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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.

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1 Answer 1

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Note that the following statements are equivalent

$$U \leq exp(x) (1-U)$$

$$U \leq \exp(x) - \exp(x)U$$

$$(1+\exp(x))U \leq \exp(x)$$

$$U \leq \frac{\exp(x)}{1+\exp(x)}$$

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  • $\begingroup$ thank you very much ^^ Mr. Siong Thye Goh $\endgroup$ Commented Jan 22, 2017 at 16:51
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    $\begingroup$ you are welcome $\endgroup$ Commented Jan 22, 2017 at 16:53

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