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Thread: Moments of Burr Distribution(tricky)

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    Moments of Burr Distribution(tricky)




    Hi all, consider the following problem:

    For the Burr Distribution given by:

    \displaystyle g(y)=\frac{abty^{t-1}}{(1+by^t )^{a+1}}\ \text{for}\ 0<y<\infty show that

    \displaystyle E(Y^k)=\frac{1}{b^{k/t}}\frac{ \Gamma \left( a-\frac{k}{t} \right) \Gamma\left( \frac{k}{t}+1 \right)} {{\Gamma \left( a \right)}} provided k<at


    Okay, the first thing I noticed is that the three gamma functions here, are similar to the beta function B(a,b)=\int_0^1 x^{a-1}\left(1-x\right)^{b-1} dx which can be written equivalently as \frac{\Gamma\left (a \right) \times \Gamma \left( b \right)}{\Gamma \left ( a+b \right )} so perhaps a simple change of variables might suffice to get the required form.

    We have then:

    E\left( Y^k \right)=\int_0^{\infty}\frac{y^{t+k-1} abt } {\left( 1 +by^t \right)^{a+1}} dy and after making the change of variables u=y^t b after some simplifying the integral becomes:

    \displaystyle \frac{1}{b^{\frac{k}{t}}}\int_0^{\infty}\frac{au^{\frac{k}{t}}}{\left( 1+u \right)^{a+1}} du and this is as far as I can go unfortunately. Nothing works from here.

    Any suggestions as to how I could proceed are greatly appreciated. Thanks.
    Last edited by JohnK; 10-10-2013 at 05:35 AM.

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    Re: Moments of Burr Distribution(tricky)


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    JohnK (10-10-2013)

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    Re: Moments of Burr Distribution(tricky)


    The last integral looks similar with one of the forms of Beta Function but not quite. The denominator is a bit different and I have double-checked my results to avoid mistakes.


    EDIT: Never mind, I figured it out, thanks.
    Last edited by JohnK; 10-10-2013 at 06:13 AM.

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