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Math Help - beta distribution problem

  1. #1
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    beta distribution problem

    Hi, I am stuck at this proof problem. Please help! Thanks.


    Let X1,,,,,Xn be a random sample from a geometric distribution with parameter θ, i.e.,
    f(xi|θ) =θ(1-θ)^xi. Show that the beta distribution

    (for ) is conjugate for θ .
    Last edited by kqed123; February 3rd 2010 at 07:16 AM. Reason: couldnt post the correct pix
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by kqed123 View Post
    Hi, I am stuck at this proof problem. Please help! Thanks.

    Let X1; : : : ;Xn be a random sample from a geometric distribution with parameter θ, i.e.,
    f(xi|θ ) = θ(1-θ )^xi . Show that the beta distribution
    f(θ  ;α,β ) =Γ(α+ β)/(Γ(α)Γ(β))*(θ^(α-1))*((1-θ)^( β-1))

    (for θ [0; 1]) is conjugate for θ .
    Please either learn to use our LaTeX system or spell out the non-ASCII charaters you are using because we can't all read them.

    CB
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  3. #3
    MHF Contributor matheagle's Avatar
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    f_1(x|\theta)=\theta(1-\theta)^x

    f_2(\theta)={\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)}\theta^{\alpha-1}(1-\theta)^{\beta-1}

    so the joint density is the product...

    f_3(x,\theta)={\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)}\theta^{\alpha}(1-\theta)^{\beta+x-1}

    next obtain the marginal of X....

    f_4(x)={\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)}\int_0^1\theta^{\alpha  }(1-\theta)^{\beta+x-1}d\theta

    making this look like a Beta, we have...

    =\left({\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)}\right)\left({\Gamma(\  alpha+1)\Gamma(\beta+x)\over \Gamma(\alpha+\beta+x+1)}\right)

    Instead of reducing, just divide...

    f_5(\theta|x)={f_3(x,\theta)\over f_4(x)}

    =\left({\Gamma(\alpha+\beta+x+1)}\over\Gamma(\alph  a+1)\Gamma(\beta+x)\right)\theta^{\alpha}(1-\theta)^{\beta+x-1}

    which is a \beta(\alpha+1,\beta+x) distribution.


    THIS was way too much typing for me.
    Last edited by matheagle; February 3rd 2010 at 05:09 PM.
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