Introduction to Probability and Statistics for Engineers and Scientists

92 Chapter 4:Random Variables and Expectation

SOLUTION The desired probability is computed as follows:

P{X> 1 }= 1 −P{X≤ 1 }

= 1 −F(1)

=e−^1

=.368 ■

4.2Types of Random Variables

As was previously mentioned, a random variable whose set of possible values is a sequence
is said to bediscrete. For a discrete random variableX, we define theprobability mass
function p(a)ofXby

p(a)=P{X=a}

The probability mass functionp(a) is positive for at most a countable number of values
ofa. That is, ifXmust assume one of the valuesx 1 ,x 2 ,..., then

p(xi)>0, i=1, 2,...

p(x)=0, all other values ofx

SinceXmust take on one of the valuesxi, we have

∑∞

i= 1

p(xi)= 1

EXAMPLE 4.2a Consider a random variableXthat is equal to 1, 2, or 3. If we know that

p(1)=^12 and p(2)=^13

then it follows (sincep(1)+p(2)+p(3)=1) that

p(3)=^16

A graph ofp(x) is presented in Figure 4.1. ■

The cumulative distribution functionFcan be expressed in terms ofp(x)by

F(a)=

∑

allx≤a

p(x)

IfXis a discrete random variable whose set of possible values arex 1 ,x 2 ,x 3 ,..., where
x 1 <x 2 <x 3 <···, then its distribution functionFis a step function. That is, the value
ofFis constant in the intervals[xi− 1 ,xi) and then takes a step (or jump) of sizep(xi)atxi.