What is a random variable?

We often care about a number given by an experiment's outcome rather than the outcome itself, such as the total of two dice, how long a customer waits or a player's winnings.

Suppose a game pays £$2$ for each head in ten tosses. With events alone, describing the winnings needs eleven of them, "win £$0$", "win £$2$", up to "win £$20$", and there is still no way to say the winnings are twice the number of heads. If we give the number of heads a name, $H$, the winnings are $2H$. A number given by the outcome in this way is called a random variable.

It is a third kind of object on a sample space, after outcomes and events.

Consider two coin tosses, with $\Omega = \{HH, HT, TH, TT\}$, and let $X$ be the number of heads. The diagram builds up each kind of object in turn, then runs the experiment.

A card's suit is not a number, but "$1$ if the card is a heart, $0$ otherwise" assigns every card a number.

So a random variable is not an outcome, an event or a single number, but a rule that assigns a number to every outcome.

A shop's records list every basket bought last week, and one basket is chosen at random. Which of these quantities are random variables?

• The items in the basket.
• The baskets that cost over £$20$.
• The total cost of the basket.
• The number of items in the basket.

Solution

Each outcome is one of last week's baskets. For each quantity we ask what it gives each basket:

A random variable must give every basket one number, which is what the middle column checks. The total cost and the number of items vary because the basket is chosen at random: the rule is fixed, the basket is not.

We've seen that a random variable assigns each outcome a number. The value $X(s)$ is the number it assigns the outcome $s$, and we find it by applying the rule to $s$.

If $X$ counts the number of heads in two coin tosses, the outcome $HT$ has one head, so $X(HT) = 1$. Applying the rule to every outcome in the sample space $\Omega = \{HH, HT, TH, TT\}$ in turn gives a table of the values of $X$, with one row per outcome.

Several random variables can be defined on the same sample space. If $Y$ counts the number of tails, its values form a second column of the same table:

A player tosses a coin three times, scoring $1$ for each head and $-1$ for each tail. Give her total score $W$ on each outcome.

Solution

We start by listing the sample space. Each outcome records the three tosses in order, so

Next we apply the rule to each outcome. For example, $HHT$ has two heads and one tail, so

Doing the same for every outcome gives the value of $W$ on each:

Suppose we toss two fair coins, and let $X$ be the number of heads. How likely is exactly one head, that is, $X = 1$?

It is tempting to write $P(X)$ or $P(1)$, but neither makes sense. $P$ gives probabilities to events, which are sets of outcomes. $X$ is a rule that gives each outcome a number, and $1$ is a number, so neither is an event. We first have to find the event that "$X = 1$" describes. The diagram shows how.

So $\{X = 1\} = \{HT, TH\}$. Its probability is written $P(X = 1)$, without the inner braces, and read as "the probability that $X$ equals $1$".

Inequalities work the same way. $\{X \ge 1\}$ is the event of at least one head, $\{HH, HT, TH\}$, and its probability is written $P(X \ge 1)$.

Two fair dice are rolled, and $M$ is the larger of the two numbers, or their shared value if they match. Find $P(M = 2)$.

Solution

We label the dice, so each outcome is an ordered pair $(a, b)$ with die $1$ showing $a$ and die $2$ showing $b$. The dice are fair, so the $36$ outcomes are equally likely.

We first list the event $\{M = 2\}$, the outcomes that $M$ assigns the value $2$. Neither die can show more than $2$, and at least one must show $2$, so

The event holds $3$ of the $36$ equally likely outcomes, so by the naive definition of probability

So $M$ takes the value $2$ with probability $\dfrac{1}{12}$.