The Poisson Distribution - USM · The Poisson Distribution. logo1...

logo1

Motivation Derivation Definition Examples Expected Value/Variance The Poisson Process

The Poisson Distribution

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science


logo1


Introduction

Some counting processes occur over time or they rely onanother continuous parameter.

1. The number of times a given web page is hit in a certainperiod of time is a random variable.

2. The number of calls to customer support of a company in acertain period of time is a random variable.

3. The number of radioactive particles registered in a Geigercounter over a certain period of time is a random variable.

But what’s the sample space and how do we assignprobabilities? We’ll actually sidestep the sample space issue(it’s complicated) and we’ll focus on the probabilities. Forvisualization, we assume the process occurs over time.



logo1


IntroductionSome counting processes occur over time or they rely onanother continuous parameter.







logo1






But what’s the sample space and how do we assignprobabilities?

We’ll actually sidestep the sample space issue(it’s complicated) and we’ll focus on the probabilities. Forvisualization, we assume the process occurs over time.



logo1






But what’s the sample space and how do we assignprobabilities? We’ll actually sidestep the sample space issue

(it’s complicated) and we’ll focus on the probabilities. Forvisualization, we assume the process occurs over time.



logo1






But what’s the sample space and how do we assignprobabilities? We’ll actually sidestep the sample space issue(it’s complicated)

and we’ll focus on the probabilities. Forvisualization, we assume the process occurs over time.



logo1






But what’s the sample space and how do we assignprobabilities? We’ll actually sidestep the sample space issue(it’s complicated) and we’ll focus on the probabilities.

Forvisualization, we assume the process occurs over time.



logo1









logo1


Take the time interval and cut it up into n subintervals.

Assumethat an event is equally likely to occur in any of the subintervals.

-s s s s s sx0 = 0 xn = Tx1 x2 x3 x4 x5 x6 · · · xn−1

That means there is a λ > 0 so that the probability that an event

occurs in a given time interval is p =λ

n. We keep this λ

independent of the number of subintervals. That way, if wedouble the number of intervals, for each interval the probabilityof an event occurring is divided by 2.

These assumptions are realistic for hits on web pages, forcustomer support calls received, for radioactive particlesregistered.



logo1


Take the time interval and cut it up into n subintervals. Assumethat an event is equally likely to occur in any of the subintervals.




