I'm studying Continuous Random Variable in the topic Universality of the Uniform.
From the book "Introduction to probability" by Joseph K. Blitzstein and Jessica Hwang page 207-208 the writer shows a calculation to verify that $ F^{-1}(U) = F(x) $ for logistic CDF: now let U ~ Uniform(0,1)
The Logistic CDF is:
$$ F(x) = \frac{e^x}{1+e^x} , x \in R $$
First we find its inverse:
$$ F^{-1}(u) = log(\frac{u}{1-u}) $$
Then we plug in U for u:
$$ F^{-1}(U) = log(\frac{U}{1-U}) $$
Now lets do calculation step by step to verify that $ F^{-1}(U) = F(x) $
$$ F^{-1}(U) = log(\frac{U}{1-U}) $$ $$ P(log(\frac{U}{1-U}) \le x) = P(\frac{U}{1-U} \le e^x) $$ $$= P(U \le e^x(1-U))$$ $$= P(U \le \frac{e^x}{1+e^x})$$ $$= \frac{e^x}{1+e^x}$$
I understand most of this except in the line that says $P(U \le e^x(1-U)) = P(U \le \frac{e^x}{1+e^x}) $ How can the $e^x(1-U) = \frac{e^x}{1+e^x}$ can someone please explain to me step by step with easy detail explanation. I'm newbie to Probability.
Note that the following statements are equivalent
$$U \leq exp(x) (1-U)$$
$$U \leq \exp(x) - \exp(x)U$$
$$(1+\exp(x))U \leq \exp(x)$$
$$U \leq \frac{\exp(x)}{1+\exp(x)}$$