Chapter 12.5, Problem 1E

Suppose that a random variable $X$ has a Poisson distribution with $\lambda =3$, as in Example 1. Compute the probabilities $p_6,p_7,p_8$.

Example 1

A Poisson Distribution

The annual number of deaths in a certain town due to a particular disease has a Poisson distribution with parameter $\lambda =3$. Verify that the associated probabilities sum to $1$.

Solution

The first few probabilities are (to four decimal places)

$p_0=e^{-3}\approx .0498,$

$p_1=\frac{3}{1}{e}^{-3}=3\cdot p_0\approx .1494,$

$p_2=\frac{3^2}{2\cdot 1}{e}^{-3}=\frac{3\cdot 3}{2\cdot 1}{e}^{-3}=\frac{3}{2}\cdot p_1\approx .2240,$

$p_3=\frac{3^3}{3\cdot 2\cdot 1}{e}^{-3}=\frac{3\cdot 3\cdot 3}{3\cdot 2\cdot 1}{e}^{-3}=\frac{3}{3}\cdot p_2\approx .2240,$

$p_4=\frac{3^4}{4\cdot 3\cdot 2\cdot 1}{e}^{-3}=\frac{3\cdot 3\cdot 3\cdot 3}{4\cdot 3\cdot 2\cdot 1}{e}^{-3}=\frac{3}{4}\cdot p_3\approx .1680,$

$p_5=\frac{3^5}{5\cdot 4\cdot 3\cdot 2\cdot 1}{e}^{-3}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{5\cdot 4\cdot 3\cdot 2\cdot 1}{e}^{-3}=\frac{3}{5}\cdot p_4\approx \mathrm{.1008.}$

Notice how each probability $p_n$ for $n\ge 1$ is computed from the preceding probability $p_n-1$.

In general, $p_n=\left(\lambda /n\right)p_{n-1}$.

To sum the probabilities for all n, use the exact values, not the decimal approximations:

$\Rightarrow e^{-3}+\frac{3}{1}{e}^{-3}+\frac{3^2}{2\cdot 1}{e}^{-3}+\frac{3^3}{3\cdot 2\cdot 1}{e}^{-3}+\frac{3^4}{4\cdot 3\cdot 2\cdot 1}{e}^{-3}+...$

$\Rightarrow e^{-3}\Rightarrow \left(1+3+\frac{1}{2!}{3}^2+\frac{1}{3!}{3}^3+\frac{1}{4!}{3}^4+...\right)$

The series inside the parentheses is the power series for $e^x$ evaluated at $x=3$ The preceding sum is $e^{-3}\cdot e^3$, which equals $1$.