Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

42 Probability and Distributions

Proof:Note that

{−∞<X≤b}={−∞<X≤a}∪{a<X≤b}.

The proof of the result follows immediately because the union on the right side of

this equation is a disjoint union.

Example 1.5.5.LetXbe the lifetime in years of a mechanical part. Assume that
Xhas the cdf

FX(x)=

{

0 x< 0

1 −e−x 0 ≤x.

The pdf ofX,dxdFX(x), is

fX(x)=

{

e−x 0 <x<∞

0elsewhere.

Actually the derivative does not exist atx= 0, but in the continuous case the next
theorem (1.5.3) shows thatP(X= 0) = 0 and we can assignfX(0) = 0 without
changing the probabilities concerningX. The probability that a part has a lifetime
between one and three years is given by

P(1<X≤3) =FX(3)−FX(1) =

∫ 3

1

e−xdx.

That is, the probability can be found byFX(3)−FX(1) or evaluating the integral.
In either case, it equalse−^1 −e−^3 =0.318.

Theorem 1.5.1 shows that cdfs are right continuous and monotone. Such func-
tions can be shown to have only a countable number of discontinuities. As the next
theorem shows, the discontinuities of a cdf have mass; that is, ifxis a point of
discontinuity ofFX,thenwehaveP(X=x)>0.

Theorem 1.5.3.For any random variable,

P[X=x]=FX(x)−FX(x−), (1.5.8)

for allx∈R,whereFX(x−) = limz↑xFX(z).

Proof:For anyx∈R,wehave

{x}=

⋂∞

n=1

(

x−

1

n

,x

]

;

that is,{x}is the limit of a decreasing sequence of sets. Hence, by Theorem 1.3.6,

P[X=x]=P

[∞

⋂

n=1

{

x−

1

n

<X≤x

}]

= lim

n→∞

P

[

x−

1

n

<X≤x

]

= lim

n→∞

[FX(x)−FX(x−(1/n))]

= FX(x)−FX(x−),