how can i find the expected value of a beta distribution to be a/(a + b)

Question

zarkon · Answer

integrate x times the pdf

zarkon · Answer

$$\int_{0}^{1}x\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\, x^{\alpha-1}(1-x)^{\beta-1}dx$$

zarkon · Answer

$$\int_{0}^{1}\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\, x^{(\alpha+1)-1}(1-x)^{\beta-1}dx$$

zarkon · Answer

$$\frac{\Gamma(\alpha+\beta)\Gamma(\alpha+1)}{\Gamma(\alpha)\Gamma(\alpha+\beta+1)}\int_{0}^{1}\frac{\Gamma(\alpha+1+\beta)}{\Gamma(\alpha+1)\Gamma(\beta)}\, x^{(\alpha+1)-1}(1-x)^{\beta-1}dx$$

zarkon · Answer

$$\int_{0}^{1}\frac{\Gamma(\alpha+1+\beta)}{\Gamma(\alpha+1)\Gamma(\beta)}\, x^{(\alpha+1)-1}(1-x)^{\beta-1}dx=1$$

zarkon · Answer

so

$$\frac{\Gamma(\alpha+\beta)\Gamma(\alpha+1)}{\Gamma(\alpha)\Gamma(\alpha+\beta+1)}\int_{0}^{1}\frac{\Gamma(\alpha+1+\beta)}{\Gamma(\alpha+1)\Gamma(\beta)}\, x^{(\alpha+1)-1}(1-x)^{\beta-1}dx$$

$$=\frac{\Gamma(\alpha+\beta)\Gamma(\alpha+1)}{\Gamma(\alpha)\Gamma(\alpha+\beta+1)}=\frac{\alpha}{\alpha+\beta}$$