Modelling a numerical result

Random variables and their distribution

So far the outcomes were objects: a die face, a card, a pair of tokens. A random variable attaches a number to each outcome — a gain, a duration, a count — so that it can be averaged, compared, measured.

The definition

A random variable $X$ is a function assigning a real number to each outcome of the sample space.

$X : \Omega \longrightarrow \mathbb{R}$

$(1;1) \longmapsto 2 \qquad (1;2) \longmapsto 3 \qquad \dots \qquad (6;6) \longmapsto 12$

When two dice are rolled, $X$ assigns their sum: the sample space $\Omega$ consists of the thirty-six pairs of faces, and $X$ sends them to the eleven integers from $2$ to $12$ .

It is written with a capital letter ( $X$ , $Y$ ), and the values it takes with lower-case ones ( $x$ , $k$ ). The event « $X$ takes the value $k$ » is written $(X = k)$ .

Experiment	A possible random variable
rolling two dice	$X$ = the sum of the two faces
playing a gambling game	$X$ = the net gain (negative if a loss)
surveying 100 people	$X$ = the number of « yes » answers
measuring parts	$X$ = the number of defective parts

The distribution of X

Giving the distribution of $X$ means stating, for each possible value, the probability of obtaining it. It is presented as a table.

Rolling a fair die, $X$ = the result:

$k$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = k)$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$

The total is indeed $6 \times \tfrac{1}{6} = 1$ .

Rolling two dice, $X$ = the sum:

$k$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
$P(X = k)$	$\dfrac{1}{36}$	$\dfrac{2}{36}$	$\dfrac{3}{36}$	$\dfrac{4}{36}$	$\dfrac{5}{36}$	$\dfrac{6}{36}$	$\dfrac{5}{36}$	$\dfrac{4}{36}$	$\dfrac{3}{36}$	$\dfrac{2}{36}$	$\dfrac{1}{36}$

The most likely value is $k = 7$ , with $P(X = 7) = \tfrac{6}{36} = \tfrac{1}{6}$ : it is the sum that can be reached in the greatest number of ways — six pairs of faces produce it, against a single one for $2$ and a single one for $12$ .

The systematic check: the probabilities in the row must add up to exactly $1$ . If they do not, the distribution is wrong, and everything that follows will be too.

The graphical display

This bar chart shows at a glance where the probability concentrates — here around $7$ , with perfect symmetry.

An example with gains

A game costs €2 per round. A die is rolled: you win €9 on a 6, nothing otherwise. Let $X$ be the net gain, that is, winnings minus stake:

Die result	$6$	$1, 2, 3, 4, 5$
Winnings	€9	€0
Net gain	$9 - 2 = 7$ €	$0 - 2 = -2$ €

Hence the distribution of $X$ :

$k$	$-2$	$7$
$P(X = k)$	$\dfrac{5}{6}$	$\dfrac{1}{6}$

The total is $\tfrac{5}{6} + \tfrac{1}{6} = 1$ .

Modelling the gain net of the stake — hence negative on a loss — is what will let us say, in the next lesson, whether the game is fair.

Summary