Discrete probability distributions

IB Mathematics: Analysis and Approaches Higher Level· Unit 4: Statistics & Probability· 35 min read

1. Discrete Random Variables & Probability Mass Functions★★☆☆☆⏱ 10 min

A random variable assigns a numerical value to each outcome of a random experiment. A discrete random variable can only take a countable number of distinct values (e.g., 0, 1, 2, ..., 10).

📘 Definition

Discrete Random Variable

A random variable that can only take a countable set of distinct possible values.

Example:

Number of red cards in a 5-card hand, number of defective items in a batch

📘 Definition

Probability Mass Function (pmf)

p(x) = P(X=x)

A function that satisfies two core conditions for all valid $x$ : $0 \leq p (x) \leq 1$ , and the sum of $p (x)$ over all possible values of $x$ equals 1.

📐 Worked Example

A discrete random variable $X$ has pmf $P (X = x) = k x$ for $x \in {1, 2, 3, 4}$ , and 0 otherwise. Find the value of $k$ .

1
Use the core property that the sum of all probabilities equals 1:
$\sum_{x=1}^4 P(X=x) = 1$
2
Substitute the pmf into the sum:
$k(1) + k(2) + k(3) + k(4) = 1 \implies 10k = 1$
3
Solve for $k$ :
$k = \frac{1}{10}$

Exam tip:

Always check that the sum of your pmf equals 1. This is a quick check to catch calculation errors in exams.

2. Cumulative Distribution Functions★★★☆☆⏱ 10 min

The cumulative distribution function (cdf) gives the probability that the random variable $X$ is less than or equal to a given value $x$ . It is useful for calculating probabilities of ranges of outcomes.

📘 Definition

Cumulative Distribution Function (cdf)

F(x) = P(X \leq x)

For discrete $X$ , $F (x)$ is the sum of the pmf for all values of $X$ that are less than or equal to $x$ . It is always non-decreasing and ranges from 0 to 1.

📐 Worked Example

Using the pmf $P (X = x) = \frac{x}{10}$ for $x = 1, 2, 3, 4$ , find the cumulative distribution function $F (x)$ .

1
For $x < 1$ , no values of $X$ are less than or equal to $x$ , so $F (x) = 0$ .
2
For $1 \leq x < 2$ , only $x = 1$ is included:
$F(x) = P(X=1) = \frac{1}{10}$
3
For $2 \leq x < 3$ , add the probability of $x = 2$ :
$F(x) = \frac{1}{10} + \frac{2}{10} = \frac{3}{10}$
4
For $3 \leq x < 4$ , add the probability of $x = 3$ :
$F(x) = \frac{3}{10} + \frac{3}{10} = \frac{3}{5}$
5
For $x \geq 4$ , add all probabilities:
$F(x) = \frac{6}{10} + \frac{4}{10} = 1$

Exam tip:

To find $P (a < X \leq b)$ for discrete $X$ , use the shortcut $F (b) - F (a)$ . This saves time compared to summing all individual probabilities.

3. Expected Value and Variance★★★☆☆⏱ 12 min

Expected value (or population mean) measures the center of the distribution, while variance measures the spread around the center. Both are core summary statistics for any discrete distribution.

📘 Definition

Expected Value

E[X] \text{ or } \mu

The expected value is the long-run average of $X$ over many repeated trials, calculated as: $E [X] = \sum_{x} x \cdot p (x)$ .

📘 Definition

Variance

\text{Var}(X) \text{ or } \sigma^2

Variance measures spread, calculated most efficiently with the shortcut formula:

\text{Var}(X) = E[X^2] - (E[X])^2, \quad \text{where } E[X^2] = \sum_x x^2 p(x)

📐 Worked Example

For the distribution $P (X = x) = \frac{x}{10}$ , $x \in {1, 2, 3, 4}$ , calculate $E [X]$ and $Var (X)$ .

1
Calculate expected value by summing $x \cdot p (x)$ :
$E[X] = (1 \cdot \frac{1}{10}) + (2 \cdot \frac{2}{10}) + (3 \cdot \frac{3}{10}) + (4 \cdot \frac{4}{10}) = \frac{30}{10} = 3$
2
Calculate $E [X^{2}]$ for the variance formula:
$E[X^2] = (1^2 \cdot \frac{1}{10}) + (2^2 \cdot \frac{2}{10}) + (3^2 \cdot \frac{3}{10}) + (4^2 \cdot \frac{4}{10}) = \frac{100}{10} = 10$
3
Apply the shortcut variance formula:
$\text{Var}(X) = 10 - (3)^2 = 1$

Exam tip:

Always use the $E [X^{2}] - (E [X])^{2}$ shortcut. It avoids expanding squared deviations and reduces calculation errors.

