Pdf of $\ln\left(\frac{Y}{1-Y}\right)$ when $Y$ follows Beta distribution

Question

The PDF of $X \sim \operatorname{Beta}(a,b)$ is given by $$f_X(x) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1} (1-x)^{b-1}, \quad 0 < x < 1.$$

Then for a monotone transformation $Y = g(X) = \log\frac{X}{1-X}$, we have

$$e^Y = \frac{X}{1-X} \rightarrow e^{-Y} = \frac{1- X}{X} = \frac{1}{X} - 1 \rightarrow 1 + e^{-Y} = \frac{1}{X} \rightarrow \frac{1}{1 + e^{-Y}} = g^{-1}(X) $$

and the derivative $ g^{-1}(x) $ in relation to y: $$\frac{dg^{-1}}{dy} = \frac{e^{-y}}{(1 + e^{-y})^2}$$

So,

$$f_Y(y) = f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right| = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} \left(\frac{1}{1 + e^{-y}}\right)^{a-1} \left(1-\frac{1}{1 + e^{-y}}\right)^{b-1} \frac{e^{-y}}{(1 + e^{-y})^2}$$

But I am stuck at this point..

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edited:

$$\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} \left(\frac{1}{1 + e^{-y}}\right)^{a-1} \left(1-\frac{1}{1 + e^{-y}}\right)^{b-1} \frac{e^{-y}}{(1 + e^{-y})^2} = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} \left(\frac{1}{1 + e^{-y}}\right)^{a} \left(1-\frac{1}{1 + e^{-y}}\right)^{b}$$

and maybe changing $a$ and $b$ to $a' -1 = a $ and $ b'- 1 = b$, I might able to get into a new beta. But I am quite confused handling the gammas functions.

Answer 1

$Y~\sim Beta(a;b)$

Let's start with the transformation

$Z=\frac{Y}{1-Y}$

$$f_Z(z)=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}z^{a-1}(1+z)^{-a-b}$$

$z \geq 0$

That is a Beta type II distribution

I do not know if $logZ$ is another generalized type of Beta that I do not know...anyway your calculations are correct

Answer 2

With $1-\frac{1}{1+e^{-y}}=\frac{e^{-y}}{1+e^{-y}}$ we can simplify this somewhat to$$f_Y(y)=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}e^{-by}(1+e^{-y})^{-a-b}.$$While this doesn't have an obvious name, with $Z:=e^{-Y}$ we then recover @tommik's notation for $Z$'s beta type II distribution (albeit with $a,\,b$ exchanged).