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FRM Exam Part I · Multivariate Random Variables

Joint and Marginal Probability Distributions Explained

Updated 11 October 2026 · Fact-checked

A joint distribution gives the probability of two random variables taking values together. A marginal distribution describes one variable alone. To get it, sum the joint probabilities over the other variable (discrete) or integrate the joint density over it (continuous). Check that the total equals 1.

Understand Joint and Marginal Probability Distributions

A single random variable has one distribution. Risk questions usually involve two or more variables at once, such as the return on a stock and the return on a bond. A joint distribution tells you how likely each combination of values is.

For discrete variables, the joint probability mass function is f(x, y) = P(X = x, Y = y). You can show it as a table. Rows are values of X and columns are values of Y. Every cell is between 0 and 1, and all cells together add up to 1.

For continuous variables, the joint probability density function f(x, y) is not a probability itself. Probability is the volume under the surface over a region. You find it with a double integral. The total volume over all values is 1.

The marginal distribution of X ignores Y. In a table, you add across a row to get P(X = x). You add down a column to get P(Y = y). These totals sit in the margins, which is where the name comes from. For continuous variables, you integrate out the other variable.

Once you have the marginals, you can compare them with the joint. If every joint probability equals the product of the marginals, the variables are independent. If even one cell fails the test, they are not independent.

Key formulas to remember

Joint PMF (discrete)
f(x, y) = P(X = x, Y = y); f(x, y) ≥ 0; Σx Σy f(x, y) = 1
All cells of the table must add to 1.
Marginal PMF
fX(x) = Σy f(x, y); fY(y) = Σx f(x, y)
Sum across the other variable.
Joint PDF (continuous)
f(x, y) ≥ 0; ∫∫ f(x, y) dx dy = 1 over the whole range
Probabilities come from integrating over a region, not from the value of f.
Marginal PDF
fX(x) = ∫ f(x, y) dy; fY(y) = ∫ f(x, y) dx
Integrate over the full range of the other variable. Watch the limits.
Independence
f(x, y) = fX(x) × fY(y) for all x, y
It must hold for every pair, not just some.
Conditional distribution
f(x | y) = f(x, y) ÷ fY(y), for fY(y) > 0
Links the joint to the marginal.

How to solve Joint and Marginal Probability Distributions questions

Use this routine for any joint or marginal distribution question.

  1. 1Identify whether the variables are discrete or continuous.
  2. 2Check the joint distribution is valid: non-negative and totals 1. If a constant is unknown, solve for it first using total = 1.
  3. 3For a table, add each row and each column to get the marginals. For a density, integrate out the other variable over its correct limits.
  4. 4Check that each set of marginals sums (or integrates) to 1.
  5. 5If asked for a probability of an event, add the matching cells or integrate over the matching region.
  6. 6If asked about independence, test f(x, y) against fX(x) × fY(y). One failure is enough to show dependence.
  7. 7If asked for conditional probability, divide the joint by the relevant marginal.

Quickest way: Margin totals first, then one cell test

When to use it: Use for discrete table questions under time pressure.

  1. Write the row and column totals in the margins straight away.
  2. Confirm the grand total is 1 (or solve for the missing cell).
  3. For independence, test a single cell: joint versus product of its two margins. A mismatch ends the question.
  4. For conditional probability, divide the cell by the margin of the given condition only.
  5. Eliminate options that are not between 0 and 1 or that ignore the margin.

Common mistakes in Joint and Marginal Probability Distributions

  • Treating a joint density value f(x, y) as a probability.

    With discrete tables, cell values are probabilities, so students carry that habit over.

    Fix: For continuous variables, always integrate over a region to get probability.

  • Summing the wrong direction when finding a marginal.

    Rows and columns are easy to confuse when X and Y labels are unclear.

    Fix: To get the marginal of X, you sum over Y, so each X value gets one total.

  • Declaring independence because the marginals look reasonable.

    Students check one or two cells that happen to match.

    Fix: Independence needs every cell to match. One mismatch proves dependence.

  • Wrong integration limits when the range is triangular.

    The support of the density depends on the other variable, such as 0 < y < x.

    Fix: Sketch the region and set limits so the inner variable's range depends on the outer one correctly.

  • Dividing by the wrong marginal in a conditional probability.

    The notation P(X | Y) is easy to flip.

    Fix: Divide by the probability of the event after the bar.

Worked examples

Example 1

The joint PMF of X (0 or 1) and Y (0 or 1) is: P(0,0) = 0.30, P(0,1) = 0.20, P(1,0) = 0.10, P(1,1) = 0.40. Find P(X = 1) and P(Y = 1 | X = 1). Are X and Y independent?

Show the solution
  1. Check the total: 0.30 + 0.20 + 0.10 + 0.40 = 1.00.
  2. Marginal of X: P(X = 1) = 0.10 + 0.40 = 0.50. P(X = 0) = 0.50.
  3. Marginal of Y: P(Y = 1) = 0.20 + 0.40 = 0.60.
  4. Conditional: P(Y = 1 | X = 1) = P(1,1) ÷ P(X = 1) = 0.40 ÷ 0.50 = 0.80.
  5. Independence test on one cell: P(X = 1) × P(Y = 1) = 0.50 × 0.60 = 0.30, but P(1,1) = 0.40. They differ.

Answer: P(X = 1) = 0.50; P(Y = 1 | X = 1) = 0.80; X and Y are not independent.

Example 2

X and Y have joint density f(x, y) = x + y for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, and 0 elsewhere. Verify it is a valid density and find the marginal density of X and P(X ≤ 0.5).

Show the solution
  1. Total: ∫0..1 ∫0..1 (x + y) dy dx. The inner integral is x·1 + 1/2 = x + 0.5.
  2. Outer integral: ∫0..1 (x + 0.5) dx = 0.5 + 0.5 = 1. The density is valid.
  3. Marginal of X: fX(x) = ∫0..1 (x + y) dy = x + 0.5, for 0 ≤ x ≤ 1.
  4. P(X ≤ 0.5) = ∫0..0.5 (x + 0.5) dx = [x²/2 + 0.5x] from 0 to 0.5.
  5. At 0.5: 0.125 + 0.25 = 0.375.

Answer: The density is valid; fX(x) = x + 0.5 on [0, 1]; P(X ≤ 0.5) = 0.375.

Exam tips

  • Expect a small joint table (2×2 or 3×3) and a question on a marginal, a conditional probability or independence.
  • Always write the margins first. Most options are built from common slips such as a joint cell used as a marginal.
  • For density questions, check the limits of integration before computing. A triangular support is a common trap.
  • If a constant k is unknown, use total probability = 1 before anything else.
  • Independence is tested against the product of marginals; zero covariance alone does not prove it.

Practice questions from Multivariate Random Variables

Joint and Marginal Probability Distributions: frequently asked questions

How do you find a marginal distribution from a joint probability table?

Add the probabilities across each row to get the marginal of the row variable. Add down each column to get the marginal of the column variable. Each set of marginals should total 1.

What is the difference between joint and marginal probability?

A joint probability is for both variables taking specific values together, such as P(X = 1, Y = 0). A marginal probability is for one variable alone, regardless of the other.

How do you know if two variables are independent from a joint table?

Compare every joint cell with the product of its row and column marginals. They must match in all cells. If one cell differs, the variables are dependent.

How is a joint density different from a joint mass function?

A mass function gives actual probabilities for discrete values. A density gives no probability by itself. You integrate it over a region to get a probability.