Raising the bar

Chapter 5. Discrete Random Variables and Probability Distributions

Unit 5.15. Poisson distribution

The derivation of the Poisson distribution from primitive assumptions would be even more difficult than for the binomial variable. Therefore we introduce it axiomatically and then start looking at real-world applications.

Taylor decomposition. The functions $latex {x^0} = 1,\,\,{x^1} = x,\,\,{x^2},...$ are called monomials. Unlike marsupials, they are considered simple. The idea is to represent other functions as linear combinations of monomials. For example, a polynomial $latex p(x) = {a_0} + {a_1}x + ... + {a_n}{x^n}$ is such a combination. Linear and quadratic functions are special cases of polynomials. If infinite linear combinations, called Taylor decompositions, are allowed, then the set of functions representable using monomials becomes much wider. We are interested in the decomposition of the exponential function

$latex {e^\lambda } = 1 + \frac{1}{{1!}}\lambda + \frac{1}{{2!}}{\lambda ^2} + ... + \frac{1}{{n!}}{\lambda ^n} + ... = \sum\limits_{x = 0}^\infty {\frac{1}{{x!}}{\lambda ^x}}. \qquad (\textrm{Eq. 5.17})$

Here $latex \lambda$ is any real number (see calculus texts for more information).

Poisson distribution. Take $latex \lambda > 0$ and divide both sides of Eq. (5.17) by $latex {e^\lambda }$ to obtain

$latex 1 = {e^{ - \lambda }} + \frac{1}{{1!}}\lambda {e^{ - \lambda }} + \frac{1}{{2!}}{\lambda ^2}{e^{ - \lambda }} + ... + \frac{1}{{n!}}{\lambda ^n}{e^{ - \lambda }} + ... = \sum\limits_{x = 0}^\infty {\frac{1}{{x!}}{\lambda ^x}{e^{ - \lambda }}}.$

We see that the quantities $latex {p_x} = \frac{1}{{x!}}{\lambda ^x}{e^{ - \lambda }}$ satisfy the probability postulates (nonnegativity and completeness axiom). Hence, we can define a random variable $latex X$ by

$latex P(X = x) = \frac{1}{{x!}}{\lambda ^x}{e^{ - \lambda }},\,\,x = 0,\,\,1,\,\,2,... \qquad (\textrm{Eq. 5.18})$

This variable is called a Poisson distribution.

Comments. (1) This is a discrete random variable with an infinite number of values. An application should be a variable which can potentially take nonnegative integer values, including very large ones. Normally, a mobile network operator, such as Verizon, has a very large number of subscribers. Let $latex X$ denote the number of all customers of Verizon who use their mobile service at a given moment. It is random, it takes nonnegative integer values and the number of customers using their mobile phones simultaneously can be very large. Finally, decisions by customers to use their cell phones can be considered independent, unless there is a disaster.

(2) Compare this definition with what is written in NCT. The section "Assumptions of the Poisson Probability Distribution" is an attempt to show how this distribution is derived from primitive assumptions. Since all the accompanying logic has been cut off, the result is a complete mystery.

Exercise 5.10. Calculate the mean and variance of the Poisson distribution.

Solution. In the definition of the mean the sum is, naturally, extended over all nonnegative integers:

$latex EX = \sum\limits_{x = 0}^\infty {x{p_x}} = \sum\limits_{x = 0}^\infty {x\frac{1}{{x!}}{\lambda ^x}{e^{ - \lambda }}}$

(the term with $latex x = 0$ is 0, so summation starts with $latex x = 1$)

$latex = \sum\limits_{x = 1}^\infty {\frac{1}{{(x - 1)!}}{\lambda ^x}{e^{ - \lambda }}}$

(we pull out $latex \lambda$, to obtain $latex x - 1$ in the power and factorial)

$latex = \lambda \sum\limits_{x = 1}^\infty {\frac{1}{{(x - 1)!}}{\lambda ^{x - 1}}{e^{ - \lambda }}}$

(replace $latex y = x - 1$, to get Poisson probabilities)

$latex = \lambda \sum\limits_{y = 0}^\infty {\frac{1}{{y!}}{\lambda ^y}{e^{ - \lambda }}} = \lambda \sum\limits_{y = 0}^\infty {{p_y}} = \lambda.$

With the view to apply the shortcut for variance, we calculate first

$latex E{X^2} = \sum\limits_{x = 0}^\infty {{x^2}{p_x}} = \sum\limits_{x = 0}^\infty {{x^2}\frac{1}{{x!}}{\lambda ^x}{e^{ - \lambda }}} = \sum\limits_{x = 1}^\infty {x\frac{1}{{(x - 1)!}}{\lambda ^x}{e^{ - \lambda }}}$

(can cancel out only one $latex x$, so we replace $latex x = x - 1 + 1$)

$latex = \sum\limits_{x = 1}^\infty {(x - 1 + 1)\frac{1}{{(x - 1)!}}{\lambda ^x}{e^{ - \lambda }}}$

$latex = \sum\limits_{x = 1}^\infty {(x - 1)\frac{1}{{(x - 1)!}}{\lambda ^x}{e^{ - \lambda }}} + \sum\limits_{x = 1}^\infty {\frac{1}{{(x - 1)!}}{\lambda ^x}{e^{ - \lambda }}}$

(the idea is again to obtain Poisson probabilities)

$latex = {\lambda ^2}\sum\limits_{x = 2}^\infty {\frac{1}{{(x - 2)!}}{\lambda ^{x - 2}}{e^{ - \lambda }}} + \lambda \sum\limits_{x = 1}^\infty {\frac{1}{{(x - 1)!}}{\lambda ^{x - 1}}{e^{ - \lambda }}} = {\lambda ^2} + \lambda.$

As a result, $latex V(X) = E{X^2} - {(EX)^2} = {\lambda ^2} + \lambda - {\lambda ^2} = \lambda$.

Conclusion. The Poisson distribution is special in that its mean and variance are the same, $latex EX = V(X) = \lambda$. This is another indication as to where it can be applied.

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