Introduction to Probability and Statistics for Engineers and Scientists

5.6Exponential Random Variables 181

with the approximation becoming exact as the number of subintervals,n, goes to∞.
However, the probability in Equation 5.6.5 is just the probability that a binomial random
variable with parametersnandp=λt/nequalsk. Hence, asnbecomes larger and larger,
this approaches the probability that a Poisson random variable with meannλt/n=λt
equalsk. Hence, from Equations 5.6.3, 5.6.4, and 5.6.5, we see upon lettingnapproach
∞that

P{N(t)=k}=e−λt

(λt)k

k!

We have shown:

PROPOSITION 5.6.2 For a Poisson process having rateλ

P{N(t)=k}=e−λt

(λt)k

k!

, k=0, 1,...

That is, the number of events in any interval of lengththas a Poisson distribution with
meanλt.
For a Poisson process, letX 1 denote the time of the first event. Further, forn>1,
letXndenote the elapsed time between (n−1)st and thenth events. The sequence
{Xn,n=1, 2,...}is called thesequence of interarrival times.For instance, ifX 1 =5 and
X 2 =10, then the first event of the Poisson process would have occurred at time 5 and
the second at time 15.
We now determine the distribution of theXn. To do so, we first note that the event
{X 1 >t}takes place if and only if no events of the Poisson process occur in the interval
[0,t] and thus,

P{X 1 >t}=P{N(t)= 0 }=e−λt

Hence,X 1 has an exponential distribution with mean 1/λ. To obtain the distribution of
X 2 , note that

P{X 2 >t|X 1 =s}=P{0 events in (s,s+t]|X 1 =s}

=P{0 events in (s,s+t]}

=e−λt

where the last two equations followed from independent and stationary increments. There-
fore, from the foregoing we conclude thatX 2 is also an exponential random variable with
mean 1/λ, and furthermore, thatX 2 is independent ofX 1. Repeating the same argument
yields:

PROPOSITION 5.6.3 X 1 ,X 2 ,...are independent exponential random variables each with
mean 1/λ.