Poisson distribution

The Poisson distribution $𝑋 \sim P o i s (𝜆)$ describes the probability of $𝑥$ events occurring in a fixed time interval, where $𝜆$ is the expected number of occurrences and the time of each event is independent from any prior ones. #m/def/prob It has the probability distribution

ℙ (𝑋 = 𝑥) = 𝑒 - 𝜆 𝜆 𝑥 𝑥!

Proof

#missing/proof

Properties

Expectation, variance: $𝔼 [𝑋] = V a r [𝑋] = 𝜆$
Moment-generating function: $𝔼 [e 𝑡 𝑋] = e 𝜆 (e 𝑡 - 1)$
Probability generating function: $𝑔 𝑋 (𝑡) = e 𝜆 (𝑡 - 1)$

Additionally

If $𝑋 \sim P o i s (𝜆)$ and $𝑌 \sim P o i s (𝜇)$ are independently distributed then $𝑋 + 𝑌 \sim P o i s (𝜆 + 𝜇)$
If $𝑋 \sim P o i s (𝜆 𝑝)$ and $𝑌 \sim P o i s (𝜆 𝑞)$ where $𝑞 = 1 - 𝑝$ are independently distributed then $𝑁 = 𝑋 + 𝑌 \sim P o i s (𝜆)$ and $𝑋 ∣ 𝑁 = 𝑛 \sim B i n (𝑛, 𝑝)$ .
Conversely, if $𝑁 \sim P o i s (𝜆)$ and $𝑋 ∣ 𝑁 = 𝑛 \sim B i n (𝑛, 𝑝)$ then $𝑋 \sim P o i s (𝜆 𝑝)$ and $𝑌 = 𝑁 - 𝑋 \sim P o i s (𝜆 𝑞)$ are independently distributed.

Proof of 1

By ^P1

𝑀 𝑋 + 𝑌 (𝑡) = 𝑀 𝑋 (𝑡) 𝑀 𝑌 (𝑡) = 𝔼 [e 𝑡 𝑋] 𝔼 [e 𝑡 𝑌] = e 𝜆 (e 𝑡 - 1) e 𝜇 (e 𝑡 - 1) = e (𝜆 + 𝜇) (e 𝑡 - 1)

as required.

Poisson paradigm

Let ${𝐴 𝑖} 𝑛 𝑖 = 1$ be independent events with $𝑝 𝑖 = ℙ (𝐴 𝑖)$ small. Then $𝑋 = \sum 𝑛 𝑗 = 1 𝐼 𝐴 𝑗$ is approximated by $𝑁 \sim P o i s (𝜆)$ where $𝜆 - \sum 𝑛 𝑖 = 1 𝑝 𝑖$ , with

| ℙ (𝑋 \in 𝐵) - ℙ (𝑁 \in 𝐵) | \leq m i n (1, 1 𝜆) 𝑛 \sum 𝑗 = 1 𝑝 2 𝑗

for any $𝐵 \subseteq ℕ$ . See also Poisson process.

Relationship to other distributions

If $𝑋 \sim P o i s (𝜆)$ and $𝑌 \sim P o i s (𝜇)$ are independently distributed, then the conditional distribution of $𝑋$ given $𝑋 + 𝑌 = 𝑛$ is $B i n (𝑛, 𝜆 𝜆 + 𝜇)$ .
As $𝑛 \to \infty$ and $𝑝 \to 0$ as $𝑛 𝑝 = 𝜆$ remains fixed $B i n (𝑛, 𝑝) ⇝ P o i s (𝜆)$ .
By the Central limits theorem $P o i s (𝑛) ⇝ N (𝑛, 𝑛)$ as $𝑛 \to \infty$

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