20  Poisson Distribution

20.1 Construction

Let \(X\) denote the number of times a certain events happening in the time interval \([0,1]\). Then the support of \(X\) is \(\{0,1,2,...\}\). Thus, \(X\) is a discrete random variable.

Divide the interval \([0,1]\) into \(n\) equal sub-intervals each of length \(\frac{1}{n}\). A sub-interval is considered a success if a phone call is received in that sub-interval. Let \(X_n\) be the number of sub-intervals in which a phone call is received. Then \(X_n\) is a Binomial random variable with parameters \((n, p)\). Note that in the time interval \([0, 1]\), \(X\) is fixed whereas \(X_n\) changes with \(n\). Clearly, \[P(X = k) = \lim_{n\to\infty} P(X_n = k)\] Now, let \(\lambda\) be the average number of events received per unit time. Then the probability that a sub-interval takes a call is \(p \approx \frac{n}{\lambda}\). Hence, \[\begin{align} P(X = k) &= \lim_{n\to\infty} {n \choose k} p^k (1-p)^{n-k} \\ &= \lim_{n\to\infty} \frac{n!}{k!(n-k)!} \frac{\lambda^k}{n^k} \bigg(1 - \frac{\lambda}{n}\bigg)^{n-k} \\ &= \lim_{n\to\infty} \frac{n(n-1)...(n-k+1)}{n^k} \frac{\lambda^k}{k!}\bigg(1 - \frac{\lambda}{n}\bigg)^n \bigg(1 - \frac{\lambda}{n}\bigg)^{-k} \\ &= e^{-\lambda}\frac{\lambda^k}{k!}\end{align}\] A random variable \(X\) is said to be a Poisson random variable with parameter \(\lambda > 0\) if its probability mass function has the form \[p_X(k) = e^{-\lambda}\frac{\lambda^k}{k!} \hskip6emk = 0,1,2,...\] and \(0\) otherwise, where \(\lambda\) indicates the average number of successes per unit time or space.

We denote \(X \sim \text{Pois}(\lambda)\) if \(X\) is said to follow a Poisson distribution with parameter \(\lambda\).

Proposition: Let \(X_i \sim \text{Pois}(\lambda_i)\) be independent Poisson random variables. Then \[X_1 + ... + X_m \sim \text{Pois}\bigg(\sum_{i=1}^m \lambda_i \bigg)\] Properties: * \(\text E[X] = \lambda\) * \(\text{Var}(X) = \lambda\) * \(\text E[X(X-1)...(X-k)] = \lambda^{k+1}\) * \(\varphi_X(t) = \exp[\lambda(e^{it}-1)]\) * \(G_X(z) = \exp[\lambda(z-1)]\) * Non-overlapping intervals are independent for a Poisson process

Example: The number of people entering a movie theater averages one every two minutes. Assuming that a Poisson distribution is appropriate. Calculate: * the probability that no people enter between 12:00 and 12:05. * the probability that at least 4 people enter between 12:00 and 12:05.

Solution: Let \(X\) be the number of people that enter between 12:00 and 12:05. We model \(X\) as a Poisson random variable with parameter λ, the average number of people that arrive during the period. Since on average, one person enters every two minutes, hence during the period of 5 minutes, on average \(2.5\) people enters. Thus \(\lambda = 2.5\).

The probability that no people enter between 12:00 and 12:05 is: \[P(X = 0) = e^{-2.5}\frac{2.5^0}{0!} = e^{-2.5} \approx 0.0821\] The probability that at least 4 people enter between 12:00 and 12:05 is: \[P(X \ge 4) = 1 - P(X\le 3) = 1 - e^{-2.5} \sum_{k=0}^3 \bigg(\frac{2.5^k}{k!}\bigg) \approx 0.2424\]

20.2 Poisson Approximation to the Binomial Distribution

The Poisson distribution can be used as an approximation of the binomial distribution if \(n\) is sufficiently large and \(p\) is sufficiently small. The Poisson distribution is a good approximation of the binomial distribution if \(n\) is at least \(20\) and \(p\) is smaller than or equal to \(0.05\), and an excellent approximation if \(n ≥ 100\) and \(np ≤ 10\).

Then, we can approximate \(B(n,p) \approx \text{Pois}(\lambda = np)\).