Xjyuk

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.

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 $X sim mathsfUniv(0,1)$ and find the CDF of $Y = g(X) = sqrtX.$ 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
$$P(Y le y) = P(g(X) le y) = P(sqrtX le y) = P(X le y^2) = y^2,$$
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 $mathsfBeta(2,1).$

Illustrating this with a random sample of $n = 10^5$ observations $X_i$ from $mathsfUnif(0,1),$ 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))

beginalign
& log(1+e^-X) le y \[12pt]
& 1+e^-X le e^y \
& textsince log text and exp \
& textare increasing functions \[12pt]
& e^-X le e^y - 1 \[12pt]
& -X le log(e^y - 1) \
& textsince log text and exp \
& textare increasing functions \[12pt]
& X ge -log(e^y - 1)
endalign

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