Chapter 3, Section 8

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Chapter 3, Section 8
The Poisson Distributions
? John J Currano, 02/18/2009
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What is the probability distribution of Y
of auto accidents at a particular intersection
during a given time period, say one week? At
first glance Y may not seem even remotely
related to a binomial random variable, but an
interesting relationship exists.
Think of the time period as being split up into
n subintervals, each so small that at most one
accident can occur in it with probability gt 0.
If we assume that the probability of an accident
in each subinterval is the same value, p, we have
(approximately) P(no accidents occur in a
subinterval) 1 p, P(one accident occurs in
a subinterval) p, and P(more than one
accident occurs in a subinterval) 0.
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P(no accidents occur in a subinterval) 1
p, P(one accident occurs in a subinterval) p,
and P(more than one accident occurs in a
subinterval) 0. Y total number of accidents
in a week
For all values of n large enough so that the
above assumptions hold, Y the total number of
subintervals that contain one accident. If the
occurrence of accidents can be regarded as
independent from interval to interval, Y
bin( n, p ). What is p? If ? is the average
number of accidents at the intersection per week,
then ?  E(Y)  np, so   , the average
number of accidents per week divided by the
number of subintervals. Now take the limit of
the binomial probability as n ? ?, to get the
probability function of Y.
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P(no accidents occur in a subinterval) 1
p, P(one accident occurs in a subinterval) p,
and P(more than one accident occurs in a
subinterval) 0. Y total number of accidents
in a week
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P(no accidents occur in a subinterval) 1
p, P(one accident occurs in a subinterval) p,
and P(more than one accident occurs in a
subinterval) 0. Y total number of accidents
in a week
Since and each of the other factors within
the
brackets has a limit of 1, we see that
for y ? 0.
A random variable with support y y ? 0 and
probability function is said to have a
Poisson distribution.
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Definition. Y is said to have the Poisson
distribution with parameter ? if ? gt 0 and
Verification. In order for p(y) to be a
probability function for a distribution, we need
to show
1. p(y) ? 0 for all y, which is clear since ?,
e, and y ! are positive and 2. Thus, p(y) is a
probability function for all ? gt 0.
the Maclaurin series for e?.
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Poisson Approximation to the Binomial. The
Binomial(n, p) distribution can be approximated
by the Poisson(np) distribution when n is large
and p is small. One rule of thumb says that the
Poisson provides a good approximation when n ? 20
and p ? 0.05 and an excellent approximation when
n ? 100 and ?  np lt 10. There is an interesting
applet illustrating this on the web at
http//www.stat.tamu.edu/jhardin/applets/signed/p
oissbin.html. One benefit of this in days past
was the ability to use Poisson tables to
calculate binomial probabilities today it has
mainly theoretical uses. However, the concept is
important, as we shall see later. What this
means is the following IF 1. X binomial (n,
p) 2. n is large and p is small as described
above and 3. Y Poisson(np) THEN P(X x)
? P(Y x), for x 0, 1, 2, . . . , n.
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• The Mean and Variance of the Poisson
  Distributions.
• Theorem. Let Y Poisson(?), so that , for y
  0, 1, 2, . . . . Then
• the mean of Y is E(Y) ?.

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• The Mean and Variance of the Poisson
  Distributions.
• Theorem. Let Y Poisson(?), so that , for y
  0, 1, 2, . . . . Then
• the variance of Y is V(Y) ?.
• Proof.

so that EY 2 EY(Y 1)Y EY(Y 1)
EY ?2 ?, and VY EY 2 E(Y)2
(?2 ?) ?2 ?.
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Example 1. Suppose Y has a Poisson
distribution with mean ? 3. Find a. P(Y
4) b. P(Y ? 4) c. P(Y ? 4 Y ? 2).
Solution. (a)
, so
We could also use Table 3 on pages 843-847
P(Y 4) P(Y ? 4) P(Y lt 4) P(Y ? 4) P(Y ?
3)
0.815 0.647 0.168. Finally. we can use
the distributions on the TI-83
0.1680313557 (b) P(Y ? 4) ?
next
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Example 1. Suppose Y has a Poisson
distribution with mean ? 3. Find a. P(Y
4) b. P(Y ? 4) c. P(Y ? 4 Y ? 2).
Solution. (b) P(Y ? 4) 1 P(Y lt 4) 1
P(Y ? 3) Using Table 3 1 0.647
0.353 Using TI-83 0.3527681107
(c) P(Y ? 4 Y ? 2)
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A Poisson random variable Y has the following
interpretation Y is the of occurrences of a
rare event that occurs in space, time, volume, or
any other dimension, in which individual
occurrences of the event are independent of each
other. In insurance, the Poisson distribution
arises naturally as the numberof claims on a
large group of policies for which the expected
number of claims is known, and claims occur
independently and infrequently. While the
binomial distribution can also be used to model
claim frequency (each policyholders claim status
being a trial), it cannot be applied in group
insurance applications where the membership of a
group frequently changes throughout the period of
consideration. The Poisson distribution can be
used provided the expected number of claims
remains fairly stable. The Poisson distribution
is also useful for modeling claim frequency on
individual policies, such as auto insurance,
where multiple claims are possible, but rare.
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Example 2. Certain sheet metal has, on the
average, five defects per 10 square feet.
Assuming a Poisson distribution, find the
probability that a 15-square foot sheet of metal
will have at least six defects. Solution. Let Y
of defects in a 15-sq. ft. sheet. We are told
that Y has a Poisson distribution. Let ? be the
mean of Y. First find ? ? (5 defects / 10
sq.ft.) ? 15 sq.ft. 5/10 ? 15 7.5. P(Y
? 6) 1 P(Y lt 6) 1 P(Y ? 5)
1 F(5) 1 0.2414
0.7586, using the TI-83 to find the cdf, F(5),
of the Poisson with ? 7.5.
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Example 2. Certain sheet metal has, on the
average, five defects per 10 square feet.
Assuming a Poisson distribution, find the
probability that a 15-square foot sheet of metal
will have at least six defects. Solution. Let Y
of defects in a 15-sq. ft. sheet. We have seen
that Y has a Poisson distribution with ? 7.5,
and we need to find P(Y ? 6) 1 P(Y lt 6)
1 P(Y ? 5) Instead of using the TI-83, we
can interpolate in Table 3 on page 845, which has
entries for ? 7.4 and ? 7.6 but not ? 7.5
since 7.5 is halfway between 7.4 and 7.6, average
the two values in the table to get
P(Y ? 5) ? (0.253  0.231)/2  0.242. Alternativel
y, we could always do it as follows P(Y ? 6)
1 P(Y lt 6) 1 P(Y ? 5) 1 p(0)
p(1) p(2) p(3) p(4) p(5) .
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