Poisson Distribution

Definition

1.The Poisson distribution is the discrete probability distribution of a count observed in a fixed region of time or space. It is used when counts in disjoint subregions are independent, the average rate is constant, and two or more occurrences in a sufficiently small subregion are unlikely. Examples include the number of calls received in ten minutes, radioactive particles detected in one second, or defects found along one meter of wire.

2.A discrete random variable $X$ has a Poisson distribution with parameter $\lambda>0,$ written $X\sim\mathrm{Poisson}(\lambda),$ if its probability mass function is:

$$ P(X=x)=p(x;\lambda)=\frac{e^{-\lambda}\lambda^x}{x!},\qquad x=0,1,2,\ldots $$

3.In this expression $x$ is the observed number of occurrences, $\lambda$ is the expected number of occurrences in the fixed interval, $e$ is Euler's number, and $x!$ is the factorial of $x.$ Thus $\lambda$ is the mean count in the chosen interval. When a process has a rate of $\rho$ occurrences per unit of time, the count in an interval of length $t$ has parameter $\lambda=\rho t.$

4.The probability mass function has non-negative values. Its values also sum to $1,$ as required for every discrete probability distribution. The Taylor series of the exponential function gives:

$$ \sum_{x=0}^{\infty}P(X=x)=e^{-\lambda}\sum_{x=0}^{\infty}\frac{\lambda^x}{x!}=e^{-\lambda}e^{\lambda}=1 $$

The Poisson process

5.A homogeneous Poisson process describes the times at which occurrences accumulate. Let $\rho>0$ be its intensity, measured in occurrences per unit of time, and let $R_1,R_2,\ldots$ be independent waiting times with the same exponential distribution:

$$ P(R_i>s)=e^{-\rho s},\qquad s\geq 0 $$

6.The time of the $n$th occurrence is the sum of the first $n$ waiting times:

$$ T_0=0,\qquad T_n=R_1+R_2+\cdots+R_n $$

7.For $t\geq0,$ the counting random variable is:

$$ N(t)=\max\{\ n\geq0\mid T_n\leq t\ \} $$

8.Thus $N(t)$ is the number of occurrences up to time $t.$ Its distribution is Poisson with parameter $\rho t:$

$$ N(t)\sim\mathrm{Poisson}(\rho t),\qquad P(N(t)=x)=\frac{e^{-\rho t}(\rho t)^x}{x!} $$

9.Counts in disjoint time intervals are independent. More precisely, for $0\leq s

$$ N(t)-N(s)\sim\mathrm{Poisson}(\rho(t-s)) $$

10.The distribution of an increment therefore depends only on the length $t-s,$ not on the position of the interval. The process has stationary and independent increments. For a short interval of length $h,$ the Poisson increment formula gives:

$$ \begin{align} P(N(t+h)-N(t)=1)&=\rho h+o(h) \\[6pt] P(N(t+h)-N(t)\geq2)&=o(h) \end{align} $$

11.Here $o(h)$ denotes a quantity whose ratio to $h$ tends to zero as $h\to0,$ in the sense of little-o notation. These relations state precisely that the rate is $\rho$ and that two or more occurrences in a very short interval have negligible probability compared with the interval length.

12.For example, suppose that $N(t)$ has intensity $\rho=4.$ To find the probability of three occurrences by time $2$ and four by time $5,$ write the event in terms of counts on disjoint intervals:

$$ \{\ N(2)=3,N(5)=4\ \}=\{\ N(2)=3,N(5)-N(2)=1\ \} $$

13.The first count has parameter $4\cdot2=8,$ while the second has parameter $4\cdot(5-2)=12.$ Independence gives:

$$ \begin{align} P(N(2)=3,N(5)=4) &=P(N(2)=3)P(N(5)-N(2)=1) \\[6pt] &=\frac{e^{-8}8^3}{3!}\frac{e^{-12}12}{1!} \\[6pt] &\approx0.0000021 \end{align} $$

Main properties

14.For $X\sim\mathrm{Poisson}(\lambda),$ the probability mass function, expected value, variance, and standard deviation are given below.

16.The equality $E(X)=\mathrm{Var}(X)$ is a necessary condition for an exact Poisson model. A persistent difference between the sample mean and sample variance can indicate that the Poisson assumptions do not fit the occurrence mechanism.

