If $X \sim \mathsf{Unif}(0,1)$, find the CDF of $Y = \ln\left(\frac{X}{1-X}\right)$

Question

Assume the random variable $X$ follows distribution $\mathsf{Unif}(0,1)$ and define the new random variable $Y = \ln\left(\frac{X}{1-X}\right)$. Find CDF of $Y.$

---

My attempt.

$F(X \leq x) = \int_{0}^{x} \ln(\frac{t}{1-t})\,dt = \ln(1-x) + x\left(\frac{x}{1-x}\right).$

Thus,

$$ F(x) =\begin{cases} 0 & x < 0 \\ \ln(1-x) + x\left(\frac{x}{1-x}\right) & 0\leq x \leq 1 \\ 1 & x > 1 \end{cases} $$

Would this be right?

Answer 1

$\mathbb{P}(Y\leq y)=\mathbb{P}(\log(\frac{X}{1-X})\leq y)=\mathbb{P}(\frac{X}{1-X}\leq e^y)=\mathbb{P}(X\leq\frac{e^y}{1+e^y})=\frac{e^y}{1+e^y}$

since $0<\frac{e^y}{1+e^y}<1$

Answer 2

The cdf of $Y$ is $F_Y(y) = P(Y \le y)$. I know you want to integrate something, but you can't yet because you don't even know the pdf of $Y$. You have to rewrite in the form:

$$P(Y \le y) = P(X \le h(y))$$

Then you can proceed as follows

$$P(X \le h(y)) = \int_{-\infty}^{h(y)} 1_{(0,1)}(x) dx = \int_{-\infty}^{h(y)} 1_{(0,1)}(x) dx$$

$$ = \int_{\mathbb R} 1_{(-\infty,h(y))}(x) 1_{(0,1)}(x) dx = \int_{\mathbb R} 1_{(-\infty,h(y)) \cap (0,1)}(x) dx$$

And I think you know how to do the intersection based on what you tried, but I'll spell it out for you anyhoo:

$$(-\infty,h(y)) \cap (0,1) = \emptyset, h(y)\le 0$$

$$(-\infty,h(y)) \cap (0,1) = (0,h(y)), h(y) \in (0,1)$$

$$(-\infty,h(y)) \cap (0,1) = (0,1), h(y) \ge 1$$

Now for the $h(y)$:

$$F_Y(y) = {P}(Y\le y)={P}(\ln(\frac{X}{1-X})\le y)={P}(\frac{X}{1-X}\le e^y)={P}(X\le\frac{e^y}{1+e^y})$$

$$=\int_{\mathbb R} 1_{(-\infty,h(y)) \cap (0,1)}(x) dx, h(y)=\frac{e^y}{1+e^y} \in (0,1)$$

$$\therefore, F_Y(y) = {P}(Y\le y) = \int_{\mathbb R} 1_{(0,\frac{e^y}{1+e^y})}(x) dx=\frac{e^y}{1+e^y}$$

Observe that all our results hold for $y \in \mathbb R$.