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Math Help - beta distribution probability density function proof

  1. #1
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    Question beta distribution probability density function proof

    Hi
    I'm having bit of trouble proving this fact.
    beta distribution probability density function proof-beta-dist-density-.jpg
    can someone help me with the proof? or at least lead me thru it?

    Thanks

    -AC
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  2. #2
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    Prove B(p,q)=\int_{0}^{1}x^{p-1}(1-x)^{q-1}dx=\frac{{\Gamma}(p){\Gamma}(q)}{{\Gamma}(p+q)}


    B(p,q)=\frac{{\Gamma}(p){\Gamma}(q)}{{\Gamma}(p+q)  }....[1]

    Start with {\Gamma}(p)=\int_{0}^{\infty}t^{p-1}e^{-t}dt

    Now, put t=y^{2}, and we get:

    {\Gamma}(p)=2\int_{0}^{\infty}y^{2p-1}e^{-y^{2}}dy....[2]

    Similarly, (the dummy integration variable can be any letter):

    {\Gamma}(p)=2\int_{0}^{\infty}x^{2q-1}e^{-x^{2}}dx

    Next, we multiply these 2 equations and change to polar coordinates:

    {\Gamma}(p){\Gamma}(q)=4\int_{0}^{\infty}\int_{0}^  {\infty}x^{2q-1}y^{2p-1}e^{-(x^{2}+y^{2})}dxdy

    =4\int_{0}^{\infty}\int_{0}^{\frac{\pi}{2}}(rcos{\  theta})^{2q-1}(rsin{\theta})^{2p-1}e^{-r^{2}}rdrd{\theta}

    =4\int_{0}^{\infty}r^{2p+2q-1}e^{-r^{2}}dr\int_{0}^{\frac{\pi}{2}}(cos{\theta})^{2q-1}(sin{\theta})^{2p-1}d{\theta}....[3]

    The r integral in [3] is \frac{1}{2}{\Gamma}(p+q) by [2].

    The {\theta} in [3] is \frac{1}{2}B(p,q) by

    B(p,q)=2\int_{0}^{\frac{\pi}{2}}(sin{\theta})^{2p-1}(cos{\theta})^{2q-1}d{\theta}

    Then, {\Gamma}(p){\Gamma}(q)=4\cdot \frac{1}{2}{\Gamma}(p+q)\cdot \frac{1}{2}B(p,q) and [1] follows.
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  3. #3
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    Wow! that was quick!
    Thanks for the help!
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