4. Linear Transformations of Discrete Random Variables★★★☆☆⏱ 8 min

Exams commonly ask for the expected value and variance of a linear transformation of a random variable, of the form $Y = a X + b$ where $a$ and $b$ are constants. There are simple rules for these calculations.

📘 Definition

Linear Transformation Rules

For any constants $a$ and $b$ , the following rules hold:

E[aX + b] = aE[X] + b \\ \text{Var}(aX + b) = a^2 \text{Var}(X)

The constant $b$ shifts the distribution but does not change its spread, so it does not affect variance.

📐 Worked Example

Given $E [X] = 3$ and $Var (X) = 1$ , find $E [2 X - 5]$ and $Var (2 X - 5)$ .

1
Apply the expected value rule:
$E[2X - 5] = 2E[X] - 5 = 2(3) - 5 = 1$
2
Apply the variance rule, squaring the coefficient 2:
$\text{Var}(2X - 5) = (2)^2 \text{Var}(X) = 4(1) = 4$

5. Common Pitfalls

Wrong move:

Forgetting to find the scaling constant $k$ to normalize the pmf

Why:

You assume the given function is already a valid pmf without checking the sum of probabilities

Correct move:

Always calculate $k$ by setting the sum of all probabilities equal to 1 before any further calculations

Wrong move:

Calculating $Var (a X + b) = a Var (X)$ instead of $a^{2} Var (X)$

Why:

Confusing the expected value rule with the variance rule

Correct move:

Remember scaling $X$ by $a$ scales the spread by $a^{2}$ , so always square the coefficient for variance

Wrong move:

Calculating $P (X < x) = F (x)$ for discrete $X$

Why:

Forgetting that $F (x)$ includes $P (X = x)$ , so it overcounts for strict inequality

Correct move:

For discrete $X$ , $P (X < x) = F (x) - P (X = x)$

Wrong move:

Treating a discrete random variable as continuous and integrating over an interval

Why:

Confusing discrete and continuous random variable probability rules

Correct move:

Always sum the probabilities of individual discrete outcomes when working with discrete distributions

Wrong move:

Ignoring a negative variance result and proceeding

Why:

You don't remember that variance can't be negative

Correct move:

A negative variance always indicates an arithmetic error. Go back and recalculate $E [X^{2}]$ and $(E [X])^{2}$

6. Quick Reference Cheatsheet

Concept	Key Formula/Rule
Discrete Random Variable	Countable distinct values
pmf $p (x) = P (X = x)$	$0 \leq p (x) \leq 1$ , $\sum p (x) = 1$
cdf $F (x)$	$P (X \leq x) = \sum_{y \leq x} p (y)$
Expected Value $E [X]$	$\sum x \cdot p (x)$
Variance $Var (X)$	$E [X^{2}] - (E [X])^{2}$ , $E [X^{2}] = \sum x^{2} p (x)$
Linear transform expectation	$E [a X + b] = a E [X] + b$
Linear transform variance	$Var (a X + b) = a^{2} Var (X)$

7. Frequently Asked

What is the difference between a pmf and a cdf?

A pmf gives the probability of $X$ being exactly equal to a value $x$ , while a cdf gives the probability of $X$ being less than or equal to $x$ . You can derive a cdf from a pmf by cumulative summation.

Can variance ever be negative?

No. Variance is the expected value of a squared deviation from the mean, so it is always non-negative. A negative variance indicates an arithmetic error in your calculation.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 2
8 mark expectation calculation
2022 · 1
6 mark pmf finding
2023 · 2
Variance of transformed variable

What's Next

Discrete probability distributions form the foundation for all named discrete distributions you will study next, and are used extensively in statistical inference later in the course. Mastery of pmf, cdf, expected value, and variance here will make working with binomial, Poisson, and other common discrete distributions much simpler. The rules for linear transformations also prepare you for working with combinations of multiple random variables, a common topic in higher-level IB exam questions.