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Help with a dirichlet function

  1. Oct 28, 2010 #1
    1. The problem statement, all variables and given/known data

    I am stuck on this problem. Can someone give me a hint on how to start?
    ZeqDK.png

    Prove fX1,X2 is a density.
    2. Relevant equations



    3. The attempt at a solution

    I think i am supposed to integrate twice to show it equals 1 to show its a density function, but I don't know how to do this...
     
    Last edited: Oct 28, 2010
  2. jcsd
  3. Oct 28, 2010 #2

    fzero

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    To integrate this, you might find the definition of the Beta function,

    [tex]B(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt, ~\text{Re}(x)> 0, \text{Re}(y)> 0,[/tex]

    useful. The Beta function also satisfies the identity

    [tex] B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.[/tex]
     
  4. Oct 29, 2010 #3
    Can you show me the next step, as I am still very lost in this question.
     
  5. Oct 29, 2010 #4

    fzero

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    You might want to look up what properties a probability density function must have. The integral that you refer to in the first post is one of them. You should try to set that integral up. You will probably need to use a change of variables and perhaps the binomial formula to express the integrals in terms of Beta functions.
     
  6. Oct 29, 2010 #5
    i can get to here
    vVkg8.png

    But I don't know how to use the beta function.
     
    Last edited: Oct 29, 2010
  7. Oct 29, 2010 #6
    i am thinking of doing change of variable, but in any example i find, there is always two new variables defined, all i can think of is one which is u=x1/(1-x2)...
     
  8. Oct 29, 2010 #7

    fzero

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    It's honestly not a very easy integral to do, but it can be done if you know certain properties of the incomplete beta function. The change of variables I'd use is u = x1+ x2, then the x2 integration will yield an incomplete beta function. You can find a series expansion for the incomplete beta function that will allow you to do the x1 integration in terms of beta functions.
     
  9. Oct 29, 2010 #8
    aww, this is too complicated for me

    the farthest i have got is

    UYCWB.png

    the right side looks like the beta identity, and somehow i need to use change of variables on the left side, the inner integral looks like the beta function, but there is another term inside (x1) that wont let me change it. How can I get rid of the X1?

    In any case if you know how to do this, can you help me get started on this...
     
  10. Oct 29, 2010 #9

    fzero

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    I made a suggestion in post #7.
     
  11. Oct 29, 2010 #10
    "The change of variables I'd use is u = x1+ x2, then the x2 integration will yield an incomplete beta function"

    I can use
    u = x1+ x2

    but what does "x2 integration" mean?
    does that mean after doing the change of variables with u=x1+x2, the integral with respect to x2 will yeild...

    Also the thing i am also not sure is, since this is 2 integrals, the way i leanred to do change of variables is get 2 new variables, i can use u1=x1+x2, but what do i make u2 equal...? If i new that i can get a new equation using the jacobian
     
  12. Oct 29, 2010 #11

    fzero

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    It means integrate over the part of the integrand that depends on x2. I didn't actually do it that way, I just checked the result on Wolfram Alpha.
     
  13. Oct 29, 2010 #12
    so now i have

    zgmXT.png

    i assume this is the incomplete beta function.

    I dont know how to find a series expansion for the incomplete beta function that will allow me to do the x1 integration in terms of beta functions.
     
  14. Oct 29, 2010 #13
    i can't figure out know how to integrate the du (inner integral) , How do you integrate it?
    If only the -x1 were not there, i can use the identity......

    I also may have done the cgange of variables wrong, if i had to use the jacobian, then, it requires 2 new variables, but i only have the u=x1+x2...
     
    Last edited: Oct 29, 2010
  15. Oct 29, 2010 #14

    fzero

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    I think all of the formulas that you need can even be found on wikipedia if you don't have access to a text that has them.

    That change of variables that I suggested isn't the only thing to attempt. I tried expanding (1-x1-x2) using the binomial formula, but that leads to a sum that I haven't been able to compute.
     
  16. Oct 29, 2010 #15
    then i don't think i can even finish this....
     
    Last edited: Oct 29, 2010
  17. Oct 30, 2010 #16

    fzero

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    There's an indirect way to compute the integral that is based on the way you derive the relationship between the beta function and gamma function. That computation is done on the wikipedia page

    http://en.wikipedia.org/wiki/Beta_function#Relationship_between_gamma_function_and_beta_function

    In your case you will rewrite [tex]\Gamma(\alpha_1)\Gamma(\alpha_2)\Gamma(\alpha_3)[/tex] using a coordinate change

    [tex] u = z x_1, ~ v = z x_2, ~ w = z(1-x_1-x_2).[/tex]

    You can then factor the triple integral into the expression for another beta function and a double integral that is the integral you're trying to compute.
     
  18. Oct 30, 2010 #17
    i am still very confused here, can you please show me the first couple of steps in latex?
     
  19. Oct 30, 2010 #18

    fzero

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    Try reproducing that calculation shown on the wikipedia page, the calculation you have to do is almost exactly the same. I can't do the work for you.
     
  20. Oct 31, 2010 #19
    why do you put a z in front of the terms?

    if i do that then i will get x1 = u/z, x2 = v/z and x3 = w/z, but in the identity in the wiki page it only had u, v, and w

    and i now have
    wdW0I.png
     
    Last edited: Oct 31, 2010
  21. Oct 31, 2010 #20

    fzero

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    It's extremely confusing to figure out what you're trying to do when you keep writing the identity that you are trying to prove over and over again without actually doing anything. The two lines that you wrote down at the bottom make no sense at all.

    Forget about the identity right now. I told you that you should compute

    [tex]\Gamma(\alpha_1)\Gamma(\alpha_2)\Gamma(\alpha_3) = \left( \int_0^\infty e^{-u} u^{\alpha_1-1} du \right)\left(\int_0^\infty e^{-v} v^{\alpha_2-1} dv \right) \left( \int_0^\infty e^{-w} w^{\alpha_3-1} dw \right)[/tex]

    using the trick on that page, namely a coordinate change

    [tex]
    u = z x_1, ~ v = z x_2, ~ w = z(1-x_1-x_2).
    [/tex]
    When you do this, you will be able to factor the new expression as a z integral that is a gamma function. The x1,x2 integral will be the integral you are trying to compute.
     
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