FRM Exam Part I · Fundamentals of Probability

Random Variables and Probability Functions for FRM Part I

Updated 11 October 2026 · Fact-checked

A random variable assigns a number to each outcome of an uncertain event. A discrete variable uses a PMF, which gives P(X = x). A continuous variable uses a PDF, where probability is the area under the curve. The CDF gives P(X ≤ x) for both. Solve questions by choosing the right function, then summing or integrating.

Understand Random Variables and Probability Functions

A random variable is a rule that turns an uncertain outcome into a number. Think of the daily loss on a portfolio, the number of defaults in a pool of 50 loans, or tomorrow's EUR/USD return. Before the event, you do not know the number. After it, you do.

A discrete random variable takes countable values, such as 0, 1, 2, 3 defaults. A continuous random variable can take any value in a range, such as a return of 1.37% or a loss of 2.4 million USD. Between any two values there are infinitely many others.

The probability mass function (PMF) works for discrete variables. It gives P(X = x) directly. Each value is between 0 and 1, and all values add up to 1.

The probability density function (PDF) works for continuous variables. A single point has probability zero, so f(x) is not a probability. It is a density. You get probability from the area under the curve over an interval. The total area is 1. A density can be greater than 1 when the range is narrow.

The cumulative distribution function (CDF) is F(x) = P(X ≤ x). It exists for both types, never decreases, and runs from 0 to 1. For a discrete variable it is a step function. For a continuous variable it is smooth. The CDF links to risk: VaR is a quantile, which is found by inverting the CDF.

Key formulas to remember

PMF conditions: 0 ≤ p(x) ≤ 1 and Σ p(x) = 1; p(x) = P(X = x). Applies to discrete variables only.
PDF conditions: f(x) ≥ 0 and ∫ f(x) dx = 1 over all x; f(x) is a density, not a probability. It can exceed 1.
Probability from a PDF: P(a ≤ X ≤ b) = ∫ from a to b of f(x) dx; For continuous X, P(X = a) = 0, so including or excluding endpoints does not matter.
CDF definition: F(x) = P(X ≤ x); Discrete: sum of p over values ≤ x. Continuous: integral of f up to x.
Interval probability from CDF: P(a < X ≤ b) = F(b) − F(a); For discrete X, endpoint a is excluded and b is included, so check the inequality signs.
Upper tail: P(X > x) = 1 − F(x); Used for loss exceedance and tail questions.
PDF from CDF: f(x) = dF(x)/dx; For continuous variables where F is differentiable.
Quantile: x_α = F⁻¹(α), where F(x_α) = α; VaR at confidence c is the quantile of the loss distribution at α = c.

How to solve Random Variables and Probability Functions questions

Use this sequence for any question on random variables, PMFs, PDFs and CDFs.

1Decide whether the variable is discrete (countable values) or continuous (any value in a range).
2Name the function you are given: PMF, PDF or CDF. Note what it measures: a probability, a density, or a cumulative probability.
3Rewrite the question as a probability statement, such as P(X ≤ 2), P(1 < X ≤ 3) or P(X > 4).
4Convert to the function you have. With a PMF, add the matching probabilities. With a PDF, find the area. With a CDF, use F(b) − F(a) or 1 − F(x).
5For discrete variables, check each inequality (< versus ≤) and include or exclude that value correctly.
6If a constant is unknown, use total probability = 1 (sum or area) to solve for it first.
7Sanity check: the answer must be between 0 and 1, and the CDF must not decrease.

Quickest way: Draw the line, shade the region

When to use it: Use for multiple-choice questions with a small PMF table, a CDF table or a simple PDF such as a uniform or triangle.

List the possible values on a number line.
Shade the values the question asks about.
Discrete: add the shaded PMF values. Or use F(top) − F(just below the bottom value).
Continuous with a flat or straight-line PDF: use geometry (rectangle or triangle area) instead of integrating.
Use 1 − F(x) for upper-tail questions.
Eliminate options outside 0 to 1 and check that all PMF values sum to 1.

