Random variables (discrete vs continuous)

What is a random variable?

A random variable is a variable whose value is determined by a random process. Rather than having a fixed value, it takes on different values with certain probabilities.

Think of rolling a die. Before you roll, the outcome is uncertain — it could be 1, 2, 3, 4, 5, or 6. A random variable $X$ represents that uncertain outcome. We write $P(X = 3)$ to mean "the probability that the die lands on 3."

Random variables come in two types: discrete and continuous.

Discrete random variables

A discrete random variable takes on a countable set of values — usually whole numbers.

Examples:

• The number of heads in 3 coin flips: $X \in \{0, 1, 2, 3\}$
• The number of customers arriving in an hour: $X \in \{0, 1, 2, \ldots\}$
• The result of rolling a die: $X \in \{1, 2, 3, 4, 5, 6\}$

For a discrete random variable, we describe its behavior with a probability mass function (PMF): a list of probabilities for each possible value. The probabilities must sum to 1:

$\sum_x P(X = x) = 1$

Example — fair die:

Continuous random variables

A continuous random variable can take any value within a range — not just whole numbers.

Examples:

• A person's height: any value in, say, $[1.4,\, 2.2]$ metres
• The time until the next bus arrives: any value $\geq 0$
• The temperature at noon tomorrow

Because there are infinitely many possible values, the probability of any single exact value is zero: $P(X = 1.756\ldots) = 0$. Instead, we talk about probability over intervals.

For continuous random variables we use a probability density function (PDF) $f(x)$. The probability that $X$ falls between $a$ and $b$ is the area under the curve:

$P(a \leq X \leq b) = \int_a^b f(x)\, dx$

The total area under the PDF equals 1:

$\int_{-\infty}^{\infty} f(x)\, dx = 1$

Expected value

For both types, the expected value $E[X]$ is the long-run average — the value you would expect on average over many repetitions.

For a discrete variable:

$E[X] = \sum_x x \cdot P(X = x)$

For the fair die: $E[X] = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + \cdots + 6 \cdot \frac{1}{6} = 3.5$

Quick comparison