Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

1.7. Continuous Random Variables 53

which is confirmed immediately by showing thatF′(x)=f(x). For the inverse of
the cdf, setu=F(x)andsolveforu. A few steps of algebra lead to

F−^1 (u)=log

{

. 2

[

(1−u)−^5 − 1

]}

, 0 <u< 1.

Thus,ξp=FX−^1 (p)=log

{

. 2

[

(1−p)−^5 − 1

]}
.The following three R functions can
be used to compute the pdf, cdf, and inverse cdf ofF, respectively. These can be
downloaded at the site listed in the Preface.

dlogF <- function(x){exp(x)/(1+5*exp(x))^(1.2)}

plogF <- function(x){1- (1+5*exp(x))^(-.2)}

qlogF <- function(x){log(.2*((1-x)^(-5) - 1))}

Once the R functionqlogFis sourced, it can be used to compute quantiles. The
following is an R script which results in the computation of the three quartiles of
X:

qlogF(.25) ; qlogF(.50); qlogF(.75)

-0.4419242; 1.824549; 5.321057

Figure 1.7.2 displays a plot of this pdf and its quartiles. Notice that this is another
example of a skewed-right distribution; i.e., the right-tail is much longer than left-
tail. In terms of the log-lifetime of mechanical parts having this distribution, it
follows that 50% of the parts survive beyond 1.83 log-units and 25% of the parts
live longer than 5.32 log-units. With the long-right tail, some parts attain a long
life.

1.7.2 Transformations

LetXbe a continuous random variable with a known pdffX. As in the discrete
case, we are often interested in the distribution of a random variableY which is
sometransformationofX,say,Y=g(X). Often we can obtain the pdf ofYby
first obtaining its cdf. We illustrate this with two examples.

Example 1.7.4.LetXbe the random variable in Example 1.7.1. Recall thatX
was the distance from the origin to the random point selected in the unit circle.
Suppose instead that we are interested in the square of the distance; that is, let
Y=X^2. The support ofYis the same as that ofX,namely,SY=(0,1). What is
the cdf ofY? By expression (1.7.5), the cdf ofXis

FX(x)=

⎧

⎨

⎩

0 x< 0

x^20 ≤x< 1

11 ≤x.

(1.7.10)

Letybe in the support ofY; i.e., 0<y<1. Then, using expression (1.7.10) and
the fact that the support ofXcontains only positive numbers, the cdf ofYis

FY(y)=P(Y≤y)=P(X^2 ≤y)=P(X≤

√

y)=FX(

√

y)=

√

y^2 =y.