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Chapter 5. Discrete Random Variables and Probability Distributions

Unit 5.16. Poisson approximation to the binomial

The Poisson distribution can be applied directly (see plenty of examples in NCT) or indirectly, via the binomial distribution. Before taking up the last point, we look at an example.

Exercise 5.11. An insurance company holds fraudulence insurance policies on 6,000 firms. In any given year the probability that any single policy will result in a claim is 0.001. Find the probability that at least three claims are made in a given year.

Solution. (a) Using the binomial.

On the exam you may not have enough time to calculate the answer. Show that you know the material by introducing the right notation and telling the main idea.

Ideas go first!

Put $latex X = 1$ if a claim is made and $latex X = 0$ if it is not. This is a Bernoulli variable, $latex P(X = 1) = 0.001 = p$. Denote $latex {S_n}$ the corresponding binomial variable. By the additivity rule

$latex P({S_n} \ge 3) = 1 - P({S_n} \le 2) = 1 - P({S_n} = 0) - P({S_n} = 1) - P({S_n} = 2).$

From this equation and the general formula Eq. (5.15) (see Distribution of the binomial variable)

$latex P({S_n} \ge 3) \qquad (\textrm{Eq. 5.19})$

$latex = 1 - {0.999^{6000}} - \frac{{6000!}}{{1!\,5999!}}0.001 \cdot {0.999^{5999}} - \frac{{6000!}}{{2!\,5998!}}{0.001^2} \cdot {0.999^{5998}}$

$latex = 1 - {0.999^{6000}} - 6000 \cdot 0.001 \cdot {0.999^{5999}} - 3000 \cdot 5999 \cdot {0.001^2} \cdot {0.999^{5998}}$

$latex = 1 - {0.999^{6000}} - 6 \cdot {0.999^{5999}} - 3 \cdot 5.999 \cdot {0.999^{5998}}.$

You will die before you finish raising 0.999 to the power 6000 by hand. The trick is to use the properties of the exponential function and natural logarithm: $latex a = {e^{\ln a}}$ and $latex \ln {a^b} = b\ln a$ for any positive number $latex a$ and real number $latex b$. In our case

$latex {0.999^{6000}} = {e^{\ln {{0.999}^{6000}}}} = {e^{6000\ln 0.999}} = {e^{6000( - 0.0010005)}} = 0.00247.$

Eventually we get $latex P({S_n} \ge 3) = 0.93812$.

(b) Using the Poisson distribution. The distinctive feature of Eq. (5.19) is that 0.999 raised to a high power may become so small that a simple calculator will identify the result with zero. This small number is multiplied by combinations $latex C_x^n$ which, when $latex n$ is very large, may be for a simple calculator the same thing as infinity. However, the product of the two numbers, one small and another large, may be manageable.

Mathematicians are reasonably lazy. When they see such numbers as in Eq. (5.19), they prefer to develop a theory than apply brute force. The device for the problem at hand sounds as follows.

Theorem 5.1 (Poisson approximation to the binomial distribution). Suppose $latex p$ is small and $latex n$ is large but their product $latex \lambda = np$ is moderate (NCT says $latex \lambda \le 7$). Then the probability of the binomial taking the value $latex x$ can be approximated by the probability of the Poisson taking the same value:

$latex P({S_n} = x) \approx P(\textrm{Poisson} = x) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}.$

This result gives a fairly good approximation $latex P({S_n} \ge 3) = 0.938031$.

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