# Derivation of CDF of a function that results in an exponential distribution

I was looking through wiki's treatment on the title topic in https://en.wikipedia.org/wiki/Random_variable and am completely stumped on this particular section: There are several specifics that elude me.

• How is the following progression derived

$$Y = \log(1 + e^{-X}) \Longrightarrow F_Y(y) = \Pr( \log(1+e^{-X}) \le y )$$

and then the next step

$$\Pr( \log(1+e^{-X}) \le y) = \Pr( X \ge -\log(e^y - 1) )$$

Thx for filling in the blanks.

• Hint: By definition, for each real number $\alpha$, the value of tthe CDF $F_Y$ at $\alpha$, usually expressed as $F_Y(\alpha)$, equals the probability that $Y$ is no larger than $\alpha$. So, knowing this, can you figure out for yourself why the implication $$Y = \log(1 + e^{-X}) \Longrightarrow F_Y(y) = \Pr(Y \leq y) = \Pr( \log(1+e^{-X}) \le y )$$ holds?? – Dilip Sarwate Jun 15 at 20:28
• Yes - ok i see that. – javadba Jun 16 at 8:17

Introduction. The so-called "CDF method" is one way to find the distribution of a the transformation $$Y = g(X)$$ of a random variable $$X$$ with a known CDF. Let's look at a simpler example first: Suppose $$X \sim \mathsf{Univ}(0,1)$$ and find the CDF of $$Y = g(X) = \sqrt{X}.$$ The support of $$X$$ is $$(0,1)$$ and it is clear that the support of $$Y$$ will also be $$(0,1).$$

The CDF of $$X$$ is $$F_X(x) = x,$$ for $$x \in (0,1).$$ Then the CDF of $$Y$$ is $$P(Y \le y) = P(g(X) \le y) = P(\sqrt{X} \le y) = P(X \le y^2) = y^2,$$ for $$y \in (0,1).$$ The last step uses $$F_X(x) = P(X \le x) = x,$$ where $$y^2 = x.$$ Thus the PDF of $$Y$$ is $$f_Y(y) = F_Y^\prime(y) = dy^2/dy = 2y,$$ which we recognize as the PDF of $$\mathsf{Beta}(2,1).$$

Illustrating this with a random sample of $$n = 10^5$$ observations $$X_i$$ from $$\mathsf{Unif}(0,1),$$ we have the following results (in R):

set.seed(615)
x = runif(10^5, 0, 1);  y = sqrt(x)
par(mfrow=c(1,2))
hist(x, prob=T, col="skyblue2", main="X ~ UNIF(0,1)")
curve(dunif(x, 0, 1), add=T, n=10001, lwd=2, col="brown")
hist(y, prob=T, col="skyblue2", main="Y ~ BETA(2,1)")
curve(dbeta(x, 2, 1), add=T, n=10001, lwd=2, col="brown")
par(mfrow=c(1,1)) Your Question. Now let's do a similar procedure for $$X$$ with CDF $$F_X(x) = P(X \le x) = (1 + e^{-x})^{-\theta},$$ for $$\theta > 0$$ and the transformation $$Y = g(X) = \log(1+e^{-X}),$$ which has support $$(0, \infty).$$

Using the CDF method again, we have:

$$F_Y(y) = P(Y\le y) = P(\log(1 + e^{-X})\le y) = P(1+e^{-X} \le e^y)\\ =P(e^{-X} \le e^y - 1) = P(-X \le \log(e^y -1))\\ = P(X \ge -\log(e^y -1)) = \cdots,$$

So, $$F_y(y) = 1-e^{-\theta y},$$ for $$y > 0,$$ as claimed.

We illustrate with a random sample of $$n = 10^5$$ observations from the original logistic distribution with $$\theta = 1.$$ This distribution can be sampled in terms of standard uniform distributions as shown in the R code; see Wikipedia, second bullet under Related Distributions.

set.seed(2019)
u = runif(10^5);  x = log(u) - log(1-u)
y = log(1 + exp(-x))
par(mfrow=c(1,2))
hist(x, prob=T, br=30, ylim=c(0,.25),  col="skyblue2", main="Logistic")
hist(y, prob=T, ylim=c(0,1), col="skyblue2", main="Exponential")
par(mfrow=c(1,1)) \begin{align} & \log(1+e^{-X}) \le y \\[12pt] & 1+e^{-X} \le e^y \\ & \text{since } \log \text{ and } \exp \\ & \text{are increasing functions} \\[12pt] & e^{-X} \le e^y - 1 \\[12pt] & {-X} \le \log(e^y - 1) \\ & \text{since } \log \text{ and } \exp \\ & \text{are increasing functions} \\[12pt] & X \ge -\log(e^y - 1) \end{align}

Question 1:

By definition, FY(y) = Pr(Y<=y)

And because we know that Y=log(1+e^−X)

Then FY(y) = Pr(log(1+e^−X) <= y)

Question 2:

All they're doing there is starting with log(1+e^−X)≤y

and simply solving for X

which entails, first doing e^ on both sides:

1+e^−X ≤ e^y

then sending the 1 to the right:

e^−X ≤ e^y - 1

then doing log on both sides:

−X ≤ log(e^y - 1)

and finally multiplying both sides by -1, which changes the direction of the ≤

So you end up with:

X≥−log(e^y−1)

which is what they got