David's book somewhere in my office, but I usually derive these things from scratch.
I can do most things faster by myself than looking it up.
Usually I can't find it or it may never have been done before.
So I just do it.

12. Thanks!!! And - I like your way better than what I found. ;-)

13. ## Got it.

Thanks!! And - I like your substitutions better than what I found. ;-)

14. You're welcome, this is what I do for a living
(if you saw that paper).
I was trying to help another person this morning and he became so rude.
I don't have to do this for free.
I gave him plenty of help and I won't ever again.
All I expect is a thank you.

15. Saw the preview of the paper on Springer - will definitely try to get a copy from the library and look at it more closely now. ;-)

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