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Table of contents
1.
Introduction
2.
Homogeneous Poisson Process
2.1.
Derivation
3.
FAQs
4.
Key Takeaways
Last Updated: Mar 27, 2024

Homogeneous Poisson Process

Author Rajat Agrawal
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Introduction

In probability theory, the Poisson process is one of the most important and commonly utilized processes. It's a popular tool for simulating random events in time or space. The Poisson Process can be used to simulate a variety of real-life scenarios. For example, the number of accidents on the road, model the number of earthquakes in a given area, etc.

Let’s learn about the Homogeneous Poisson Process in-depth.

Homogeneous Poisson Process

To understand the Poisson process, we first need to understand about counting process.

Counting Process: A counting process, N(t), is any integer-valued process with the following properties:-

1.)  N(0) = 0.

2.) N(t + s) ≥ N(t), ∀s ≥ 0.

Poisson Process is derived as a counting process here. Assume we're tracking the number of times a specific event occurs over a given period of time. (In this case, we're using time as an example.) We may also think of things like space and so on).

We can classify them as Poisson Process events if they meet the following criteria.

1.) The frequency of occurrences over disjoint time intervals is independent.

2.) The probability of a single occurrence in a short time span is proportional to the interval's length.

3.) The probability of multiple occurrences during a small time interval can be ignored.

If we use the symbol X(t) to represent the number of occurrences across a time interval of length t, then.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mfenced><mrow><mi>X</mi><mfenced><mi>t</mi></mfenced><mo>=</mo><mi>n</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup><msup><mfenced><mrow><mi>&#x3BB;</mi><mi>t</mi></mrow></mfenced><mi>n</mi></msup></mrow><mrow><mi>n</mi><mo>!</mo></mrow></mfrac></math>, where <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>&#x3BB;</mi></math> is the rate of occurrence.

Derivation

Let’s prove our claim that if X(t) be the number of occurrence in an interval of length t, then.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mfenced><mrow><mi>X</mi><mfenced><mi>t</mi></mfenced><mo>=</mo><mi>n</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup><msup><mfenced><mrow><mi>&#x3BB;</mi><mi>t</mi></mrow></mfenced><mi>n</mi></msup></mrow><mrow><mi>n</mi><mo>!</mo></mrow></mfrac></math>, where <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>&#x3BB;</mi></math> is the rate of occurrence.

The statement will be proven using mathematical induction. We begin by expressing the above assumptions mathematically. According to criteria 3 in a small time interval h.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mfenced><mrow><mi>X</mi><mfenced><mi>h</mi></mfenced><mo>&gt;</mo><mn>1</mn></mrow></mfenced><mo>=</mo><mo>&#xA0;</mo><mi>o</mi><mfenced><mi>h</mi></mfenced></math>, where <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>o</mi><mfenced><mi>h</mi></mfenced></mrow><mi>h</mi></mfrac></math> tends to zero as h tends to zero or.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>-</mo><mi>P</mi><mfenced><mrow><mi>X</mi><mfenced><mi>h</mi></mfenced><mo>=</mo><mn>0</mn></mrow></mfenced><mo>-</mo><mi>P</mi><mfenced><mrow><mi>X</mi><mfenced><mi>h</mi></mfenced><mo>=</mo><mn>1</mn></mrow></mfenced><mo>=</mo><mo>&#xA0;</mo><mi>o</mi><mfenced><mi>h</mi></mfenced></math>

If <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>&#x3BB;</mi></math> is the rate of occurrence, then according to criteria 2, we get,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mfenced><mrow><mi>X</mi><mfenced><mi>h</mi></mfenced><mo>=</mo><mn>1</mn></mrow></mfenced><mo>=</mo><mi>&#x3BB;</mi><mi>h</mi></math>

Take a small interval (t, t+h) and an interval (0, t). P(X(t)=n) will be abbreviated as <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>n</mi></msub><mfenced><mi>t</mi></mfenced></math> . As a result, the equations above can be represented as,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>-</mo><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>h</mi></mfenced><mo>-</mo><msub><mi>P</mi><mn>1</mn></msub><mfenced><mi>h</mi></mfenced><mo>=</mo><mi>o</mi><mfenced><mi>h</mi></mfenced><mspace linebreak="newline"/><mi>o</mi><mi>r</mi><mspace linebreak="newline"/><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>h</mi></mfenced><mo>=</mo><mn>1</mn><mo>-</mo><mi>&#x3BB;</mi><mi>h</mi><mo>-</mo><mi>o</mi><mfenced><mi>h</mi></mfenced></math>

