Question

Proving (Mathematical Statistics)

Answer 1

The beta function is typically defined as
\[B(\alpha,\beta)=\int_0^1u^{\alpha-1}(1-u)^{\beta-1}\,du\]
Here's a substitution we'll find useful:
\[u=\frac{s}{1+s}~~\iff~~\color{red}{\frac{u}{1-u}=s}~~\implies~~du=\frac{ds}{(s+1)^2}\]
Let's start with the \(s\) integral and try to derive the \(u\) integral:
\[\begin{align*}
B(\alpha,\beta)&=\int_0^\infty \frac{s^{\alpha-1}}{(1+s)^{\alpha+\beta}}\,ds\\\\
&=\int_0^\infty \frac{s^{\alpha}}{s(1+s)^{\alpha}(1+s)^{\beta}}\,ds\\\\
&=\int_0^1 \frac{u^{\alpha}\color{red}{(1-u)}}{\color{red}{u}\left(1+\dfrac{u}{1-u}\right)^{\beta-2}}\,du\\\\
&=\int_0^1 \frac{u^{\alpha-1}(1-u)}{\left(\dfrac{1-u+u}{1-u}\right)^{\beta-2}}\,du\\\\
&=\int_0^1 \frac{u^{\alpha-1}(1-u)}{\left(1-u\right)^{2-\beta}}\,du\\\\
&=\int_0^1 u^{\alpha-1}(1-u)^{\beta-1}\,du\end{align*}\]
as desired.