n. We keep this λ





logo1



-

s s s s s sx0 = 0 xn = Tx1 x2 x3 x4 x5 x6 · · · xn−1



n. We keep this λ





logo1



-s

s s s s sx0 = 0 xn = Tx1 x2 x3 x4 x5 x6 · · · xn−1



n. We keep this λ





logo1



-s s

s s s sx0 = 0 xn = Tx1 x2 x3 x4 x5 x6 · · · xn−1



n. We keep this λ





logo1



-s s s

s s sx0 = 0 xn = Tx1 x2 x3 x4 x5 x6 · · · xn−1



n. We keep this λ





logo1



-s s s s

s sx0 = 0 xn = Tx1 x2 x3 x4 x5 x6 · · · xn−1



n. We keep this λ





logo1



-s s s s s

sx0 = 0 xn = Tx1 x2 x3 x4 x5 x6 · · · xn−1



n. We keep this λ





logo1



-s s s s s s

x0 = 0 xn = Tx1 x2 x3 x4 x5 x6 · · · xn−1



n. We keep this λ





logo1



-s s s s s sx0 = 0

xn = Tx1 x2 x3 x4 x5 x6 · · · xn−1



n. We keep this λ





logo1



-s s s s s sx0 = 0 xn = T

x1 x2 x3 x4 x5 x6 · · · xn−1



n. We keep this λ





logo1



-s s s s s sx0 = 0 xn = Tx1

x2 x3 x4 x5 x6 · · · xn−1



n. We keep this λ





logo1



-s s s s s sx0 = 0 xn = Tx1 x2

x3 x4 x5 x6 · · · xn−1



n. We keep this λ





logo1



-s s s s s sx0 = 0 xn = Tx1 x2 x3

x4 x5 x6 · · · xn−1



n. We keep this λ





logo1



-s s s s s sx0 = 0 xn = Tx1 x2 x3 x4

x5 x6 · · · xn−1



n. We keep this λ





logo1



-s s s s s sx0 = 0 xn = Tx1 x2 x3 x4 x5

x6 · · · xn−1



n. We keep this λ





logo1



-s s s s s sx0 = 0 xn = Tx1 x2 x3 x4 x5 x6

· · · xn−1



n. We keep this λ





logo1



-s s s s s sx0 = 0 xn = Tx1 x2 x3 x4 x5 x6 · · ·

xn−1



n. We keep this λ





logo1






n. We keep this λ





logo1






n.

We keep this λ





logo1






n. We keep this λ

independent of the number of subintervals.

That way, if wedouble the number of intervals, for each interval the probabilityof an event occurring is divided by 2.




logo1






n. We keep this λ





logo1






n. We keep this λ


These assumptions are realistic

for hits on web pages, forcustomer support calls received, for radioactive particlesregistered.



logo1






n. We keep this λ


These assumptions are realistic for hits on web pages

, forcustomer support calls received, for radioactive particlesregistered.



logo1






n. We keep this λ


These assumptions are realistic for hits on web pages, forcustomer support calls received

, for radioactive particlesregistered.



logo1






n. We keep this λ





logo1


We will also assume that no two events can happen in the sameinterval.

For a sufficiently small time scale (large n) this is realistic in theabove examples.

-s s s s ssx0 = 0 x1 x2 x3 x4 x5 x6 · · · xn−1 xn = T

Plus, we could argue that within time intervals of a certainlength only one event can be registered.



logo1




-s s s s ss

x0 = 0 x1 x2 x3 x4 x5 x6 · · · xn−1 xn = T




logo1




-s s s s ssx0 = 0

x1 x2 x3 x4 x5 x6 · · · xn−1 xn = T




logo1




-s s s s ssx0 = 0 x1

x2 x3 x4 x5 x6 · · · xn−1 xn = T




logo1




-s s s s ssx0 = 0 x1 x2

x3 x4 x5 x6 · · · xn−1 xn = T




logo1




-s s s s ssx0 = 0 x1 x2 x3

x4 x5 x6 · · · xn−1 xn = T




logo1




-s s s s ssx0 = 0 x1 x2 x3 x4

x5 x6 · · · xn−1 xn = T




logo1




-s s s s ssx0 = 0 x1 x2 x3 x4 x5

x6 · · · xn−1 xn = T




logo1




-s s s s ssx0 = 0 x1 x2 x3 x4 x5 x6

· · · xn−1 xn = T




logo1




-s s s s ssx0 = 0 x1 x2 x3 x4 x5 x6 · · ·

xn−1 xn = T




logo1




-s s s s ssx0 = 0 x1 x2 x3 x4 x5 x6 · · · xn−1

xn = T




logo1




-s s s s ssx0 = 0 x1 x2 x3 x4 x5 x6 · · · xn−1 xn = T




logo1


Under the assumptions that

the interval we investigate has nsubintervals so that the probability an event occurs in a

subinterval is p =λ

nand that at most one event occurs in each

subinterval, the probability that x events occur overall is givenby the binomial distribution.

b(x;n,p) =(

nx

)px(1−p)n−x =

n!(n− x)!x!

λ x

nx

(1− λ

n

)n−x

=n!

(n− x)!nxλ x

x!

((1− λ

n

)n) n−xn

n→∞−→ 1 · λx

x!· e−λ

(We let n→ ∞ because cutting up the original time interval wasjust a way to get the model.)



logo1


Under the assumptions that the interval we investigate has nsubintervals

so that the probability an event occurs in a




b(x;n,p) =(

nx

)px(1−p)n−x =

n!(n− x)!x!

λ x

nx

(1− λ

n

)n−x

=n!

(n− x)!nxλ x

x!