So we have to prove that,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>n</mi></msub><mfenced><mi>t</mi></mfenced><mo>&#xA0;</mo><mo>=</mo><mfrac><mrow><mo>&#xA0;</mo><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup><msup><mfenced><mrow><mi>&#x3BB;</mi><mi>t</mi></mrow></mfenced><mi>n</mi></msup></mrow><mrow><mi>n</mi><mo>!</mo></mrow></mfrac></math>

We'll start by proving the result for n=0 and n=1. Then we'll verify that if the result is correct for n=m, it'll be correct for n=m+1.

Take the interval (0, t+h). Now, 

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mn>0</mn></msub><mfenced><mrow><mi>t</mi><mo>+</mo><mi>h</mi></mrow></mfenced><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>t</mi></mfenced><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>h</mi></mfenced></math>

(since occurrences in the intervals (0, t) and (t, t+h) are independent) or,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mn>0</mn></msub><mfenced><mrow><mi>t</mi><mo>+</mo><mi>h</mi></mrow></mfenced><mo>=</mo><mo>&#xA0;</mo><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>t</mi></mfenced><mfenced><mrow><mn>1</mn><mo>-</mo><mi>&#x3BB;</mi><mfenced><mi>h</mi></mfenced><mo>-</mo><mi>o</mi><mfenced><mi>h</mi></mfenced></mrow></mfenced><mspace linebreak="newline"/><mi>o</mi><mi>r</mi><mspace linebreak="newline"/><mfrac><mrow><msub><mi>P</mi><mn>0</mn></msub><mfenced><mrow><mi>t</mi><mo>+</mo><mi>h</mi></mrow></mfenced><mo>-</mo><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>t</mi></mfenced></mrow><mi>h</mi></mfrac><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mo>-</mo><mi>&#x3BB;</mi><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>t</mi></mfenced><mo>-</mo><mfrac><mrow><mi>o</mi><mfenced><mi>h</mi></mfenced></mrow><mi>h</mi></mfrac></math>

Taking the limit as h tends to zero, we get, 

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><msup><mi>P</mi><mo>'</mo></msup><mn>0</mn></msub><mfenced><mi>t</mi></mfenced><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mo>-</mo><mi>&#x3BB;</mi><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>t</mi></mfenced></math>

The solution of the above differential equation is, 

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>t</mi></mfenced><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>c</mi><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup></math>

taking the initial condition, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mn>0</mn></msub><mfenced><mn>0</mn></mfenced><mo>=</mo><mn>1</mn></math> we evaluate c=0. Hence, 

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>t</mi></mfenced><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup></math>, So our claim is valid for n=0. 

Now we try to prove it for n=1. 

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mn>1</mn></msub><mfenced><mrow><mi>t</mi><mo>+</mo><mi>h</mi></mrow></mfenced><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><msub><mi>P</mi><mn>1</mn></msub><mfenced><mi>t</mi></mfenced><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>h</mi></mfenced><mo>+</mo><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>t</mi></mfenced><msub><mi>P</mi><mn>1</mn></msub><mfenced><mi>h</mi></mfenced></math>

(We use the fact that the occurrence must be in either of the interval (0, t) and (t, t+h)), or 

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mn>1</mn></msub><mfenced><mrow><mi>t</mi><mo>+</mo><mi>h</mi></mrow></mfenced><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><msub><mi>P</mi><mn>1</mn></msub><mfenced><mi>t</mi></mfenced><mfenced><mrow><mn>1</mn><mo>-</mo><mi>&#x3BB;</mi><mi>h</mi><mo>-</mo><mi>o</mi><mfenced><mi>h</mi></mfenced></mrow></mfenced><mo>+</mo><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup><mfenced><mrow><mi>&#x3BB;</mi><mi>h</mi></mrow></mfenced><mspace linebreak="newline"/><mi>o</mi><mi>r</mi><mspace linebreak="newline"/><mfrac><mrow><msub><mi>P</mi><mn>1</mn></msub><mfenced><mrow><mi>t</mi><mo>+</mo><mi>h</mi></mrow></mfenced><mo>-</mo><msub><mi>P</mi><mn>1</mn></msub><mfenced><mi>t</mi></mfenced></mrow><mi>h</mi></mfrac><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mo>-</mo><mi>&#x3BB;</mi><msub><mi>P</mi><mn>1</mn></msub><mfenced><mi>t</mi></mfenced><mo>-</mo><mi>&#x3BB;</mi><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup><mo>-</mo><mfrac><mrow><mi>o</mi><mfenced><mi>h</mi></mfenced></mrow><mi>h</mi></mfrac></math>

