Poisson Distribution Explained

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

Definition

• A discrete random variable $X$ has a Poisson distribution with parameter $\lambda > 0$.
• Probability mass function (PMF): $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$, $k = 0, 1, 2,...$
  • $e$ 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 in the interval.

Mean and Variance

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

Example

• average goals in a soccer match = 2.
• The chance of 5 goals being scored is 0.0361.
• $P(X = 5) = \frac{e^{-2} 2^5}{5!} = 0.0361$

Poisson Sums

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

Example

• Two email accounts exist.
• The number of emails arriving to the first account follows a Poisson distribution at a rate of 3 emails per hour.
• The number of emails arriving to the second account follows a Poisson distribution at a rate of 5 emails per hour.
• The probability of receiving 10 emails total in one hour is 0.09926.
• 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 $Y \sim Poisson(5)$.
• $X + Y \sim Poisson(3 + 5) = Poisson(8)$
• $P(X + Y = 10) = \frac{e^{-8} 8^{10}}{10!} = 0.09926$