((1− λ

n

)n) n−xn

n→∞−→ 1 · λx

x!· e−λ




logo1


Under the assumptions that the interval we investigate has nsubintervals so that the probability an event occurs in a


n

and that at most one event occurs in eachsubinterval, the probability that x events occur overall is givenby the binomial distribution.

b(x;n,p) =(

nx

)px(1−p)n−x =

n!(n− x)!x!

λ x

nx

(1− λ

n

)n−x

=n!

(n− x)!nxλ x

x!

((1− λ

n

)n) n−xn

n→∞−→ 1 · λx

x!· e−λ




logo1





subinterval,

the probability that x events occur overall is givenby the binomial distribution.

b(x;n,p) =(

nx

)px(1−p)n−x =

n!(n− x)!x!

λ x

nx

(1− λ

n

)n−x

=n!

(n− x)!nxλ x

x!

((1− λ

n

)n) n−xn

n→∞−→ 1 · λx

x!· e−λ




logo1






b(x;n,p) =(

nx

)px(1−p)n−x =

n!(n− x)!x!

λ x

nx

(1− λ

n

)n−x

=n!

(n− x)!nxλ x

x!

((1− λ

n

)n) n−xn

n→∞−→ 1 · λx

x!· e−λ




logo1






b(x;n,p)

=(

nx

)px(1−p)n−x =

n!(n− x)!x!

λ x

nx

(1− λ

n

)n−x

=n!

(n− x)!nxλ x

x!

((1− λ

n

)n) n−xn

n→∞−→ 1 · λx

x!· e−λ




logo1






b(x;n,p) =(

nx

)px(1−p)n−x

=n!

(n− x)!x!λ x

nx

(1− λ

n

)n−x

=n!

(n− x)!nxλ x

x!

((1− λ

n

)n) n−xn

n→∞−→ 1 · λx

x!· e−λ




logo1






b(x;n,p) =(

nx

)px(1−p)n−x =

n!(n− x)!x!

λ x

nx

(1− λ

n

)n−x

=n!

(n− x)!nxλ x

x!

((1− λ

n

)n) n−xn

n→∞−→ 1 · λx

x!· e−λ




logo1






b(x;n,p) =(

nx

)px(1−p)n−x =

n!(n− x)!x!

λ x

nx

(1− λ

n

)n−x

=n!

(n− x)!nxλ x

x!

((1− λ

n

)n) n−xn

n→∞−→ 1

· λx

x!· e−λ




logo1






b(x;n,p) =(

nx

)px(1−p)n−x =

n!(n− x)!x!

λ x

nx

(1− λ

n

)n−x

=n!

(n− x)!nxλ x

x!

((1− λ

n

)n) n−xn

n→∞−→ 1 · λx

x!

· e−λ




logo1






b(x;n,p) =(

nx

)px(1−p)n−x =

n!(n− x)!x!

λ x

nx

(1− λ

n

)n−x

=n!

(n− x)!nxλ x

x!

((1− λ

n

)n) n−xn

n→∞−→ 1 · λx

x!· e−λ




logo1


Definition.

A random variable is said to have a Poissondistribution with parameter λ > 0 if and only if its probability

mass function is p(x;λ ) =e−λ λ x

x!

This really is a probability mass function, because of the series

representation of the exponential function: eλ =∞

∑x=0

λ x

x!.



logo1


Definition. A random variable is said to have a Poissondistribution with parameter λ > 0

if and only if its probability


x!



∑x=0

λ x

x!.



logo1


Definition. A random variable is said to have a Poissondistribution with parameter λ > 0 if and only if its probability


x!



∑x=0

λ x

x!.



logo1




x!


representation of the exponential function

: eλ =∞

∑x=0

λ x

x!.



logo1




x!



∑x=0

λ x

x!.



logo1


Example.

The number of calls a certain customer supportdepartment receives in any given 5 minute interval is Poissondistributed with λ = 2. What is the probability that at most 3calls are received in the next 5 minutes?

p(0;2)+p(1;2)+p(2;2)+p(3;2) = 0.857

(Used a Poisson distribution table to look up p(x≤ 3;2), whichis just a value of the cumulative distribution function.)



logo1


Example. The number of calls a certain customer supportdepartment receives in any given 5 minute interval is Poissondistributed with λ = 2.

