The Poisson Distribution

$X \sim \text{Po}(\lambda)$

$P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}$

$E(X) = \lambda$, $\text{Var}(X) = \lambda$.

• Events occur independently.
• Events occur at a constant average rate.
• Two events cannot occur simultaneously.

A shop averages 3 customers per hour. $X \sim \text{Po}(3)$.

$P(X = 5) = \frac{e^{-3} \cdot 243}{120} \approx 0.1008$.

$P(X \leq 2) = P(0) + P(1) + P(2) = e^{-3}(1 + 3 + 4.5) = 0.0498 \times 8.5 \approx 0.423$.

If $X \sim \text{Po}(\lambda_1)$ and $Y \sim \text{Po}(\lambda_2)$ are independent: $X + Y \sim \text{Po}(\lambda_1 + \lambda_2)$.