Variance of beta distribution (fastest way) Suppose a Random variable $X \sim \mathrm{Beta}(a,b)$ 
Find the $\mathrm{Var}( \frac{X}{1-X} ) $
My initial approach is to calculate  $\mathrm{E}( \frac{X}{1-X} ) $ and  $\mathrm{E}( [\frac{X}{1-X}]^2 ) $ 
And then find it :
$\mathrm{Var}( \frac{X}{1-X} )  =  \mathrm{E}[ (\frac{X}{1-X})^2 ]  - \mathrm{E}[(\frac{X}{1-X})]^2  $
Where   $\mathrm{E}( \frac{X}{1-X} )  = \int\limits_0^1 \frac{X}{1-X} f(x)\, \mathrm{d}x=\ldots $
My question first of all is if this approach is correct.
Also,
Is there a quicker way to calculate the $\mathrm{Var}( \frac{X}{1-X} ) $? 
Or could you suggest an alternative way?
 A: It looks correct--but there are some ways to do the integration without much work.
Let's begin by listing the simplest and most obvious facts about Beta distributions.


*

*It is apparent from almost any characterization of these distributions that when a random variable $X$ has a Beta$(a,b)$ distribution, then $Y=1-X$ has a Beta$(b,a)$ distribution.

*The unnormalized density function $f_{a,b}$ of a Beta$(a,b)$ distribution is a multiple of $x^{a};$ that is, it can be expressed as $$f_{a,b}(x) = x^{a} g_{b}(x).$$

*The normalizing constant of a Beta$(a,b)$ distribution is 
$$B(a,b) = \int f_{a,b}(y)\mathrm{d}y = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
where the gamma function has the property that for any $z$ that is not zero or a negative integer,
$$\Gamma(z+1) = z\Gamma(z).$$
(If you didn't know this you could still compute the variance as shown below, but you wouldn't obtain a simplified formula for it.)
These suffice, along with the variance formula
$$\operatorname{Var}(X) = E(X^2) - E(X)^2$$
and the (easily proven) fact that $\operatorname{Var}(X+\lambda) = \operatorname{Var}(X)$ for any constant $\lambda,$ to obtain an answer simply.  Here's how it goes:
First, algebra gives
$$Z=\frac{X}{1-X} = \frac{1}{1-X} -1 = \frac{1}{Y}-1$$
where $Y=1-X$ has a Beta$(b,a)$ distribution by $(1).$  Thus, we obtain moments of $Y$ from $(2),$ without any calculation, as
$$\eqalign{
\mu_k(Y) 
&= \frac{1}{B(b,a)}\int y^k f_{b,a}(y)\mathrm{d}y \\
&= \frac{1}{B(b,a)}\int y^k y^{b} g_{a}(y)\mathrm{d}y \\
&= \frac{1}{B(b,a)}\int y^{b+k} g_{a}(y)\mathrm{d}y \\
&= \frac{1}{B(b,a)}\int f_{b+k,a}(y)\mathrm{d}y \\
&= \frac{B(b+k,a)}{B(b,a)}.
}$$

Consequently the variance is $$\eqalign{
\operatorname{Var}(Z) &=\operatorname{Var}(Y^{-1}-1)= \operatorname{Var}(Y^{-1}) = E\left((Y^{-1})^2\right) - E\left(Y^{-1}\right)^2 \\
&= \mu_{-2}(Y)-(\mu_{-1}(Y))^2 \\
&= \frac{B(b-2,a)}{B(b,a)} -\left(\frac{B(b-1,a)}{B(b,a)}\right)^2.
}
$$

This simplifies to
$$\operatorname{Var}\left(\frac{X}{1-X}\right)=\frac{B(b-2,a)}{B(b,a)} -\left(\frac{B(b-1,a)}{B(b,a)}\right)^2 = \frac{a(a+b-1)}{(b-2)(b-1)^2}.$$
The only algebra needed to achieve the last step--and therefore the only calculation in the entire derivation--comes from the relation $(3)$ and the assumption that everything is well-defined, which it will be when $b-2\gt 0$ (for otherwise the expectation of $Y^{-2}$ diverges).
