Beginner20 min

Random Variables

Mapping outcomes to numbers, discrete vs continuous, and the support of a random variable.

Assumes you know

Sample Spaces and the Axioms of Probability

Random Variables

Intuition first

Events are sets — "the roll is even", "the email is spam". Sets are awkward to do arithmetic with. You cannot average a set, or compute its variance, or feed it to a model.

A random variable fixes that by attaching a number to every outcome. Roll two dice: the outcome is a pair like $(3, 5)$ , but the random variable "sum" turns it into $8$ . Now you can ask for the average sum, the spread of sums, the chance the sum exceeds 9 — all numerical questions.

The key thing to get straight, and it trips up nearly everyone at first: a random variable is a function, not a number. It has no value until an outcome occurs. The notation $X = 8$ does not mean "X equals 8"; it is shorthand for the event consisting of all outcomes that the function $X$ maps to 8.

Definition

A random variable is a function from the sample space to the real numbers:

X : \Omega \to \mathbb{R}

Notation used in this lesson
Symbol	Meaning	Read aloud
X, Y, Z	Random variables — capital letters	big X
x, y, z	Particular values a random variable can take	little x
{X = x}	The event that X takes the value x	the event X equals x
P(X = x)	Probability of that event	P of X equals x
Sₓ	Support — the set of values X can actually take	support of X

A worked mapping

Toss a fair coin three times. The sample space has eight equally likely outcomes:

\Omega = \{HHH,\, HHT,\, HTH,\, HTT,\, THH,\, THT,\, TTH,\, TTT\}

Define $X$ = number of heads. The function maps:

Outcome	$X$
$HHH$	3
$HHT$ , $HTH$ , $THH$	2
$HTT$ , $THT$ , $TTH$	1
$TTT$	0

The support is $S_X = \{0, 1, 2, 3\}$ . Now collect probabilities by counting outcomes in each pre-image:

\Prob(X = 0) = \tfrac{1}{8}, \quad \Prob(X = 1) = \tfrac{3}{8}, \quad \Prob(X = 2) = \tfrac{3}{8}, \quad \Prob(X = 3) = \tfrac{1}{8}

These sum to 1, as they must, because $X$ assigns every outcome to exactly one value.

A random variable is a function from outcomes to numbers. Several outcomes can map to the same value — three sequences give X = 2 — and the probability of that value is the total probability of its pre-image.

Discrete versus continuous

Discrete — the support is finite or countably infinite. Counts, categories, number of arrivals. Individual values carry positive probability.

Continuous — the support is an interval. Heights, times, temperatures. Here something unintuitive happens: $\Prob(X = x) = 0$ for every single value.

Why continuous variables assign zero probability to every pointAdvanced

Suppose $X$ is uniform on $[0, 1]$ and every single value had the same probability $p > 0$ . The interval contains uncountably many points, so any positive $p$ would give total probability exceeding 1 — indeed exceeding any bound. So $p$ must be 0.

Yet some value occurs. Both statements are consistent because probability for continuous variables lives in intervals, not points:

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

and the integral over a single point is zero. A density $f(x)$ is therefore not a probability — it is probability per unit length, and can exceed 1 without contradiction. A uniform distribution on $[0, 0.5]$ has density 2 everywhere on that interval.

A practical consequence: for continuous variables the inequalities do not matter,

\Prob(X \leq a) = \Prob(X < a)

because the single point $a$ contributes nothing. For discrete variables they differ, and conflating them is a standard source of off-by-one errors.

Functions of random variables

If $X$ is a random variable and $g$ is a function, then $g(X)$ is also a random variable — a composition of functions is a function. This is why expressions like $X^2$ , $e^X$ and $X_1 + X_2$ make sense, and it is what makes random variables useful for building models.

Solved problem 1 · Two random variables on the same sample space

Two fair dice are rolled. Define $S$ = sum of the dice and $M$ = maximum of the two dice. Find the distribution of each, and $\Prob(S = 7)$ and $\Prob(M = 5)$ .

Step 1 — the sample space

Ordered pairs, so $\lvert \Omega \rvert = 36$ , each with probability $1/36$ .

Step 2 — outcomes giving S = 7

(1,6),\, (2,5),\, (3,4),\, (4,3),\, (5,2),\, (6,1)

Six outcomes, so

\Prob(S = 7) = \frac{6}{36} = \frac{1}{6}

Step 3 — outcomes giving M = 5

The maximum is 5 when at least one die shows 5 and neither shows 6. Count directly: pairs where the larger value is exactly 5.

