7/27/2019 Poisson 215

1/4

7/27/2019 Poisson 215

2/4

the coin or special opportunities. Such a parameter does not exist with thePoisson distribution, since there are no special opportunities.

Instead of the probability p, we have a different parameter that describes onaverage, how many events we should expect in that interval. We traditionallyuse the Greek letter lambda: , which you already encountered in Lagrange mul-tipliers (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 ip was independent of theothers. In the Poisson distribution, we assumed that an occurence happeningat one time interval is independent of another occurence happening at a differenttime interval.

This characterizes the examples given at the beginning of the notes.

2 The distribution formulaJust 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 thenumber occurring in that interval, and denotes the average number (if youlike, E (X )).

Then the probability that X is equal to some whole number k is

P r (X = k) = k

k !e .

Remember that e is our favorite number about 2 .718282, and ex is our favoritefunction 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 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

10!e 20 = .0058

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?

2

7/27/2019 Poisson 215

3/4

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 equal to 3 (the average number of orders in a day). We would like to nd the probability of X > 5, which (in principle) involves adding upthe probability for X = 6 to the probability for X = 7 , and adding this to the probabililty for X = 8 , and so on, ad innitum.

I dont want to add an innite number of things together! But there is a good trick to remember. I can instead nd 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

0!e 3 +

31

1!e 3 +

32

2!e 3 +

33

3!e 3 +

34

4!e 3 +

35

5!e 3

= e 3 + 3 e 3 +92

e 3 +92

e 3 +278

e 3 +8140

e 3

= .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

3 Mean and variance

As already mentioned, if X is a Poisson random variable with parameter , then

E (X ) = V ar (X ) =

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 numberof shoplifting incidents is not a whole number. Well, we never said it had to bea whole number. In fact, its not a number of incidents; its an average number

3

7/27/2019 Poisson 215

4/4

of incidents. An average of 0.2 is possible, if we get one shoplifting incident perday.

The second comment is that there are two time intervals to worry abouthere. The average is given in terms of how many shoplifting incidents there arein a single day. The question is phrased, though, in terms of how many thereare 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 = (0 .2)(4) = 0 .8.

If X is the number of shoplifting incidents in the four-day interval, then

P r (X = 2) =0.82

2!e 0 . 8 = .144

5 Approximately PoissonRecall that the difference between Poisson and Binomial random variables wasthat in the Poisson random variable, the occurrence could happen at any time,while in the binomial random variable, there are special times when the oc-curence 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 iscannot happen at an arbitrary time; it can only happen each time the typisttypes a letter!

Nevertheless, there are so many letters on a page that we can pretend an errorcan 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 averagenumber to occur in the given interval is of moderate size (like 0.1).

So the binomial and the Poisson random variables are approximately similarwhen the number of occurences is large and the probability is small.

Solution 4 We imagine this is a Poisson random variable, with = 0 .1; if X is the number of errors on a given page,

P r (X = 0) =0.10

0!e 0 . 1 = e 0 . 1 = 0.9048

4