Results 1 to 6 of 6

Math Help - Variance in Inverse Gamma Distribution

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    30

    Variance in Inverse Gamma Distribution

    Can someone help me how to
    show Var(Y)=\frac{\beta^2}{(\alpha-1)^2(\alpha-2)}, in Inverse Gamma Distribution.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Aug 2009
    Posts
    27
    To compute the expectation value, note that

    <br />
EX = \int_0^{\infty} x \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{-\alpha - 1} e^{-\beta / x} dx<br />

    <br />
EX = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty} x^{-(\alpha - 1) - 1} e^{-\beta / x} dx<br />


    Completing the term inside the integral,

    <br />
EX = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \int_0^\infty x^{-(\alpha - 1) - 1} \frac{\beta^{\alpha - 1}}{\Gamma(\alpha - 1)} e^{-\beta/x} dx \frac{\Gamma(\alpha - 1)}{\beta^{\alpha - 1}}<br />

    <br />
EX = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \frac{\Gamma(\alpha - 1)}{\beta^{\alpha - 1}}<br />

    Since the term inside the integral is now an inverse Gamma with parameters

     \alpha - 1 ; \beta

    So,

    <br /> <br />
\frac{\beta}{\alpha - 1}<br />

    Do the same for  EX^2 and find the variance.
    Last edited by gustavodecastro; August 31st 2009 at 08:34 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    30
    Thanks Gustavodecastro, can you retype the code, because i can't read it properly?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2009
    Posts
    27
    Sorry for that. Now I think it is okay. You have to remember that

    <br />
\Gamma(n) = (n-1)!<br />
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    Posts
    30
    I'm not sure how finding EX^2, will help me find the variance?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Aug 2009
    Posts
    27
    To compute  EX^2 , note that

    <br />
Ex = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \int_0^\infty x^{-(\alpha - 2) - 1} \frac{\beta^{\alpha - 2}}{\Gamma(\alpha - 2)} e^{-\beta/x} dx \frac{\Gamma(\alpha - 2)}{\beta^{\alpha - 2}}<br />

    And so you have an inverse gamma with parameters  \alpha - 2 ; \beta inside the integral. Do the same as before and find  EX^2 .

    <br />
EX^2 = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \frac{\Gamma(\alpha - 2)}{\beta^{\alpha - 2}} = \frac{\beta^2}{(\alpha - 2)(\alpha - 1)}<br />

    Then, as  VarX = EX^2 - \left( EX \right)^2 , find the variance.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. cumulative distribution function using a gamma distribution
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: September 11th 2010, 11:05 AM
  2. part of inverse gamma
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: September 11th 2009, 10:08 PM
  3. Inverse Gamma Distribution
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: August 28th 2009, 12:29 AM
  4. Replies: 3
    Last Post: February 6th 2009, 04:59 AM
  5. Replies: 0
    Last Post: March 30th 2008, 01:44 PM

Search Tags


/mathhelpforum @mathhelpforum