Given beta distribution as:
$$
\mathcal{B}(x;a,b) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1} (1-x)^{b-1}
$$
I am trying to show:
$$
\int_0^1 x^{a-1} (1-x)^{b-1}\,dx = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}
$$
which can be considered as the normalizing constant.
Here is what I have done so far (using the hints at Bishop's). Write $\Gamma(a)\Gamma(b)$ and change variable $t=y+x$ fixing $x$:
\begin{align*}
\newcommand{\ints}{\int_0^\infty\!\!\int_0^\infty}
\Gamma(a)\Gamma(b) &= \int_0^\infty \exp(-x)x^{a-1} \, dx \int_0^\infty \exp(-y)y^{b-1} \, dy \\
&= \ints \exp(-x)x^{a-1} \exp(-y)y^{b-1} \, dx\,dy \\
&= \ints e^{-t} x^{a-1}(t-x)^{b-1} \, dt\,dx
\end{align*}
Now if we make the variable change $x=\mu t$
\begin{align*}
\ints e^{-t} x^{a-1}(t-x)^{b-1} \, dt\,dx &= \ints e^{-t} (\mu t)^{a-1} (t-\mu t)^{b-1} t \, dt\, d\mu \\
&= \ints e^{-t} t^{a+b-1} \mu^{a-1} (1-\mu)^{b-1} \, d\mu \, dt
\end{align*}
We can split the integrals:
\begin{align*}
&= \underbrace{\int_0^\infty e^{-t} t^{a+b-1} \,dt}_{\Gamma(a+b)} \underbrace{\int_0^\infty \mu^{a-1} (1-\mu)^{b-1} \, d\mu}_{\text{limits are not from 0 to 1}}
\end{align*}
What is wrong with the limits? Is it my mistake? Or can it be substituted to from $0$ to $1$?
PS. sorry for the big space between double integrals, I don't know the right way to do them.
Best Answer
The problem is in the last step here:
As you made change of variables $t=x+y$, naturally $t \geqslant x$, thus the last integral should be $$ \int_0^\infty \exp(-t)\left( \int_0^t x^{a-1} (t-x)^{b-1} \mathrm{d}x\right) \mathrm{d}t \stackrel{x=\mu t}{=} \int_0^\infty \exp(-t) \left( t^{a+b-1} \int_0^t \mu^{a-1} (1-\mu)^{b-1} \mathrm{d}\mu\right) \mathrm{d}t = \\ \int_0^\infty t^{a+b-1}\mathrm{e}^{-t} \mathrm{d}t \cdot \int_0^1 \mu^{a-1} (1-\mu)^{b-1} \mathrm{d} \mu $$