The Poisson distribution models the number of successes ("arrivals") in a fixed-length time interval $\tau$. Events occur [[Independence of Events|independently]] and at a constant average rate $\lambda$. **Probability Mass Function:** The probability that exactly $k$ events occur during the interval $\tau$ is given by the [[Probability Mass Function|PMF]]. $ \mathbf P(k, \tau)=p_{k, \tau}(k, \tau)=\frac{(\lambda \tau)^k}{k!}*e^{-\lambda \tau} $ When we assume $\tau=1$, this simplifies to: $ \mathbf P(k)=p_{k}(k)=\frac{\lambda^k}{k!}*e^{-\lambda} $ **Expectation and Variance:** The [[Expectation]] can be derived in the usual manner, by iterating over all $k$ multiplied by their respective probability mass, with the caveat of an infinite sum. $ \mathbb E[X_\tau]=\sum_{k=0}^\infty k * \mathbf P(k, \tau)$ Alternatively, using the [[Poisson Approximation of Binomial Distribution]], we simply plug-into the formula for [[Binomial Distribution#Expectation|expectation]] and also for the [[Binomial Distribution#Variance|variance]] of the Binomial, where: $ \delta=\frac{\tau}{n}, \quad n=\frac{\tau}{\delta}, \quad p=\lambda \delta$ Expectation: $ \mathbb E[X_\tau]=np=\frac{\tau}{\delta} * \lambda\delta=\lambda \tau$ Variance: $ \mathrm{Var}(X_\tau) = np(1-p) = \frac{\tau}{\delta}*\lambda\delta*(1-\lambda \delta)=\lambda\tau $ **Interpretation of lambda:** We can view $\lambda$ as the success rate per unit of time, since: $ \mathbb{E}[X_\tau]=\lambda\tau \implies \lambda = \frac{\mathbb{E}[X_\tau]}{\tau}$