Mean of the Poisson distribution

17.The expected value of $X$ is defined by:

$$ E(X)=\sum_{x=0}^{\infty}xP(X=x) $$

18.Substitution of the probability mass function gives:

$$ E(X)=\sum_{x=0}^{\infty}x\frac{e^{-\lambda}\lambda^x}{x!} $$

19.The term with $x=0$ is zero. For every $x\geq1,$ the identity $x/x!=1/(x-1)!$ gives:

$$ E(X)=e^{-\lambda}\lambda\sum_{x=1}^{\infty}\frac{\lambda^{x-1}}{(x-1)!} $$

20.With $k=x-1,$ the exponential series yields:

$$ E(X)=e^{-\lambda}\lambda\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}=e^{-\lambda}\lambda e^{\lambda}=\lambda $$

21.Thus the expected count in the fixed interval is the parameter $\lambda.$

Variance of the Poisson distribution

22.The variance can be computed from the second moment by using:

$$ \mathrm{Var}(X)=E(X^2)-[E(X)]^2 $$

23.The identity $x^2=x(x-1)+x$ separates the second moment into two sums:

$$ E(X^2)=\sum_{x=0}^{\infty}x(x-1)\frac{e^{-\lambda}\lambda^x}{x!}+\sum_{x=0}^{\infty}x\frac{e^{-\lambda}\lambda^x}{x!} $$

24.The second sum is $E(X)=\lambda.$ In the first sum the terms with $x=0$ and $x=1$ are zero. Cancelling $x(x-1)$ against $x!$ and setting $k=x-2$ gives:

$$ \sum_{x=2}^{\infty}\frac{e^{-\lambda}\lambda^x}{(x-2)!}=\lambda^2e^{-\lambda}\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}=\lambda^2 $$

25.Hence $E(X^2)=\lambda^2+\lambda,$ and therefore:

$$ \mathrm{Var}(X)=\lambda^2+\lambda-\lambda^2=\lambda $$

26.The standard deviation is the positive square root $\sqrt{\lambda}.$

Cumulative Poisson distribution

27.For a non-negative integer $r,$ the cumulative distribution function gives the probability of at most $r$ occurrences:

$$ F(r;\lambda)=P(X\leq r)=e^{-\lambda}\sum_{x=0}^{r}\frac{\lambda^x}{x!} $$

28.The same function gives upper-tail and interval probabilities. For integers $1\leq a\leq b,$ we have:

$$ \begin{align} &P(X\geq a)=1-F(a-1;\lambda) \\[6pt] &P(a\leq X\leq b)=F(b;\lambda)-F(a-1;\lambda) \end{align} $$

29.Cumulative Poisson tables list precomputed values of $F(r;\lambda)$ for selected values of $r$ and $\lambda.$ A small portion of such a table is shown below.

30.The entry in row $\lambda$ and column $r$ is $P(X\leq r).$ Blank cells in this shortened table correspond to values that round to $1.000$ at three decimal places.

Computing a Poisson probability

31.Suppose a call center receives an average of five calls every ten minutes. We find the probability that exactly eight calls arrive during a ten-minute interval, assuming independent arrivals at a constant rate.

32.Let $X$ count the calls received in ten minutes. Then $X\sim\mathrm{Poisson}(5),$ and direct substitution gives:

$$ P(X=8)=\frac{e^{-5}5^8}{8!}=\frac{e^{-5}390625}{40320}\approx0.0653 $$

33.The probability is approximately $0.0653,$ or $6.53\%.$

34.The same probability can be recovered from a cumulative table. For $\lambda=5,$ suppose the table gives:

$$ \begin{align} P(X\leq8)&=0.9319 \\[6pt] P(X\leq7)&=0.8666 \end{align} $$

35.The probability of exactly eight calls is the difference between these cumulative probabilities:

$$ P(X=8)=P(X\leq8)-P(X\leq7)=0.9319-0.8666=0.0653 $$

From the binomial to the Poisson distribution

36.The Poisson distribution is the limit of the binomial distribution when the number of trials tends to infinity and the probability of success tends to zero while the expected number of successes remains fixed. Divide a fixed interval into $n$ subintervals and assume that:

38.Under these assumptions the total count $X$ has a binomial distribution with parameters $n$ and $p.$ For a fixed non-negative integer $x,$ its probability is:

$$ P(X=x)=\binom{n}{x}p^x(1-p)^{n-x} $$

39.After substituting $p=\lambda/n$ and expanding the binomial coefficient, the probability becomes:

$$ \begin{align} P(X=x) &=\binom{n}{x}\left(\frac{\lambda}{n}\right)^x\left(1-\frac{\lambda}{n}\right)^{n-x} \\[6pt] &=\frac{n!}{x!(n-x)!}\frac{\lambda^x}{n^x}\left(1-\frac{\lambda}{n}\right)^n\left(1-\frac{\lambda}{n}\right)^{-x} \end{align} $$

40.As $n\to\infty,$ with $x$ and $\lambda$ fixed, the factors that depend on $n$ have the limits:

$$ \frac{n!}{n^x(n-x)!}\to1,\qquad \left(1-\frac{\lambda}{n}\right)^n\to e^{-\lambda},\qquad \left(1-\frac{\lambda}{n}\right)^{-x}\to1 $$

41.It follows that:

$$ \lim_{n\to\infty}P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!} $$

42.For finite $n,$ this limit supports the Poisson approximation to a binomial count when $n$ is large, $p$ is small, and $np=\lambda.$