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Math Help - [SOLVED] beta function

  1. #1
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    [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!
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  2. #2
    MHF Contributor matheagle's Avatar
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    I'm having trouble with the subscripts of S_N

    Are these two order stats that differ by 2?
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  3. #3
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    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!
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  4. #4
    MHF Contributor matheagle's Avatar
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    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<z<y<1)

    for the joint pdf of the 2 order stats.
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  5. #5
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    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

    <br />
s = X_{(k)} - X_{(k-2)}<br />


    and

    <br />
t= X_{(k)}<br />

    I then get that

    <br />
X_{(k-2)} = t - s<br />


    and

    <br />
X_{(k)} = t<br />


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

    <br />
\frac{n!} {(k-3)!(n-k)!} \ (t-s)^{k-3} *(s) * (1-t)^{n-k}<br />

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

    Thanks!
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  6. #6
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    integration

    Hi,

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

    <br />
\frac{n! * s} {(k-3)!(n-k)!} \int (t-s)^{k-3} * (1-t)^{n-k} * dt<br />

    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!

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  7. #7
    MHF Contributor matheagle's Avatar
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    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
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  8. #8
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    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.
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  9. #9
    MHF Contributor matheagle's Avatar
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    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.
    Last edited by matheagle; February 15th 2010 at 09:46 PM.
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  10. #10
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    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
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  11. #11
    MHF Contributor matheagle's Avatar
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    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<z<y<1)

    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<t<s+t<1)

    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.
    Last edited by matheagle; February 15th 2010 at 10:54 PM.
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  12. #12
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    Thanks!!! And - I like your way better than what I found. ;-)
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  13. #13
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    Got it.

    Thanks!! And - I like your substitutions better than what I found. ;-)
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  14. #14
    MHF Contributor matheagle's Avatar
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    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.
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  15. #15
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    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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