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Math Help - mean and variance of a beta distribution

  1. #1
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    mean and variance of a beta distribution

    Let X have a beta distribution f(x)=\frac{\gamma(\alpha+\beta)}{\gamma(\alpha)\ga  mma(\beta)}x^{\alpha-1}(1-x)^{\beta-1},0<x<1. Show that the mean and variance of X are \mu=\frac{\alpha}{\alpha+\beta} and \sigma^2=\frac{\alpha\beta}{(\alpha+\beta+1)(\alph  a+\beta)^2}.

    \mu=\int_0^1x\frac{\gamma(\alpha+\beta)}{\gamma(\a  lpha)\gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}dx=\int_0^1\frac{\gamma(\alpha+\beta)}{\gamma(\a  lpha)\gamma(\beta)}x^{\alpha}(1-x)^{\beta-1}dx I can't figure out how to integrate this. Can I get some help please?
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  2. #2
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    Quote Originally Posted by dori1123 View Post
    Let X have a beta distribution f(x)=\frac{\gamma(\alpha+\beta)}{\gamma(\alpha)\ga  mma(\beta)}x^{\alpha-1}(1-x)^{\beta-1},0<x<1. Show that the mean and variance of X are \mu=\frac{\alpha}{\alpha+\beta} and \sigma^2=\frac{\alpha\beta}{(\alpha+\beta+1)(\alph  a+\beta)^2}.

    \mu=\int_0^1x\frac{\gamma(\alpha+\beta)}{\gamma(\a  lpha)\gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}dx=\int_0^1\frac{\gamma(\alpha+\beta)}{\gamma(\a  lpha)\gamma(\beta)}x^{\alpha}(1-x)^{\beta-1}dx I can't figure out how to integrate this. Can I get some help please?
    \mu=\int_0^1x\frac{\gamma(\alpha+\beta)}{\gamma(\a  lpha)\gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}dx


    =\int_0^1\frac{\gamma(\alpha+\beta)}{\gamma(\alpha  )\gamma(\beta)}x^{\color{red}{(\alpha+1)-1}} (1-x)^{\beta-1}dx


    =\frac{\gamma(\alpha+\beta)}{\gamma(\alpha)\gamma(  \beta)}\cdot \frac{1}{\frac{\gamma(\alpha+1+\beta)}{\gamma(\alp  ha+1)\gamma(\beta)}}\int_0^1 \underbrace{\frac{\gamma(\alpha+1+\beta)}{\gamma(\  alpha+1)\gamma(\beta)} x^{(\alpha+1)-1}(1-x)^{\beta-1}}dx \color{red}{\text{This is the pdf of a } \beta(\alpha+1,\beta) \text{ r.v.}}


    =\frac{\gamma(\alpha+\beta)}{\gamma(\alpha)\gamma(  \beta)}\cdot \frac{1}{\frac{\gamma(\alpha+1+\beta)}{\gamma(\alp  ha+1)\gamma(\beta)}}\cdot 1

    Apply the same technique to obtain the second moment.

    aNon1
    Last edited by Anonymous1; May 9th 2010 at 03:50 PM.
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