Poisson Distribution: Definition and Examples

The Poisson Distribution

• The Poisson distribution is a discrete probability distribution.
• It expresses the probability of a given number of events occurring in a fixed interval of time or space.
• These events occur with a known constant mean rate.
• These events occur independently of the time since the last event.

Definition of Poisson Distribution

• A discrete random variable $X$ has a Poisson distribution with parameter $\lambda > 0$ if its probability mass function (PMF) is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $k = 0, 1, 2,...$
• $e$ is Euler's number ($e \approx 2.71828$).
• $k$ is the number of occurrences of an event.
• $k!$ is the factorial of $k$.
• $\lambda$ is a positive real number, equal to the expected number of occurrences during the interval.

Mean and Variance of Poisson Distribution

• If $X \sim Poisson(\lambda)$, then:
  • $E[X] = \lambda$ (Mean)
  • $Var(X) = \lambda$ (Variance)

Example of Poisson Distribution

• Scenario: Average goals in a soccer match = 2.
• Question: What is the chance of 5 goals being scored in a match?
• Solution:
  • $P(X = 5) = \frac{e^{-2} 2^5}{5!} = 0.0361$

Poisson Sums

• Let $X_1, X_2,..., X_n$ be independent Poisson random variables with means $\lambda_1, \lambda_2,..., \lambda_n$, respectively.
• Then $X_1 + X_2 +... + X_n$ is a Poisson random variable with mean $\lambda_1 + \lambda_2 +... + \lambda_n$.

Example of Poisson Sums

• Scenario: Two email accounts. The number of emails arriving to the first account follows a Poisson distribution with a rate of 3 emails per hour. The number of emails arriving to the second account follows a Poisson distribution with a rate of 5 emails per hour.
• Question: What is the probability that you will receive a total of 10 emails in one hour?
  • Let X be the number of emails arriving to the first account, and Y be the number of emails arriving to the second account. Then $X \sim Poisson(3)$ and $Y \sim Poisson(5)$.
  • $X + Y \sim Poisson(3 + 5) = Poisson(8)$
  • $P(X + Y = 10) = \frac{e^{-8} 8^{10}}{10!} = 0.09926$