5.2The Poisson Random Variable 153
The Poisson approximation result can be shown to be valid under even more general
conditions than those so far mentioned. For instance, suppose thatnindependent trials
are to be performed, with theith trial resulting in a success with probabilitypi,i=1,...,n.
Then it can be shown that ifnis large and eachpiis small, then the number of successful
trials is approximately Poisson distributed with mean equal to
∑n
i= 1 pi. In fact, this
result will sometimes remain true even when the trials are not independent, provided that
their dependence is “weak.” For instance, consider the following example.
EXAMPLE 5.2e At a partynpeople put their hats in the center of a room, where the
hats are mixed together. Each person then randomly chooses a hat. IfXdenotes the
number of people who select their own hat then, for largen, it can be shown thatXhas
approximately a Poisson distribution with mean 1. To see why this might be true, let
Xi={
1 if theith person selects his or her own hat
0 otherwiseThen we can expressXas
X=X 1 +···+Xn
and soXcan be regarded as representing the number of “successes” inn“trials” where trial
iis said to be a success if theith person chooses his own hat. Now, since theith person is
equally likely to end up with any of thenhats, one of which is his own, it follows that
P{Xi= 1 }=1
n(5.2.2)Suppose now thati =jand consider the conditional probability that theith person
chooses his own hat given that thejth person does — that is, considerP{Xi= 1 |Xj= 1 }.
Now given that thejth person indeed selects his own hat, it follows that theith individual
is equally likely to end up with any of the remainingn−1, one of which is his own. Hence,
it follows that
P{Xi= 1 |Xj= 1 }=1
n− 1(5.2.3)Thus, we see from Equations 5.2.2 and 5.2.3 that whereas the trials are not independent,
their dependence is rather weak [since, if the above conditional probability were equal to
1/nrather than 1/(n−1), then trialsiandjwould be independent]; and thus it is not
at all surprising thatXhas approximately a Poisson distribution. The fact thatE[X]= 1
follows since
E[X]=E[X 1 +···+Xn]
=E[X 1 ]+···+E[Xn]=n(
1
n)
= 1