# Solved – 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.

Contents

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:

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

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")   curve(exp(-x)/(1+exp(-x))^2, add=T, lwd=2, col="brown")  hist(y, prob=T, ylim=c(0,1), col="skyblue2", main="Exponential")   curve(dexp(x,1), add=T, lwd=2, n=10001, col="brown") par(mfrow=c(1,1)) ``

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