Common mistakes in Random Variables and Probability Functions

Treating the PDF value f(x) as a probability.
The PMF gives probabilities directly, so the two functions get mixed up.
Fix: For continuous variables, probability is always an area. Only the PMF value at a point is a probability.
Saying P(X = x) is positive for a continuous variable.
It feels natural that a specific outcome has some chance.
Fix: For a continuous variable, P(X = x) = 0. Only intervals have positive probability.
Using F(b) − F(a) for a discrete variable when the question says P(a ≤ X ≤ b).
The formula is learned for continuous variables, where endpoints do not matter.
Fix: For discrete X, P(a ≤ X ≤ b) = F(b) − F(a − 1 step), that is, F(b) minus the CDF at the value just below a.
Forgetting that the total must equal 1 when finding an unknown constant.
Students try to match the formula to an answer option instead of using the condition.
Fix: Set the sum or integral equal to 1 and solve for the constant first.
Reading the CDF as a density or thinking it can decrease.
Graphs of PDF and CDF are confused.
Fix: The CDF is cumulative, so it only rises or stays flat, from 0 to 1. The PDF is its slope.
Using F(x) for P(X > x).
Students forget the CDF measures the lower tail.
Fix: Write P(X > x) = 1 − F(x) before calculating.

Worked examples

Example 1

The number of loan defaults X in a small portfolio has this PMF: P(X=0) = 0.50, P(X=1) = 0.30, P(X=2) = 0.15, P(X=3) = 0.05. What is P(1 ≤ X ≤ 2), and what is F(2)?

Show the solution

The variable is discrete, so add PMF values.
Check the total: 0.50 + 0.30 + 0.15 + 0.05 = 1.00. The PMF is valid.
P(1 ≤ X ≤ 2) = P(X=1) + P(X=2) = 0.30 + 0.15 = 0.45.
F(2) = P(X ≤ 2) = 0.50 + 0.30 + 0.15 = 0.95, or equivalently 1 − P(X=3) = 0.95.

Answer: P(1 ≤ X ≤ 2) = 0.45 and F(2) = 0.95.

Example 2

A loss L (in USD million) has PDF f(l) = k for 0 ≤ l ≤ 4, and f(l) = 0 elsewhere. Find k, then P(L > 3) and the CDF value F(1).

Show the solution

The area under the PDF must equal 1. The area is a rectangle: k × 4 = 1, so k = 0.25.
P(L > 3) is the rectangle from 3 to 4: 0.25 × (4 − 3) = 0.25.
F(1) = P(L ≤ 1) = 0.25 × 1 = 0.25.
Check: P(L > 3) = 1 − F(3) = 1 − 0.75 = 0.25. This matches.

Answer: k = 0.25, P(L > 3) = 0.25 and F(1) = 0.25.

Exam tips

First identify discrete or continuous. Many wrong options come from using the wrong function.
For discrete CDF questions, read the inequality signs carefully. This is where marks are lost.
For simple PDFs, use geometry for area instead of integration. It is faster and less error-prone.
Link CDF questions to VaR: a 95% VaR is the loss level x where F(x) = 0.95.
If a PMF or PDF has an unknown constant, solve it first using total probability = 1.

Practice questions from Fundamentals of Probability

Random Variables and Probability Functions in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

CA FoundationRandom Variables and Probability Distributions

Random Variables and Probability Functions: frequently asked questions

What is the difference between PMF and 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 CDF?

Use P(a < X ≤ b) = F(b) − F(a) and P(X > x) = 1 − F(x). For a discrete variable, check whether the endpoints are included before subtracting.

Can a PDF be greater than 1?

Yes. The PDF is a density, not a probability. Only the total area under it must equal 1. A narrow range can produce density values above 1.

Is the probability of an exact value zero?

For a continuous random variable, yes: P(X = x) = 0. For a discrete random variable, the PMF can give a positive probability at each possible value.