# Poisson distribution ¶

Go back

It's called the loi des événements/phénomènes rares in French. The Poisson distribution $\mathbb{P}(\lambda)$ require that the probability $p$ is relatively smaller than $n$, while the parameter $\lambda$ is equals to $n*p$.

• The mass function is $\mathbb{P}(X=k) = \frac{\lambda^k * e^{-\lambda}}{k!}$
• $E(X) = \lambda$
• $V(X) = \lambda$

## Demonstration ¶

You need to understand this demonstration because you are going to do something like this a lot, and it's easy

Demonstration of $E[X]=\lambda$

With the expected value formula, we have

• $\sum_{k \in \mathbb{N}} k * P(X=k)$
• $= \sum_{k \in \mathbb{N}} k * \frac{\lambda^k * e^{-\lambda}}{k!}$
• $e$ is a constant, so we can extract it
• $= e^{-\lambda} * \sum_{k \in \mathbb{N}} \frac{\lambda^k}{(k-1)!}$
• I'm taking one $\lambda$ out (it's a constant too)
• $= \lambda e^{-\lambda} * \sum_{k \in \mathbb{N}} \frac{\lambda^{k-1}}{(k-1)!}$

Here we are starting the magic. There are well-known series that can be replaced by a function.

Check them here, but this is the only one that we will use: $\sum_{k \in \mathbb{N}} \frac{\lambda^{k}}{k!} = e^k$

• so using the exponential series development
• $= \lambda e^{-\lambda} * e^{\lambda}$
• $= \lambda e^{-\lambda+\lambda}$
• $= \lambda e^{0}$
• $= \lambda * 1$
• $= \lambda$

Remember the magic trick, it's useful.