Introduction to Probability and Statistics for Engineers and Scientists

180 Chapter 5: Special Random Variables

0 t

n

2 t

n

3 t

n t =

nt

n

t

(n – 1)n

FIGURE 5.10

and (e) state that in a small interval of lengthh, the probability of one event occurring is
approximatelyλh, whereas the probability of 2 or more is approximately 0.
We will now show that these assumptions imply that the number of events occurring
in any interval of lengthtis a Poisson random variable with parameterλt. To be precise,
let us call the interval [0,t] and denote byN(t) the number of events occurring in that
interval. To obtain an expression forP{N(t)=k}, we start by breaking the interval [0,t]
intonnonoverlapping subintervals each of lengtht/n(Figure 5.10). Now there will bek
events in [0,t] if either

(i) N(t) equalskand there is at most one event in each subinterval;

(ii) N(t) equalskand at least one of the subintervals contains 2 or more events.

Since these two possibilities are clearly mutually exclusive, and since Condition (i) is
equivalent to the statement thatkof thensubintervals contain exactly 1 event and the
othern−kcontain 0 events, we have that

P{N(t)=k}=P{kof thensubintervals contain exactly 1 event (5.6.3)

and the othern−kcontain 0 events}+P{N(t)=k

and at least 1 subinterval contains 2 or more events}

Now it can be shown, using Condition (e), that

P{N(t)=kand at least 1 subinterval contains 2 or more events}

−→0asn→∞ (5.6.4)

Also, it follows from Conditions (d) and (e) that

P{exactly 1 event in a subinterval}≈

λt

n

P{0 events in a subinterval}≈ 1 −

λt

n

Hence, since the numbers of events that occur in different subintervals are independent
[from Condition (b)], it follows that

P{kof the subintervals contain exactly 1 event and the othern−kcontain 0 events}

≈

(n

k

)(λt

n

)k(

1 −

λt

n

)n−k

(5.6.5)