What is the probability that at most 3calls are received in the next 5 minutes?

p(0;2)+p(1;2)+p(2;2)+p(3;2) = 0.857




logo1


Example. The number of calls a certain customer supportdepartment receives in any given 5 minute interval is Poissondistributed with λ = 2. What is the probability that at most 3calls are received in the next 5 minutes?

p(0;2)+p(1;2)+p(2;2)+p(3;2) = 0.857




logo1



p(0;2)

+p(1;2)+p(2;2)+p(3;2) = 0.857




logo1



p(0;2)+p(1;2)

+p(2;2)+p(3;2) = 0.857




logo1



p(0;2)+p(1;2)+p(2;2)

+p(3;2) = 0.857




logo1



p(0;2)+p(1;2)+p(2;2)+p(3;2)

= 0.857




logo1



p(0;2)+p(1;2)+p(2;2)+p(3;2) = 0.857




logo1


For large values of n, the Poisson distribution can be used toapproximate binomial probabilities.

Example. Suppose the probability of a missing page in anindividual book a certain manufacturer makes is 0.1. What isthe probability that a batch of 200 books has at most 10 bookswith a page missing?Using a Poisson distribution, we have thatλ = np = 200 ·0.1 = 20.Now from a Poisson distribution table, p(x≤ 10;20)≈ 0.011.

With a CAS we obtain B(10;200,0.1)≈ 0.008.



logo1



Example.

Suppose the probability of a missing page in anindividual book a certain manufacturer makes is 0.1. What isthe probability that a batch of 200 books has at most 10 bookswith a page missing?Using a Poisson distribution, we have thatλ = np = 200 ·0.1 = 20.Now from a Poisson distribution table, p(x≤ 10;20)≈ 0.011.




logo1



Example. Suppose the probability of a missing page in anindividual book a certain manufacturer makes is 0.1.

What isthe probability that a batch of 200 books has at most 10 bookswith a page missing?Using a Poisson distribution, we have thatλ = np = 200 ·0.1 = 20.Now from a Poisson distribution table, p(x≤ 10;20)≈ 0.011.




logo1



Example. Suppose the probability of a missing page in anindividual book a certain manufacturer makes is 0.1. What isthe probability that a batch of 200 books has at most 10 bookswith a page missing?

Using a Poisson distribution, we have thatλ = np = 200 ·0.1 = 20.Now from a Poisson distribution table, p(x≤ 10;20)≈ 0.011.




logo1



Example. Suppose the probability of a missing page in anindividual book a certain manufacturer makes is 0.1. What isthe probability that a batch of 200 books has at most 10 bookswith a page missing?Using a Poisson distribution, we have that

λ = np = 200 ·0.1 = 20.Now from a Poisson distribution table, p(x≤ 10;20)≈ 0.011.




logo1



Example. Suppose the probability of a missing page in anindividual book a certain manufacturer makes is 0.1. What isthe probability that a batch of 200 books has at most 10 bookswith a page missing?Using a Poisson distribution, we have thatλ

= np = 200 ·0.1 = 20.Now from a Poisson distribution table, p(x≤ 10;20)≈ 0.011.




logo1



Example. Suppose the probability of a missing page in anindividual book a certain manufacturer makes is 0.1. What isthe probability that a batch of 200 books has at most 10 bookswith a page missing?Using a Poisson distribution, we have thatλ = np

= 200 ·0.1 = 20.Now from a Poisson distribution table, p(x≤ 10;20)≈ 0.011.




logo1



Example. Suppose the probability of a missing page in anindividual book a certain manufacturer makes is 0.1. What isthe probability that a batch of 200 books has at most 10 bookswith a page missing?Using a Poisson distribution, we have thatλ = np = 200 ·0.1

= 20.Now from a Poisson distribution table, p(x≤ 10;20)≈ 0.011.




logo1



Example. Suppose the probability of a missing page in anindividual book a certain manufacturer makes is 0.1. What isthe probability that a batch of 200 books has at most 10 bookswith a page missing?Using a Poisson distribution, we have thatλ = np = 200 ·0.1 = 20.

