mean and variance of a beta distribution

**dori1123** · May 9th 2010, 10:51 AM

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?

**Anonymous1** · May 9th 2010, 01:47 PM

Originally Posted by dori1123

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

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