Download Math 215 Lecture notes for 10/29/98: Poisson Distribution 1

Math 215 Lecture notes for 10/29/98:
Poisson Distribution
The Poisson distribution is used for circumstances like the following:
• The number of airline crashes next year
• The number of people arriving at a drive-through window in the next hour
• The number of lightbulbs removed from inventory today
• The number of cars passing under a bridge in the next minute
• The number of people arriving at your store in the next hour
• The number of earthquakes of magnitude 4 or larger in Malibu in the next
year
• The number of typographical errors in our textbook on a given page
They are all in terms of “number” of a certain occurence in a certain interval
of time (or space). They are all described by a random variable, in that there
are many possible answers, and we don’t know which one it is.
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Compare with binomial
It is helpful to compare and contrast this distribution to the one you have
already learned: the binomial distribution.
Like the binomial distribution, these numbers are whole numbers, and such
a random variable is a discrete random variable. (The number of airline crashes
in the next year is a whole number, not a fraction.)
Unlike the binomial distribution, it is an infinite random variable. That is
because (for instance) there is no upper limit to the number of cars that will
pass under a bridge in the next minute.
In both the binomial and the Poisson distribution, you ask how many times
a certain occurence happens.
In the binomial distribution, you (for instance) flip a coin a certain number
of times and ask how many times heads appeared. Or more generally, there are
a certain number of opportunities for the occurence to happen.
In the Poisson distribution, a plane crash (for instance) may happen at any
time. There is no such thing as a special “opportunity” when a plane might or
might not crash. As opposed to the binomial distribution, when you flip a coin,
and there is a special “opportunity” when a coin may come out heads or not:
the moment you flip the coin.
In the binomial distribution, we had two parameters: n and p that determined what kind of a problem we had. Now n was the number of “flips of
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the coin” or “special opportunities”. Such a parameter does not exist with the
Poisson distribution, since there are no “special opportunities.”
Instead of the probability p, we have a different parameter that describes on
average, how many events we should expect in that interval. We traditionally
use the Greek letter lambda: λ, which you already encountered in Lagrange multipliers (this subject, by the way, has nothing to do with Lagrange multipliers.
We just ran out Greek letters).
In the binomial distribution, we assumed each “flip” was independent of the
others. In the Poisson distribution, we assumed that an “occurence” happening
at one time interval is independent of another occurence happening at a different
time interval.
This characterizes the examples given at the beginning of the notes.
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The distribution formula
Just as with the binomial distribution, we have a formula for the probability of
each outcome.
Suppose we have a Poisson process like one described above: the number of
a certain occurence happening in a given time interval. Suppose X denotes the
number occurring in that interval, and λ denotes the average number (if you
like, E(X)).
Then the probability that X is equal to some whole number k is
P r(X = k) =
λk −λ
e .
k!
Remember that e is our favorite number about 2.718282, and ex is our favorite
function whose derivative is itself.
Example 1 Cars enter the 101S onramp at Las Virgines at the average rate of
20 per minute. What is the probability that in the next minute, we will have 10
cars entering the 101S onramp at Las Virgines?
Solution 1 Let X be the number of cars entering the 101S onramp at Las
Virgines. It is true that X is a Poisson random variable. The parameter λ
is equal to the average number that enter in a minute, which is 20. So the
probability that X = 10 is
P r(X = 10) =
2010 −20 ∼
e
= .0058
10!
Example 2 You manage a manufacturing plant, and orders for your product
come in at random times, where the number in at time is independent of the
number at another time. On average, you get 3 orders in a day. If you get
more than 5 orders in a day, you experience a backlog. What is the probability
of experiencing a backlog today?
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Solution 2 We are actually asking, “what is the probability that the number of
orders is more than 5?”
Let X be the number of orders today. It is a Poisson random variable, with
parameter λ equal to 3 (the average number of orders in a day). We would
like to find the probability of X > 5, which (in principle) involves adding up
the probability for X = 6 to the probability for X = 7, and adding this to the
probabililty for X = 8, and so on, ad infinitum.
I don’t want to add an infinite number of things together! But there is a good
trick to remember. I can instead find the probability of X ≤ 5, which involves
only adding 6 terms (see below). It is the probability of getting 0 orders, plus
the probability of getting one, and so on up to the probability of getting 5 orders.
P r(X ≤ 5)
= P r(X = 0) + P r(X = 1) + P r(X = 2) + P r(X = 3) + P r(X = 4) + P r(X = 5)
30 −3 31 −3 32 −3 33 −3 34 −3 35 −3
e + e + e + e + e + e
=
0!
1!
2!
3!
4!
5!
9
9
27
81
= e−3 + 3e−3 + e−3 + e−3 + e−3 + e−3
2
2
8
40
∼
= .6671
This is the probability of not getting a backlog. Thus, the probability of getting
a backlog is one minus this probability, or
P r(X > 5) = 1 − .6671 = .3329
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Mean and variance
As already mentioned, if X is a Poisson random variable with parameter λ, then
E(X)
V ar(X)
= λ
= λ
√
λ
SD(X) =
4
Changing time intervals
Suppose we had the following problem:
Example 3 On average, a certain store has 0.2 shoplifting incidents per day.
What is the probability that there are two shoplifting incidents in a 4-day period?
There are two comments worth making at the start. First, the average number
of shoplifting incidents is not a whole number. Well, we never said it had to be
a whole number. In fact, it’s not a number of incidents; it’s an average number
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of incidents. An average of 0.2 is possible, if we get one shoplifting incident per
day.
The second comment is that there are two time intervals to worry about
here. The average is given in terms of how many shoplifting incidents there are
in a single day. The question is phrased, though, in terms of how many there
are in a 4-day interval. Those time intervals are of different length.
Solution 3 If there are 0.2 incidents in one day on average, then surely there
are (0.2)(4) incidents in four days, on average. Thus λ = (0.2)(4) = 0.8.
If X is the number of shoplifting incidents in the four-day interval, then
P r(X = 2) =
5
0.82 −0.8 ∼
e
= .144
2!
Approximately Poisson
Recall that the difference between Poisson and Binomial random variables was
that in the Poisson random variable, the occurrence could happen at any time,
while in the binomial random variable, there are special times when the occurence could happen. In some cases, the distinction is minimal:
Example 4 On average, a typist produces 0.1 errors per page. What is the
probability of the typist making no errors on a given page?
Here we are counting how many errors occur. An occurence of an error is
cannot happen at an arbitrary time; it can only happen each time the typist
types a letter!
Nevertheless, there are so many letters on a page that we can pretend an error
can happen at any time. This approximation works well since the probability of
an error on each letter is so small (on average, there are only 0.1 errors per page;
imagine how few errors there are per letter!) in such a way that the average
number to occur in the given interval is of moderate size (like 0.1).
So the binomial and the Poisson random variables are approximately similar
when the number of occurences is large and the probability is small.
Solution 4 We imagine this is a Poisson random variable, with λ = 0.1; if X
is the number of errors on a given page,
P r(X = 0) =
0.10 −0.1
e
= e−0.1 ∼
= 0.9048
0!
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