The Poisson distribution is different from the Binomial and Geometric distributions because it is not represented by a sequence of independent Bernoulli random variables.

A Poisson random variable is a random variable \(X\) with pmf \[p(k) = \frac{\lambda^k}{k!}e^{-\lambda}\] for all integers \(k \geq 0\) where \(\lambda > 0\) is a constant. We write this distribution as Poisson\((\lambda).\)

A Poisson random variable can be used to represent the number of events that occur over a certain time interval. For example, the number of phone calls over a one hour time interval or the number of customers in a day.

Let \(X\) be Poisson\((\lambda)\). The expected value of \(X\) is \begin{align} E[X] & = \sum_{k = 0}^\infty k \cdot \frac{\lambda^k}{k!}e^{-\lambda} \\ & = \sum_{k = 1}^\infty k \cdot \frac{\lambda^k}{k!}e^{-\lambda} \\ & = \sum_{k = 1}^\infty \frac{\lambda^k}{(k-1)!}e^{-\lambda} \\ & = \lambda e^{-\lambda} \sum_{k = 1}^\infty \frac{\lambda^{k-1}}{(k-1)!} \\ & = \lambda e^{-\lambda} \sum_{k = 0}^\infty \frac{\lambda^k}{k!} \\ & = \lambda e^{-\lambda} e^\lambda \\ & = \lambda \\ \end{align} Next we compute \(E[X^2]\) so we can find the variance of \(X.\) \begin{align} E[X^2] & = \sum_{k = 0}^\infty k^2 \cdot \frac{\lambda^k}{k!}e^{-\lambda} \\ & = \sum_{k = 0}^\infty (1 + k - 1)k \cdot \frac{\lambda^k}{k!}e^{-\lambda} \\ & = \sum_{k = 0}^\infty k\frac{\lambda^k}{k!}e^{-\lambda} + (k - 1)k \cdot \frac{\lambda^k}{k!}e^{-\lambda} \\ & = E[X] + \sum_{k = 0}^\infty (k - 1)k \cdot \frac{\lambda^k}{k!}e^{-\lambda} \\ & = \lambda + \sum_{k = 2}^\infty (k - 1)k \cdot \frac{\lambda^k}{k!}e^{-\lambda} \\ & = \lambda + \sum_{k = 2}^\infty \frac{\lambda^k}{(k-2)!}e^{-\lambda} \\ & = \lambda + \lambda^2 e^{-\lambda}\sum_{k = 2}^\infty \frac{\lambda^{k-2}}{(k-2)!} \\ & = \lambda + \lambda^2 e^{-\lambda}\sum_{k = 0}^\infty \frac{\lambda^k}{k!} \\ & = \lambda + \lambda^2 e^{-\lambda}e^\lambda \\ & = \lambda + \lambda^2 \\ \end{align} So, \[\text{Var}(X) = E[X^2] - E[X]^2 = \lambda + \lambda^2 - \lambda^2 = \lambda\] In summary, if \(X\) is Poisson\((\lambda)\) then \[E[X] = \lambda, \text{Var}(X) = \lambda\]