Again taking the limit as h tends to zero, 

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><msub><mo>'</mo><mn>1</mn></msub><mfenced><mi>t</mi></mfenced><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mo>-</mo><mi>&#x3BB;</mi><msub><mi>P</mi><mn>1</mn></msub><mfenced><mi>t</mi></mfenced><mo>-</mo><mi>&#x3BB;</mi><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup></math>

This is a first-order linear differential equation, and the solution is, 

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mn>1</mn></msub><mfenced><mi>t</mi></mfenced><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>&#x3BB;</mi><mi>t</mi><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup><mo>+</mo><msub><mi>c</mi><mn>1</mn></msub></math>, where c1 is a constant. 

Since, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mn>1</mn></msub><mfenced><mn>0</mn></mfenced><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mn>0</mn></math>. We get, 

c1 = 0 .

Hence 

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mn>1</mn></msub><mfenced><mi>t</mi></mfenced><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>&#x3BB;</mi><mi>t</mi><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup><mspace linebreak="newline"/><mi>o</mi><mi>r</mi><mo>&#xA0;</mo><mspace linebreak="newline"/><msub><mi>P</mi><mn>1</mn></msub><mfenced><mi>t</mi></mfenced><mo>&#xA0;</mo><mo>=</mo><mfrac><mrow><mo>&#xA0;</mo><msup><mfenced><mrow><mi>&#x3BB;</mi><mi>t</mi></mrow></mfenced><mn>1</mn></msup><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup></mrow><mrow><mn>1</mn><mo>!</mo></mrow></mfrac></math>

So our claim is valid for n=1. We assume that our claim is right for n=m. 

We will show that it is true for n=m+1. So, 

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mrow><mi>t</mi><mo>+</mo><mi>h</mi></mrow></mfenced><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><msub><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mi>t</mi></mfenced><msub><mi>P</mi><mn>0</mn></msub><mfenced><mi>h</mi></mfenced><mo>+</mo><msub><mi>P</mi><mi>m</mi></msub><mfenced><mi>t</mi></mfenced><msub><mi>P</mi><mn>1</mn></msub><mfenced><mi>h</mi></mfenced><mo>+</mo><munderover><mo>&#x2211;</mo><mrow><mi>j</mi><mo>=</mo><mn>1</mn></mrow><mi>m</mi></munderover><msub><mi>P</mi><mrow><mi>m</mi><mo>-</mo><mi>j</mi></mrow></msub><mfenced><mi>t</mi></mfenced><msub><mi>P</mi><mrow><mi>j</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mi>h</mi></mfenced></math>

(We suppose that the m+1 occurrence can occur in a variety of ways, including m+1 occurrences in (0, t) and no occurrence in (t, t+h), or m occurrences in (0, t) and 1. an occurrence in (t, t+h), or m-j occurrences in (0, t) and j+1 occurrence in (t, t+h) for j=1 to m).

So, 

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mrow><mi>t</mi><mo>+</mo><mi>h</mi></mrow></mfenced><mo>=</mo><msub><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mi>t</mi></mfenced><mfenced><mrow><mn>1</mn><mo>-</mo><mi>&#x3BB;</mi><mi>h</mi><mo>-</mo><mi>o</mi><mfenced><mi>h</mi></mfenced></mrow></mfenced><mo>+</mo><mfrac><mrow><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup><msup><mfenced><mrow><mi>&#x3BB;</mi><mi>t</mi></mrow></mfenced><mi>m</mi></msup></mrow><mrow><mi>m</mi><mo>!</mo></mrow></mfrac><mi>&#x3BB;</mi><mi>h</mi><mo>+</mo><munderover><mo>&#x2211;</mo><mrow><mi>j</mi><mo>=</mo><mn>1</mn></mrow><mi>m</mi></munderover><msub><mi>P</mi><mrow><mi>m</mi><mo>-</mo><mi>j</mi></mrow></msub><mfenced><mi>t</mi></mfenced><mi>o</mi><mfenced><mi>h</mi></mfenced></math>

