FRM Exam Part I · Random Variables

Probability Mass and Density Functions for FRM Part I

Q: What is the difference between a PMF and a PDF?

A PMF applies to discrete variables and gives the probability of each exact value. A PDF applies to continuous variables and gives a density. You get probability by finding the area under the PDF over an interval.

Q: How do I calculate probability from a PDF?

Integrate the PDF between the lower and upper limits of the interval. For simple shapes such as rectangles or triangles, compute the area with geometry instead. The total area under the PDF must equal 1.

Q: Can a PDF value be greater than 1?

Yes. A PDF value is a density, not a probability. Only the area under the curve must be at most 1. A PMF value cannot exceed 1.

Q: Why is the probability of an exact value zero for a continuous variable?

A single point has no width, so the area above it is zero. This is why P(X ≤ a) and P(X < a) are the same for continuous variables.

Updated 11 October 2026 · Fact-checked

A PMF gives the probability that a discrete random variable equals each exact value, and the values sum to 1. A PDF describes a continuous variable; probability is the area under the curve over an interval, found by integration, and any single point has probability zero.

Understand Probability Mass and Density Functions

A random variable assigns a number to each outcome of an uncertain event. A bond's default flag (0 or 1), the number of defaults in a portfolio, and tomorrow's loss in USD are all random variables.

A discrete random variable takes a countable set of values, such as 0, 1, 2, 3. Its probability mass function (PMF) gives P(X = x) for each value. Each probability is between 0 and 1, and all of them add up to 1. You read probabilities straight off the PMF.

A continuous random variable can take any value in a range, such as a return or an interest rate. Its probability density function (PDF), f(x), is not a probability. It is the height of a curve. Probability is the area under the curve between two points. The total area under the whole curve is 1.

This leads to a point that is tested often. For a continuous variable, P(X = a) = 0 for any single value a. So P(a ≤ X ≤ b) equals P(a < X < b). For a discrete variable, the endpoints matter, so check whether the inequality is strict.

A PDF value can exceed 1. If X is uniform on [0, 0.5], then f(x) = 2 on that range. That is fine because the area is still 1. A PMF value can never exceed 1.

Key formulas to remember

PMF definition: p(x) = P(X = x); Each p(x) ≥ 0 for a discrete variable.
PMF total probability: Σ p(x) = 1; Sum over every possible value of X. Use this to find a missing probability or constant.
Discrete probability of a set: P(a ≤ X ≤ b) = Σ p(x) for all x from a to b; Include or exclude endpoints exactly as the inequality says.
PDF non-negativity: f(x) ≥ 0; A density can exceed 1. It cannot be negative.
PDF total area: ∫ f(x) dx = 1 over all x; Use this to solve for an unknown constant in f(x).
Continuous probability: P(a ≤ X ≤ b) = ∫ f(x) dx from a to b; Area under the curve. For a rectangle or triangle, use geometry instead of integrating.
Single point: P(X = a) = 0 for continuous X; So P(X ≤ a) = P(X < a).

How to solve Probability Mass and Density Functions questions

Use this routine for any PMF or PDF question. It keeps you from mixing up the discrete and continuous rules.

1Decide whether X is discrete (countable values) or continuous (any value in a range).
2Write down the function given: a table or list for a PMF, a formula or graph for a PDF.
3If a constant or probability is missing, set the total to 1: sum the PMF, or set the PDF area equal to 1, and solve.
4Translate the question into a range: for example, X > 2 or 1 ≤ X ≤ 3. Watch strict versus non-strict signs for discrete X.
5Discrete: add the PMF values in that range. Continuous: find the area under f(x) over the range, using geometry or integration.
6Use the complement when it is shorter: P(X > a) = 1 − P(X ≤ a).
7Check that the answer is between 0 and 1 and makes sense.

Quickest way: Total-equals-one and shape shortcuts

When to use it: Use when the PDF is a flat line, a triangle or a simple polynomial, or when the PMF is a short table.

For a PMF table, use the sum-to-1 rule first to fill the missing entry.
For a flat or triangular PDF, draw it and use area = base × height (× ½ for a triangle). Skip the integral.
Find the unknown height from total area = 1.
Use the complement to cut the number of terms you add.
For a polynomial PDF, integrate term by term: ∫ x^n dx = x^(n+1) ÷ (n+1).

Common mistakes in Probability Mass and Density Functions

Reading a PDF height as a probability.
The PMF value is a probability, so students assume the PDF value is too.
Fix: For continuous X, only area is probability. The height f(x) is a density and can exceed 1.
Giving P(X = a) a positive value for a continuous variable.
It feels like a specific outcome should have some chance.
Fix: A single point has zero width and so zero area. Probability exists only over intervals.
Treating endpoints as irrelevant for a discrete variable.
The continuous rule is applied everywhere.
Fix: For discrete X, P(X < 3) and P(X ≤ 3) differ by p(3). Read the inequality carefully.
Forgetting to set total probability to 1 when finding a constant.
The question seems to give no equation, so the student does not know where to start.
Fix: Whenever a constant k is unknown, use Σ p(x) = 1 or ∫ f(x) dx = 1.
Integrating over the wrong limits.
The student integrates over the whole support instead of the interval asked about.
Fix: Mark the interval from the question on the axis first. Integrate only between those limits.
Dropping the support when integrating a PDF defined in pieces.
The formula is used outside the range where it applies.
Fix: The PDF is zero outside its stated range. Cut the interval to fit inside the support.

Worked examples

Example 1

A discrete random variable X, the number of loan defaults in a month, has P(X=0) = 0.40, P(X=1) = 0.30, P(X=2) = 0.20, and P(X=3) = k. Find P(X ≥ 2).

Show the solution

Total probability must be 1: 0.40 + 0.30 + 0.20 + k = 1.
So k = 1 − 0.90 = 0.10.
P(X ≥ 2) = P(X=2) + P(X=3) = 0.20 + 0.10.
P(X ≥ 2) = 0.30.

Answer: 0.30

Example 2

A continuous random variable X has PDF f(x) = k x on 0 ≤ x ≤ 2, and 0 elsewhere. Find P(X ≤ 1).

Show the solution

Set total area to 1: ∫ k x dx from 0 to 2 = k × (2² ÷ 2) = 2k.
2k = 1, so k = 0.5.
P(X ≤ 1) = ∫ 0.5 x dx from 0 to 1 = 0.5 × (1² ÷ 2).
P(X ≤ 1) = 0.25.

Answer: 0.25

Exam tips

First decide discrete or continuous. Many wrong options are built from using the wrong rule.
If an option says a continuous variable has a positive probability at one exact value, it is wrong.
Use the sum-to-1 or area-to-1 rule to find missing constants before anything else.
Use geometry for flat or triangular densities. It saves time over 100 questions in 4 hours.
Read inequality signs closely on discrete questions: < versus ≤ changes the answer.

Practice questions from Random Variables

Probability Mass and Density Functions: frequently asked questions

What is the difference between a PMF and a PDF?

A PMF applies to discrete variables and gives the probability of each exact value. A PDF applies to continuous variables and gives a density. You get probability by finding the area under the PDF over an interval.

How do I calculate probability from a PDF?

Integrate the PDF between the lower and upper limits of the interval. For simple shapes such as rectangles or triangles, compute the area with geometry instead. The total area under the PDF must equal 1.

Can a PDF value be greater than 1?

Yes. A PDF value is a density, not a probability. Only the area under the curve must be at most 1. A PMF value cannot exceed 1.

Why is the probability of an exact value zero for a continuous variable?

A single point has no width, so the area above it is zero. This is why P(X ≤ a) and P(X < a) are the same for continuous variables.