Discrete Random Variables — CAPE Applied notes

Unit 1 built discrete random variables from scratch: a table of probabilities, then $E(X)$ and $\operatorname{Var}(X)$ from that table. This topic is about the situations that come up so often they have named distributions, with formulae you can quote instead of tabulating.

Named discrete distributions

Distribution	Situation	$P(X = x)$	$E(X)$	$\operatorname{Var}(X)$
Binomial $B(n,p)$	$x$ successes in $n$ fixed trials	$\binom{n}{x}p^x q^{n-x}$	$np$	$npq$
Geometric $\text{Geo}(p)$	the first success is on trial $x$	$q^{x-1}p$	$\dfrac1p$	$\dfrac{q}{p^2}$
Poisson $\text{Po}(\lambda)$	$x$ events in a fixed interval	$\dfrac{e^{-\lambda}\lambda^x}{x!}$	$\lambda$	$\lambda$

with $q = 1 - p$ throughout.

The geometric distribution

Trials are repeated until the first success. $X$ is the trial number on which it occurs, so $X = 1, 2, 3, \dots$ with no upper limit.

$P(X = x) = q^{x-1}p$

The reasoning is direct: $x - 1$ failures, then a success.

15% of tyres from a production line are faulty. Tyres are tested one at a time until a faulty one is found.

$P(X = 4) = 0.85^3 \times 0.15 = 0.0921$

$E(X) = \frac{1}{0.15} = 6.67 \qquad \operatorname{Var}(X) = \frac{0.85}{0.15^2} = 37.8$

The tail probability has a shortcut worth memorising — "more than $n$ trials" means the first $n$ all failed:

$P(X > n) = q^n \qquad\text{so}\qquad P(X > 5) = 0.85^5 = 0.4437$

and therefore $P(X \le 3) = 1 - 0.85^3 = 0.3859$ .

The Poisson distribution

Events occur at random, independently, at a constant average rate $\lambda$ per interval. Then

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

A switchboard receives calls at an average of 3 per five-minute period.

$P(X = 2) = \frac{e^{-3}\,3^2}{2!} = 0.2240$

$P(X \ge 1) = 1 - P(X = 0) = 1 - e^{-3} = 0.9502$

$P(X \le 2) = e^{-3}\left(1 + 3 + \tfrac92\right) = 0.4232$

Changing the interval, and adding variables

Two properties make the Poisson unusually convenient.

Scaling. If events occur at $\lambda$ per interval, then over $k$ intervals the count is $\text{Po}(k\lambda)$ . Three five-minute periods give $\lambda = 9$ :

$P(X = 10) = \frac{e^{-9}\,9^{10}}{10!} = 0.1186$

Addition. If $X \sim \text{Po}(\lambda_1)$ and $Y \sim \text{Po}(\lambda_2)$ are independent, then

$X + Y \sim \text{Po}(\lambda_1 + \lambda_2)$

Emails arriving at 4 per hour and texts at 6 per hour give a combined $\text{Po}(10)$ , so $P(\text{8 messages in an hour}) = e^{-10}10^8/8! = 0.1126$ .

Poisson as an approximation to the binomial

When $n$ is large and $p$ is small, $B(n,p) \approx \text{Po}(np)$ . The usual working rule is $n > 50$ and $np < 5$ .

A component has a 1% failure rate; 200 are installed.

Here $\lambda = np = 2$ , so

$P(X = 3) \approx \frac{e^{-2}2^3}{3!} = 0.1804$

The exact binomial value is $0.1814$ — an error of about half a percent, for a fraction of the arithmetic.

$P(X \ge 2) = 1 - e^{-2}(1 + 2) = 0.5940$

Choosing the right model

Before writing anything down, ask three questions.

Is there a fixed number of trials? Yes → binomial.
Are you counting trials until the first success? Yes → geometric.
Are you counting events in an interval, with no natural maximum? Yes → Poisson.