Let $ f(x) = x^{ \alpha-1}(1+x)^{- \alpha-\beta}\frac{Γ(α+β)}{Γ(α)Γ(β)},$ where $Γ(x) = (x − 1)Γ(x-1)$ and $α,β > 1$. Let $ f(x) =  x^{ \alpha-1}(1+x)^{- \alpha-\beta}\frac{Γ(α+β)}{Γ(α)Γ(β)},$ where $Γ(x) = (x − 1)Γ(x-1)$ and $α,β > 1$. Show that
$E(X) = \frac{α}{β-1}.$
I've tried in a couple of ways, but couldn't see the direction.
I've used moment generating function to obtain $E(X)$, but couldn't see a way to simplify.
Also, I've tried converting the part $f(x) = x^{ \alpha-1}(1+x)^{- \alpha-\beta} $ into some form similar to
$ B(α,β)=\int_0^1 x^{α−1}(1−x)^{β−1}dx$ so that I could create some gamma functions, which eventually could lead to do a cross out with the given gamma functions $\frac{Γ(α+β)}{Γ(α)Γ(β)}.$
In class, we did not really cover the basics of the gamma function, but mostly on expected values. Can someone suggest what topic or section I should read in the textbook for some ideas? E.g. proof of $E(X)$ in terms of a gamma functions, etc.
Thank you.
 A: $$E[X]=\int_{0}^{\infty}x\cdot f(x)\,dx =\frac{1}{B(\alpha,\beta)}\int_{0}^{\infty} x \frac{x^{\alpha-1}}{(1+x)^{\alpha+\beta}}\,dx =\\\frac{1}{B(\alpha,\beta)}\int_{0}^{\infty}\frac{x^{\alpha}}{(1+x)^{\alpha+\beta}}\,dx $$
$$=\frac{B(\alpha+1,\beta-1)}{B(\alpha,\beta)}=\frac{\Gamma(\alpha+1)\Gamma(\beta-1)}{\Gamma(\alpha+\beta)}\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}=\\\frac{\alpha\cdot\Gamma(\alpha)\cdot\Gamma(\beta-1)}{(\beta-1)\cdot\Gamma(\alpha)\cdot\Gamma(\beta-1)}=\frac{\alpha}{\beta -1}$$
These are just using the properties of Gamma and Beta functions.
$\displaystyle B(\alpha,\beta)=\int_{0}^{\infty}\frac{x^{\alpha-1}}{(1+x)^{\beta+\alpha}}\,dx=\int_{0}^{1}x^{\alpha-1}(1-x)^{\beta-1}\,dx$
Substitute $t=\frac{1}{1+x}$ above to see the equivalence.
And $B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$
Have a read here and Here
Also you should write the pdf along with the domain of $x$. For example $f(x)=\frac{1}{B(\alpha,\beta)}x^{\alpha-1}(1+x)^{-\alpha-\beta}$ when $x>0$ .
So infact you should say that $f(x)=\begin{cases} \frac{1}{B(\alpha,\beta)}x^{\alpha-1}(1+x)^{-\alpha-\beta}\,,x>0\\ 0\,,\text{otherwise}\end{cases}$
A: Let $$B(\alpha,\beta) = \int_{x=0}^1 x^{\alpha - 1} (1-x)^{\beta-1} \, dx = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}. \tag{1}$$  This is the usual definition of the Beta function.
Then with the substitution
$$u = \frac{x}{1-x}, \quad du = \frac{1}{(1-x)^2} \, dx, \tag{2}$$
or equivalently,
$$x = \frac{u}{u+1}, \quad dx = \frac{1}{(u+1)^2} \, du, \tag{3}$$
we obtain
$$\begin{align}
B(\alpha,\beta) &= \int_{u=0}^\infty \left(\frac{u}{u+1}\right)^{\alpha-1} \left(1 - \frac{u}{u+1}\right)^{\beta - 1} \frac{1}{(u+1)^2} \, du \\
&= \int_{u=0}^\infty u^{\alpha-1} (u+1)^{1-\alpha-2} \left(\frac{1}{u+1}\right)^{\beta-1} \, du \\
&= \int_{u=0}^\infty u^{\alpha-1} (u+1)^{-\alpha-1-\beta+1} \, du\\
&= \int_{u=0}^\infty u^{\alpha-1} (u+1)^{-\alpha-\beta} \, du, \tag{4}
\end{align}$$
which is your target integral.  Therefore, $(1)$ and $(4)$ are equivalent.
From this, we know that
$$\int_{x=0}^\infty f_X(x) \, dx = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \int_{x=0}^\infty x^{\alpha-1}(x+1)^{-\alpha-\beta} \, dx = 1 \tag{5}$$ for any choice of parameters $\alpha, \beta > 0$.
Consequently,
$$\begin{align}
\operatorname{E}[X]
&= \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \int_{x=0}^\infty x \cdot x^{\alpha-1}(x+1)^{-\alpha-\beta} \, dx \\
&= \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \int_{x=0}^\infty x^{(\alpha+1) - 1} (x+1)^{-(\alpha+1)-(\beta-1)} \, dx \\
&= \frac{\Gamma(\alpha+1) \Gamma(\beta - 1)}{\Gamma(\alpha)\Gamma(\beta)} \cdot \frac{\Gamma((\alpha+1)+(\beta-1))}{\Gamma(\alpha+1)\Gamma(\beta-1)} \int_{x=0}^\infty x^{(\alpha+1) - 1} (x+1)^{-(\alpha+1)-(\beta-1)} \, dx \tag{6} \\
&= \frac{\Gamma(\alpha+1) \Gamma(\beta-1)}{\Gamma(\alpha)\Gamma(\beta)} \tag{7} \\
&= \frac{\alpha}{\beta-1}, \tag{8}
\end{align}$$
for $\alpha > 0$ and $\beta > 1$.  Note that the condition $\beta > 1$ arises from the rewriting of the integrand in the form of $(4)$ with the substitution $\alpha \to \alpha + 1$, $\beta \to \beta-1$, hence the integral in $(6)$ converges only if $\beta - 1 > 0$.  In $(7)$ we exploit $(5)$, except with $\alpha + 1$ and $\beta - 1$ in place of $\alpha$ and $\beta$.  In $(8)$ we use the recursive property of the gamma function $\Gamma(z+1) = z \Gamma(z)$.