(1,5),\, (2,5),\, (3,5),\, (4,5),\, (5,5),\, (5,4),\, (5,3),\, (5,2),\, (5,1)

Nine outcomes:

\Prob(M = 5) = \frac{9}{36} = \frac{1}{4}

Step 4 — a general formula for M, as a check

The maximum is at most $m$ exactly when both dice are at most $m$ , which happens in $m^2$ of the 36 outcomes. So

\Prob(M \leq m) = \frac{m^2}{36} \quad\Longrightarrow\quad \Prob(M = m) = \frac{m^2 - (m-1)^2}{36} = \frac{2m - 1}{36}

For $m = 5$ : $(2 \times 5 - 1)/36 = 9/36 \;\checkmark$

Step 5 — full distributions

\Prob(S = s) = \frac{6 - \lvert s - 7 \rvert}{36}, \qquad s = 2, \dots, 12

\Prob(M = m) = \frac{2m - 1}{36}, \qquad m = 1, \dots, 6

Check that each sums to 1. For $M$ : $1 + 3 + 5 + 7 + 9 + 11 = 36$ , so $36/36 = 1 \;\checkmark$

Answer

$\Prob(S = 7) = 1/6$ and $\Prob(M = 5) = 1/4$ . Two different random variables on the same sample space, each summarising the outcome differently — and note that $S$ and $M$ are not independent, since knowing the maximum constrains the possible sums.

Indicator random variables

The simplest and most useful special case. For an event $A$ :

\mathbb{1}_A(\omega) = \begin{cases} 1 & \text{if } \omega \in A \\ 0 & \text{otherwise}\end{cases}

Its expectation is the probability of the event, $\E[\mathbb{1}_A] = \Prob(A)$ , which turns counting problems into expectation problems. Writing a count as a sum of indicators and using linearity of expectation is one of the most powerful techniques in probability, and it appears repeatedly from here on.

python

import numpy as np

rng = np.random.default_rng(0)
rolls = rng.integers(1, 7, size=(200_000, 2))   # two dice, many trials

S = rolls.sum(axis=1)     # the random variable "sum"
M = rolls.max(axis=1)     # the random variable "maximum"

print(f"P(S=7) empirical {np.mean(S == 7):.4f}  theory {1/6:.4f}")
print(f"P(M=5) empirical {np.mean(M == 5):.4f}  theory {9/36:.4f}")

# Full distribution of M against the formula (2m-1)/36.
for m in range(1, 7):
    print(f"  P(M={m}) empirical {np.mean(M == m):.4f}  theory {(2*m - 1)/36:.4f}")

Exercise 1

A coin is tossed until the first head, at most 4 times. Let $X$ be the number of tosses. Give the distribution of $X$ and verify it sums to 1.

Show solution

The process stops at the first head, or after 4 tosses regardless.

\Prob(X = 1) = \Prob(H) = \tfrac12

\Prob(X = 2) = \Prob(TH) = \tfrac14

\Prob(X = 3) = \Prob(TTH) = \tfrac18

For $X = 4$ the toss happens whether or not it is a head, since we stop after four either way. So $X = 4$ covers both $TTTH$ and $TTTT$ :

\Prob(X = 4) = \Prob(TTT) = \tfrac18

Verify:

\tfrac12 + \tfrac14 + \tfrac18 + \tfrac18 = \tfrac{4 + 2 + 1 + 1}{8} = 1 \;\checkmark

The subtlety is the final value. A truncated geometric distribution puts the leftover probability mass on the last value, so $\Prob(X = 4) = 1/8$ rather than $1/16$ .

Exercise 2

Explain why $\Prob(X = 3)$ can be non-zero for a discrete random variable but must be zero for a continuous one, and what replaces it.

Show solution

For a discrete variable, $\{X = 3\}$ is a genuine event — a set of outcomes with positive total probability. The three-coin example has $\Prob(X = 3) = 1/8$ , the single outcome $HHH$ .

For a continuous variable the support is uncountable. If every point had positive probability $p$ , then summing over uncountably many points would exceed 1, so $p = 0$ necessarily. There is no contradiction with "some value occurs", because probability is assigned to intervals via an integral, and the integral over a degenerate interval $[3,3]$ is zero.

What replaces it is the density $f(x)$ , which gives probability per unit length:

\Prob(3 - \tfrac{\epsilon}{2} \leq X \leq 3 + \tfrac{\epsilon}{2}) \approx f(3)\,\epsilon

So $f(3)$ tells you how probable values near 3 are relative to elsewhere, and multiplying by a width gives an actual probability. Densities can exceed 1; probabilities cannot.

Next: PMF, PDF and CDF, the three standard ways to describe a random variable's distribution.