2.5 ( ) www In this exercise, we prove that the beta distribution, given by (2.13), is correctly normalized, so that (2.14) holds. This is equivalent to showing that

 1 0

μa−1(1 − μ)b−1 dμ =

Γ(a)Γ(b)

Γ(a +

b) . (2.265)

From the definition (1.141) of the gamma function, we have

Γ(a)Γ

(b) =

 ∞

0 exp(−x)xa−1 dx

 ∞

0 exp(−y)yb−1 dy. (2.266)

Use this expression to prove (2.265) as follows. First bring the integral over y inside the integrand of the integral over x, next make the change of variable t = y + x where x is fixed, then interchange the order of the x and t integrations, and finally make the change of variable x = tμ where t is fixed.