Since, 

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mrow><mi>j</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mi>h</mi></mfenced><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>o</mi><mfenced><mi>h</mi></mfenced></math> for j>=1. 

or, 

<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><msub><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mrow><mi>t</mi><mo>+</mo><mi>h</mi></mrow></mfenced><mo>-</mo><msub><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mi>t</mi></mfenced></mrow><mi>h</mi></mfrac><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mo>-</mo><mi>&#x3BB;</mi><msub><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mi>t</mi></mfenced><mo>+</mo><mfrac><mrow><msup><mi>&#x3BB;</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msup><msup><mi>t</mi><mi>m</mi></msup></mrow><mrow><mi>m</mi><mo>!</mo></mrow></mfrac><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup><mo>+</mo><mfrac><mrow><mi>o</mi><mfenced><mi>h</mi></mfenced></mrow><mi>h</mi></mfrac><munderover><mo>&#x2211;</mo><mrow><mi>j</mi><mo>=</mo><mn>1</mn></mrow><mi>m</mi></munderover><msub><mi>P</mi><mrow><mi>m</mi><mo>-</mo><mi>j</mi></mrow></msub><mfenced><mi>t</mi></mfenced><mo>-</mo><mfrac><mrow><mi>o</mi><mfenced><mi>h</mi></mfenced></mrow><mi>h</mi></mfrac><msub><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mi>t</mi></mfenced></math>

Taking the limit as h goes to zero, we have, 

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><msub><mo>'</mo><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mi>t</mi></mfenced><mo>=</mo><mo>-</mo><mi>&#x3BB;</mi><msub><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mi>t</mi></mfenced><mo>+</mo><mfrac><mrow><msup><mi>&#x3BB;</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msup><msup><mi>t</mi><mi>m</mi></msup></mrow><mrow><mi>m</mi><mo>!</mo></mrow></mfrac><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup></math>

This is a first-order differential equation with a solution that looks like this.

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mi>t</mi></mfenced><mo>&#xA0;</mo><mo>=</mo><mfrac><mrow><mo>&#xA0;</mo><msup><mfenced><mrow><mi>&#x3BB;</mi><mi>t</mi></mrow></mfenced><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msup><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup></mrow><mrow><mfenced><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mo>!</mo></mrow></mfrac><mo>+</mo><msub><mi>c</mi><mn>2</mn></msub></math>

If we assume that <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mn>0</mn></mfenced><mo>=</mo><mn>1</mn></math> we get c2 = 0. 

So the final result is, 

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msub><mfenced><mi>t</mi></mfenced><mo>&#xA0;</mo><mo>=</mo><mfrac><mrow><mo>&#xA0;</mo><msup><mfenced><mrow><mi>&#x3BB;</mi><mi>t</mi></mrow></mfenced><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></msup><msup><mi>e</mi><mrow><mo>-</mo><mi>&#x3BB;</mi><mi>t</mi></mrow></msup></mrow><mrow><mfenced><mrow><mi>m</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mo>!</mo></mrow></mfrac></math>

Hence the result is proved.

Thus, we have derived the pmf of no. of occurrences in a Poisson Process, a Poisson Distribution with parameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>&#x3BB;</mi></math>.

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FAQs

1.) Differentiate between homogeneous and non-homogeneous Poisson processes.

A non-homogeneous Poisson process is identical to a regular Poisson process, except that the average rate of arrivals might fluctuate over time. Non-homogeneous processes are more accurately approximated in many applications that create random points in time.

2.) Mention one use of the Poisson Process.

In a large population, the Poisson distribution is used to characterize the distribution of unusual events. For example, there is a probability that a single cell among a massive population of cells will acquire a mutation at any given time.

3.) What is Poisson Experiment?

A Poisson experiment is a statistical test with the following characteristics: The experiment's outcome might be judged as either a success or a failure. It is possible to calculate the average number of successes in a specific region.

Key Takeaways

In this article, we have extensively discussed the Homogeneous Poisson Process and its uses. If you want to learn more, check out our article on Partial Differential Equations and System of Linear Equations.

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