# Poisson Random Variable Theory and Poisson Calculation in MATLAB

A random X variable that
takes the values 0, 1, 2, ... is called a Poisson random variable with a
parameter λ if for an λ> 0,

Equation defines a mass function of probability cause

The distribution of Poisson Probability was introduced by S.D. Poisson in a book he wrote about applying the theory of opportunity in the matter of lawsuits, criminal trials, and the like. This book, which was published in 1837, was given the title "

*Recherches Sur La Probabilite Des Jugements En Maliere Criminelle Et En Matiere Civile"*

Poisson random variables have a very wide range of
applications in various fields because they can be used as a set of binomic
random variables with parameters (

*n, p*) the value of*n*is large and*p*is small enough so that np is medium. To see this, suppose*X*is a binom random variable with parameters (*n, p*) and suppose that*λ = np*. Then
Now, for n correct and λ being medium

Therefore, for large

*n*and λ
In other words, if we do n actions that are free, each of which results in "success" with probability

*p*, then, for large*n*and*p*is small enough so that*np*is moderate, the amount of success that occurs can be approached by a Poisson random variable with parameter λ =*np*. This value of λ (which will later be shown to be the same as the expected value of success) is usually determined empirically.
Some examples of random variables that usually obey the law of opportunity Poisson

**Equation 1**are- Many print errors on a page (or a number of pages) of a book.
- The number of residents in a society reaches the age of 100 years.
- The number of telephone calls was wrong one day.
- Lots of delicious packets of food sold in a shop every day.
- The number of customers entering the post office one day
- The number of alpha particles scattered over a certain period of time from a radioactive particle

Each random variable above, and many other random variables, are approached by Poisson random variables for the same reason, because

**Poisson random variables**approach**random binomic variables**. For example, we can assume that each letter typed has the opportunity to type a typing*p*. So the number of typos on a page can be approached by Poisson with λ =*np*, with*n*stating the number of letters on one page. Similarly, we can assume that every member of a community has a certain opportunity to reach the age of 100. Also, anyone who enters a store can be imagined to have a certain opportunity to buy a cat food package, and so on.**Example calculation of Random Poisson Variable 1**

Suppose the number of typographical errors on one page of this book follows a Poisson distribution with the parameter λ = 1/2. Count the Probability for at least one error on this page.

Solution: If it is assumed that

*X*is the number of errors on this page, then
P {

*X≥1*}: At least one error
Probability Calculation of Poisson Typing Errors Using MATLAB.

**Example of a Poisson Random**

**Variable**

**2**

Suppose that the probability of a product defect produced by a machine is 0.1. Calculate the probability that an example of 10 products will contain at most 1 defective product

Solution: Because Poisson random variables can be used as a set for binomic random variables, the calculation is as follows,

Probability Calculation of Binomic Typing Errors Using MATLAB.

% Know n = 10 (number of products)

% Opportunity for Product Defects p = 0.1

% Calculate Opportunities At least 1 defective P {X <= 1}?

% P {X <= 1} = P {X = 0} + P {X = 1}

P = binopdf (0,10,0.1) + binopdf (1,10,0.1)

P =

0.7361

Whereas if the Poisson random variable calculation is done, the answer is obtained

Probability Calculation of Poisson Typing Errors Using MATLAB.

% Know n = 10 (number of
products)

% Opportunity for Product
Defects p = 0.1

% Calculate Opportunities
At least 1 defective P {X <= 1}?

% Because Calculate
Opportunities At least 1 defective then Lamda = 1

% P {X <= 1} = P {X = 0}
+ P {X = 1}

P=poisspdf(0,1)+poisspdf(1,1)

**Example of a Poisson Random Variable 3**

In an experiment, enumeration of the number of α particles was emitted in a time interval of 1 second by 1 gram of radioactive. If from past experience it was known that on average emitted 3.2 α-particles, give a possibility for the probability that no more than 2 α-particles would appear?