Now from a Poisson distribution table, p(x≤ 10;20)≈ 0.011.




logo1



Example. Suppose the probability of a missing page in anindividual book a certain manufacturer makes is 0.1. What isthe probability that a batch of 200 books has at most 10 bookswith a page missing?Using a Poisson distribution, we have thatλ = np = 200 ·0.1 = 20.Now from a Poisson distribution table, p(x≤ 10;20)≈ 0.011.




logo1


Theorem.

Let X be a Poisson distributed random variable.Then E(X) = λ and V(X) = λ .

Proof.

E(X) =∞

∑x=0

xe−λ λ x

x!=

∞

∑x=1

e−λ λ x

(x−1)!

= λe−λ∞

∑x=1

λ x−1

(x−1)!= λe−λ

∞

∑j=0

λ j

j!

= λe−λ eλ = λ



logo1


Theorem. Let X be a Poisson distributed random variable.

Then E(X) = λ and V(X) = λ .

Proof.

E(X) =∞

∑x=0

xe−λ λ x

x!=

∞

∑x=1

e−λ λ x

(x−1)!

= λe−λ∞

∑x=1

λ x−1

(x−1)!= λe−λ

∞

∑j=0

λ j

j!

= λe−λ eλ = λ



logo1


Theorem. Let X be a Poisson distributed random variable.Then E(X) = λ and V(X) = λ .

Proof.

E(X) =∞

∑x=0

xe−λ λ x

x!=

∞

∑x=1

e−λ λ x

(x−1)!

= λe−λ∞

∑x=1

λ x−1

(x−1)!= λe−λ

∞

∑j=0

λ j

j!

= λe−λ eλ = λ



logo1



Proof.

E(X)

=∞

∑x=0

xe−λ λ x

x!=

∞

∑x=1

e−λ λ x

(x−1)!

= λe−λ∞

∑x=1

λ x−1

(x−1)!= λe−λ

∞

∑j=0

λ j

j!

= λe−λ eλ = λ



logo1



Proof.

E(X) =∞

∑x=0

xe−λ λ x

x!

=∞

∑x=1

e−λ λ x

(x−1)!

= λe−λ∞

∑x=1

λ x−1

(x−1)!= λe−λ

∞

∑j=0

λ j

j!

= λe−λ eλ = λ



logo1



Proof.

E(X) =∞

∑x=0

xe−λ λ x

x!=

∞

∑x=1

e−λ λ x

(x−1)!

= λe−λ∞

∑x=1

λ x−1

(x−1)!= λe−λ

∞

∑j=0

λ j

j!

= λe−λ eλ = λ



logo1



Proof.

E(X) =∞

∑x=0

xe−λ λ x

x!=

∞

∑x=1

e−λ λ x

(x−1)!

= λe−λ∞

∑x=1

λ x−1

(x−1)!

= λe−λ∞

∑j=0

λ j

j!

= λe−λ eλ = λ



logo1



Proof.

E(X) =∞

∑x=0

xe−λ λ x

x!=

∞

∑x=1

e−λ λ x

(x−1)!

= λe−λ∞

∑x=1

λ x−1

(x−1)!= λe−λ

∞

∑j=0

λ j

j!

= λe−λ eλ = λ



logo1



Proof.

E(X) =∞

∑x=0

xe−λ λ x

x!=

∞

∑x=1

e−λ λ x

(x−1)!

= λe−λ∞

∑x=1

λ x−1

(x−1)!= λe−λ

∞

∑j=0

λ j

j!

= λe−λ eλ

= λ



logo1



Proof.

E(X) =∞

∑x=0

xe−λ λ x

x!=

∞

∑x=1

e−λ λ x

(x−1)!

= λe−λ∞

∑x=1

λ x−1

(x−1)!= λe−λ

∞

∑j=0

λ j

j!

= λe−λ eλ = λ

