Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

40 Probability and Distributions

F(x)

x

1 2 3 4 5 6

1.0

0.5

(0, 0)

Figure 1.5.1:Distribution function for Example 1.5.3.

The following example discusses the cdf for the continuous random variable
discussed in Example 1.5.2.

Example 1.5.4 (Continuation of Example 1.5.2). Recall thatXdenotes a real
number chosen at random between 0 and 1. We now obtain the cdf ofX.First,if
x<0, thenP(X≤x)=0. Next,ifx≥1, thenP(X≤x) = 1. Finally, if 0<x<
1, it follows from expression (1.5.3) thatP(X≤x)=P(0<X≤x)=x−0=x.
Hence the cdf ofXis

FX(x)=

⎧

⎨

⎩

0ifx< 0

x if 0≤x< 1

1ifx≥ 1.

(1.5.7)

AsketchofthecdfofXis given in Figure 1.5.2. Note, however, the connection
betweenFX(x) and the pdf for this experimentfX(x), given in Example 1.5.2, is

FX(x)=

∫x

−∞

fX(t)dt, for allx∈R,

anddxdFX(x)=fX(x), for allx∈R, except forx=0andx=1.

LetXandY be two random variables. We say thatXandY areequal in

distributionand writeXD=Y if and only ifFX(x)=FY(x), for allx∈R.It
is important to note whileXandY may be equal in distribution, they may be
quite different. For instance, in the last example define the random variableY as
Y=1−X.ThenY =X. But the space ofY is the interval (0,1), the same asX.
Further, the cdf ofY is 0 fory<0; 1 fory≥1; and for 0≤y<1, it is

FY(y)=P(Y≤y)=P(1−X≤y)=P(X≥ 1 −y)=1−(1−y)=y.

Hence,Yhas the same cdf asX, i.e.,Y
D
=X, butY =X.