Solution: If we imagine that the radioactive material consists of a large number of n atoms which each have a 3.2 /

*n*probability to decay and release an α-particle in that interval of 1 second, then the number of α particles released can be approached by a Poisson random variable with the parameter λ = 3.2. So, the opportunity sought is
Probability for Release of Radioactive Materials Using MATLAB Calculations

% average emitted 3.2
?-particles , Lamda= 3.2

% the probability that no
more than 2 ?-particles would appear

%
P{X<=2}=P{X=0}+P{X=1}+P{X=2}

P=Poisspdf(0,3.2)+
Poisspdf(1,3.2) + Poisspdf(2,3.2)

Poisson's distribution of Probability can also be applied to situations where "events" occur at certain times, such as earthquakes, another possibility is "the event" in the form of someone entering a particular building (bank, post office, gas station, etc. so); another possibility is "what happened" in the form of a war. Suppose that "events" occur at certain times (which are random), and suppose that for a certain positive constant some of the following assumptions are fulfilled.

The probability of occurring exactly 1 event in any interval whose length h is equal to λ

*h*+*o(h)*, in this case*o(h)*expresses any function*f(h)*which is . [As, for example,*f(h)*=*h*^{2}is o(h), while*f(h)*=*h*is not]
Probability for the occurrence of 2 or more events in any interval whose length is equal to

*o(h)*.
For any integer

*n, j*, and for any n intervals that are set aside, if we define_{1}, j_{2}, ..., j_{n}*E*as the occurrence of the exact occurrence of events in the interval_{1}*j*, then the events E_{i}*, E*_{1}*, ..., E*_{2}*are free from each other.*_{n}
In words, Assumptions 1 and 2 say that for a small value of h, the probability of occurring exactly 1 event in a interval whose length is equal to

*λh*plus something relatively small compared to*h*, while the probability of occurrence of 2 or more events is relatively small compared to*h*, Assumption 3 says that what happens inside a hose has no probability effect on what will happen in the other hose which is set aside by the hose.
Under Assumptions 1, 2 and 3, it will now be shown that the number of events that occur in any interval whose length is t is a

**Poisson random variable**with the parameter*λt*. To be more precise, let us call the interval [0,*t*] and the number of events that occur in the interval we represent as*N(t)*. To get the expression for*P{N(t) = k}*, we start by dividing the interval [0,*t*] into n-hose children who set aside each other whose lengths are*t / n*.
Now

*P{N(t) = k}*=

*P*{

*k*of

*n*-hose the child contains exactly the event whereas the other

*n-k*contains 0 events} +

*P{N(t) = k*and at least 1 child-interval contains 2 or more events. .................. Equation 2.

This is due to the event on the left side of

**Equation 2**, which is the event {*N(t) = k*}, obviously the same as the combination of the two events which are set aside on the right hand side of**Equation 2**. If the two events that are set aside on the right hand side of**Equation 2**are denoted as A and B, then
Next, for any

*t, t / n*⇾0 if*n⇾∞*, so that*o(t / n) / (t / n)*⇾0 if*n⇾∞*according to
P(B)⇾0 If n⇾∞ .........

**Equation 3**

On the other hand, because
Assumptions 1 and 2 have implications that

(the number of two functions both

*o(h)*also*o(h)*) then from the assumption of freedom (Assumption 3) is obtained that
However, because

So, based on the same argument that verifies the almost Poisson for binom,

So, from

**Equation 2**,**Equation 3**, and**Equation 4**we get, by taking*n*⇾∞,
Therefore, if Assumptions 1, 2, and 3 are fulfilled, the number of events that occur in a hose remains arbitrary, the length

*t*is a Poisson random variable with the parameter λ*t*, and we say that these events occur following a Poisson process with the rate of λ. The value of λ, which can be indicated by the rate of occurrence of events per unit time, is a constant that must be determined empirically.
The discussion above explains why a Poisson random variable is usually a good possibility for the following phenomena:

- The number of earthquakes during a certain period of time
- Number of wars per year
- The number of electrons emitted from a cathode during a certain time interval
- Many insurance policy holders die within a certain period of time

**Example of a Poisson Random Variable 4**

Suppose an earthquake occurs in the western region of Indonesia in accordance with assumptions 1, 2, and 3 with λ = 2 and as a unit of time taken 1 week. (That is, earthquakes occur following the three assumptions above at a rate of 2 times per week).

1. Calculate opportunities for at least 3 earthquakes in the next 2 weeks

From

**Equation 5**we obtain,
2. Calculate the distribution of time opportunities, from now until the next earthquake occurs.

Suppose

*X*states how long (in weeks) until the next earthquake occurs. Because*X*will be greater than*t*if and only if there is no earthquake at all in time unit*t*, then from**Equation 5**we get,
so the F probability distribution function for random variable X is

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