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Lets say I have a Random Variable:

$ \Pi \sim\mathcal U[\frac{1}{2},1)$

And we also know that

$\Xi = \ln\left(\dfrac{\Pi}{1-\Pi}\right)$

How can I find the cumulative distribution or the density function of $\Xi$?

thank you

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If $\Xi = \ln (\Pi/(1-\Pi))$, then $\Pi = 1/(1+e^{-\Xi})$

( Further we note that $\Xi$ is a monotonic increasing and invertible function of $\Pi$ over the support interval. )

If $\Pi\sim\mathcal U[\tfrac 12, 1)$ then $\mathsf P(\Pi\leq y)= 2y\cdot\mathbf 1_{y\in[1/2;1)}+\mathbf 1_{y\in[1;\infty)}$, and so: $$\mathsf P(\Xi\leq x) ~{= \mathsf P(\Pi\leq 1/(1+e^{-x})) \\ =\lower{1ex}\ldots}$$

Find the pdf by taking the derivative with respect to $x$.

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Let $f : \mathbb{R} \to \mathbb{R}$ be continuous and bounded. Then $$\begin{align*} E(f(\Xi)) &= E(f(\log(\Pi/(1 - \Pi))))\\ &= \int_{\mathbb{R}} f\left(\log\frac{u}{1 - u}\right) f_\Pi(u) du\\ &= 2\int_{1/2}^1 f\left(\log\frac{u}{1 - u}\right)du\\ &= 2\int_{-\infty}^0 f(s)\frac{e^s}{(1 + e^s)^2} ds \end{align*}$$ By the substitution $s = \log\frac{u}{1 - u}$, i.e. $$e^s = \frac{u}{1- u} \Leftrightarrow (1 - u)e^s = u \Leftrightarrow e^s = u(1 + e^s) \Leftrightarrow u = \frac{e^s}{1 + e^s}$$ and $$ds = \frac{1 - u}{u} \frac{1}{(1 - u)^2}du = \frac{1}{u(1-u)}du = \frac{(e^s + 1)^2}{e^s}du$$ Hence your density is given by $$\boxed{f_\Xi(s) = 2\frac{e^s}{(1 + e^s)^2}\chi_{(-\infty,0]}(s).}$$

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$P(Xi<ln(x/1-x)) = P(Pi<x)$

Letting $y = ln(x/1-x)$

$P(Xi<y) = P(Pi<(e^{-y}+1)^{-1})$

Let $C_p$ be the CDF of Pi and $C_x$ be the CDF of Xi.

$C_p(x) =\begin{cases}0, \ x\leq1/2 \\2x-1, \ 1/2<x<1 \\1, \ x\geq1 \end{cases}$

$C_x(x) =\begin{cases}0, \ x\leq0 \\(2(e^{-x}+1)^{-1})-1, \ x>0 \end{cases}$

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Hint:

$$\Xi>x\iff \frac{\Pi}{1-\Pi}>e^x\iff \Pi>\frac{e^x}{1+e^x}$$

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