12.1  Poisson Distributions
12.1  Poisson DistributionsSituation
Let the discrete random variable \(X\) denote the number of times an event occurs in an interval of time (or space). Then \(X\) may be a Poisson random variable with \(x=0, 1, 2, \ldots\)
Examples 121
 Let \(X\) equal the number of typos on a printed page. (This is an example of an interval of space — the space being the printed page.)
 Let \(X\) equal the number of cars passing through the intersection of Allen Street and College Avenue in one minute. (This is an example of an interval of time — the time being one minute.)
 Let \(X\) equal the number of Alaskan salmon caught in a squid driftnet. (This is again an example of an interval of space — the space being the squid driftnet.)
 Let \(X\) equal the number of customers at an ATM in 10minute intervals.
 Let \(X\) equal the number of students arriving during office hours.
 Poisson Random Variable

If \(X\) is a Poisson random variable, then the probability mass function is:
\(f(x)=\dfrac{e^{\lambda} \lambda^x}{x!}\)
for \(x=0, 1, 2, \ldots\) and \(\lambda>0\), where \(\lambda\) will be shown later to be both the mean and the variance of \(X\).
Recall that the mathematical constant e is the unique real number such that the value of the derivative (slope of the tangent line) of the function \(f(x)=e^x\) at the point \(x=0\) is equal to 1. It turns out that the constant is irrational, but to five decimal places, it equals:
\(\mathbf{e} = 2.71828\)
Also, note that there are (theoretically) an infinite number of possible Poisson distributions. Any specific Poisson distribution depends on the parameter \(\lambda\).
"Derivation" of the p.m.f.
Let \(X\) denote the number of events in a given continuous interval. Then \(X\) follows an approximate Poisson process with parameter \(\lambda>0\) if:
 The number of events occurring in nonoverlapping intervals are independent.
 The probability of exactly one event in a short interval of length \(h=\frac{1}{n}\) is approximately \(\lambda h = \lambda \left(\frac{1}{n}\right)=\frac{\lambda}{n}\).
 The probability of exactly two or more events in a short interval is essentially zero.
With these conditions in place, here's how the derivation of the p.m.f. of the Poisson distribution goes:
Now, let's make the intervals even smaller. That is, take the limit as \(n\) approaches infinity \(n\rightarrow \infty\) for fixed \(x\). Doing so, we get: