Q. 5.26 - A First Course in Probability

Question

If $\mathrm{x}$ is a beta random variable with parameters $\mathrm{a}$ and $\mathrm{b}$, show that

$\mathrm{E}\left[\mathrm{X}\right]=\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}}$

$Var\left(\mathrm{X}\right)=\frac{\mathrm{ab}}{(\mathrm{a}+\mathrm{b}{)}^2(\mathrm{a}+\mathrm{b}+1)}$

Step-by-Step Solution

Answer

In general, find the $\mathrm{k}$th moment and use it to calculate the mean and variance.

Step 1: Beta distribution

Density function,

$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta })}{\mathrm{\Gamma }\left(\mathrm{\alpha }\right)\mathrm{\Gamma }\left(\mathrm{\beta }\right)}\times {\mathrm{x}}^{\mathrm{\alpha }-1}(1-\mathrm{x}{)}^{\mathrm{\beta }-1}$

$\mathrm{x}\in \left[0,1\right]$is to be zero.

$\mathrm{E}\left({\mathrm{X}}^{\mathrm{k}}\right)={\int }_{\mathrm{ℝ}}{\mathrm{x}}^{\mathrm{k}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta })}{\mathrm{\Gamma }\left(\mathrm{\alpha }\right)\mathrm{\Gamma }\left(\mathrm{\beta }\right)}{\int }_0^1{\mathrm{x}}^{\mathrm{k}}\times {\mathrm{x}}^{\mathrm{\alpha }-1}(1-\mathrm{x}{)}^{\mathrm{\beta }-1}\mathrm{dx}$

$=\frac{\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta })}{\mathrm{\Gamma }\left(\mathrm{\alpha }\right)\mathrm{\Gamma }\left(\mathrm{\beta }\right)}{\int }_0^1{\mathrm{x}}^{\mathrm{k}+\mathrm{\alpha }-1}(1-\mathrm{x}{)}^{\mathrm{\beta }-1}\mathrm{dx}$

So,

$\mathrm{E}\left({\mathrm{X}}^{\mathrm{k}}\right)=\frac{\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta })}{\mathrm{\Gamma }\left(\mathrm{\alpha }\right)\mathrm{\Gamma }\left(\mathrm{\beta }\right)}\times \mathrm{B}(\mathrm{k}+\mathrm{\alpha },\mathrm{\beta })$

$=\frac{\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta })}{\mathrm{\Gamma }\left(\mathrm{\alpha }\right)\mathrm{\Gamma }\left(\mathrm{\beta }\right)}\times \frac{\mathrm{\Gamma }(\mathrm{k}+\mathrm{\alpha })\mathrm{\Gamma }\left(\mathrm{\beta }\right)}{\mathrm{\Gamma }(\mathrm{k}+\mathrm{\alpha }+\mathrm{\beta })}$

Step 2: Explanation

For$\mathrm{k}=1$

$\mathrm{E}\left(\mathrm{X}\right)=\frac{\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta })}{\mathrm{\Gamma }\left(\mathrm{\alpha }\right)}\times \frac{\mathrm{\Gamma }(\mathrm{\alpha }+1)}{\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta }+1)}$

$=\frac{\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta })}{\mathrm{\Gamma }\left(\mathrm{\alpha }\right)}\times \frac{\mathrm{\alpha \Gamma }\left(\mathrm{\alpha }\right)}{(\mathrm{\alpha }+\mathrm{\beta })\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta })}=\frac{\mathrm{\alpha }}{\mathrm{\alpha }+\mathrm{\beta }}$

For $\mathrm{k}=2$

$\mathrm{E}\left({\mathrm{X}}^2\right)=\frac{\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta })}{\mathrm{\Gamma }\left(\mathrm{\alpha }\right)}\times \frac{\mathrm{\Gamma }(\mathrm{\alpha }+2)}{\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta }+2)}$

$=\frac{\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta })}{\mathrm{\Gamma }\left(\mathrm{\alpha }\right)}\times \frac{(\mathrm{\alpha }+1)\mathrm{\alpha \Gamma }\left(\mathrm{\alpha }\right)}{(\mathrm{\alpha }+\mathrm{\beta }+1)(\mathrm{\alpha }+\mathrm{\beta })\mathrm{\Gamma }(\mathrm{\alpha }+\mathrm{\beta })}$

So,

$Var\left(\mathrm{X}\right)=\mathrm{E}\left({\mathrm{X}}^2\right)-\mathrm{E}(\mathrm{X}{)}^2$

$=\frac{(\mathrm{\alpha }+1)\mathrm{\alpha }}{(\mathrm{\alpha }+\mathrm{\beta }+1)(\mathrm{\alpha }+\mathrm{\beta })}-\frac{{\mathrm{\alpha }}^2}{(\mathrm{\alpha }+\mathrm{\beta }{)}^2}$

$=\frac{(\mathrm{\alpha }+1)\mathrm{\alpha }(\mathrm{\alpha }+\mathrm{\beta })-{\mathrm{\alpha }}^2(\mathrm{\alpha }+\mathrm{\beta }+1)}{(\mathrm{\alpha }+\mathrm{\beta }+1)(\mathrm{\alpha }+\mathrm{\beta }{)}^2}$

$=\frac{\mathrm{\alpha \beta }}{(\mathrm{\alpha }+\mathrm{\beta }+1)(\mathrm{\alpha }+\mathrm{\beta }{)}^2}$