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FRM Exam Part I · Multivariate Random Variables

Conditional Distributions and Independence of Random Variables

Updated 11 October 2026 · Fact-checked

A conditional distribution gives the probabilities of one variable once you know the value of another: f(x | y) = f(x, y) ÷ f(y). Two variables are independent if the joint distribution equals the product of the marginals for every pair of values. Check one cell at a time; a single failure proves dependence.

Understand Conditional Distributions and Independence

A joint distribution describes two random variables together. The marginal distribution of X is what you get by summing (or integrating) the joint over all values of Y. It ignores Y.

A conditional distribution answers a narrower question: what does X look like if we know Y = y? You take the row or column of the joint table for that y and rescale it so it sums to 1. The rescaling divides by the marginal probability of y. That is why f(x | y) = f(x, y) ÷ f(y), defined only when f(y) > 0.

The conditional expectation E(X | Y = y) is the mean of that rescaled distribution. If you do not fix y, E(X | Y) is itself a random variable, because it changes with Y. The law of iterated expectations says that averaging it over Y gives back the ordinary mean: E(X) = E[E(X | Y)].

X and Y are independent if knowing one tells you nothing about the other. Formally, f(x, y) = f(x) × f(y) for all x and y. Equivalently, f(x | y) = f(x) for every y with f(y) > 0. Independence implies zero covariance and zero correlation (when they exist), but the reverse is false. Uncorrelated only rules out a linear relationship. Y = X² with X symmetric around 0 is uncorrelated with X but fully dependent. The exception is the bivariate normal, where zero correlation does imply independence.

Key formulas to remember

Conditional probability (mass or density)
f(x | y) = f(x, y) ÷ f(y)
Valid only when f(y) > 0. Each conditional distribution sums (or integrates) to 1.
Marginal from joint
f(x) = Σ f(x, y) over y (discrete); f(x) = ∫ f(x, y) dy (continuous)
Sum across the row or column to get the marginal.
Independence
f(x, y) = f(x) × f(y) for all x, y
Must hold for every pair. One failure means dependent.
Independence, conditional form
f(x | y) = f(x) for all y with f(y) > 0
Equivalent to the product rule.
Conditional expectation (discrete)
E(X | Y = y) = Σ x × f(x | y)
Mean of the conditional distribution.
Law of iterated expectations
E(X) = E[E(X | Y)]
Weight each conditional mean by P(Y = y) and add.
Conditional variance
Var(X | Y = y) = E(X² | Y = y) − [E(X | Y = y)]²
Use conditional probabilities throughout.
Independence and moments
If independent: E(XY) = E(X)E(Y) and Cov(X, Y) = 0
The converse is false in general.

How to solve Conditional Distributions and Independence questions

Use this routine for any question on conditional distributions, conditional expectations or independence.

  1. 1Write down the joint table or joint function and check that the probabilities sum to 1.
  2. 2Compute the marginals by summing each row and each column.
  3. 3Identify the condition (for example Y = 2) and take the matching row or column of the joint.
  4. 4Divide each joint entry by the marginal of the condition to get the conditional probabilities. Check they sum to 1.
  5. 5For a conditional mean or variance, apply the usual formulas using the conditional probabilities.
  6. 6For an independence test, compare f(x, y) with f(x) × f(y) cell by cell. Stop at the first mismatch: the variables are dependent.
  7. 7For E(X), weight each conditional mean by the marginal of the condition (law of iterated expectations).
  8. 8Reread the question: it may ask about independence, correlation or the conditional mean, and these are different.

Quickest way: Single-cell independence check and row rescaling

When to use it: Use it for discrete joint tables when time is short.

  1. To test independence, pick the cell with the most unusual-looking probability and check f(x, y) = f(x)f(y). One failure ends the test.
  2. If any cell has zero joint probability but both marginals are positive, the variables are dependent immediately.
  3. For a conditional distribution, take the row, divide each entry by the row total.
  4. For a conditional mean, compute Σ x × (entry) over the row, then divide by the row total. This skips writing the conditional probabilities.
  5. Eliminate options that are not plausible: a conditional probability must be between 0 and 1.

Common mistakes in Conditional Distributions and Independence

  • Dividing by the wrong marginal, or not dividing at all.

    The joint entry looks like the answer, and the condition is easy to mix up.

    Fix: Divide by the marginal of the variable you are conditioning on, the one after the bar.

  • Concluding independence from zero correlation.

    Independence implies zero correlation, and the reverse feels natural.

    Fix: Zero correlation only rules out linear dependence. Test the product rule. Only for a bivariate normal does zero correlation imply independence.

  • Checking only one or two cells and declaring independence.

    Matching cells look convincing.

    Fix: Independence needs every cell to match. Passing a few cells proves nothing, while one failure proves dependence.

  • Treating E(X | Y) as a number rather than a random variable.

    The notation resembles E(X | Y = y).

    Fix: E(X | Y) varies with Y. Averaging it over Y gives E(X).

  • Using the unconditional mean inside conditional variance.

    Students reuse E(X) from an earlier part.

    Fix: Compute both E(X² | Y = y) and E(X | Y = y) from the conditional probabilities.

  • Using a simple average of conditional means for E(X).

    Forgetting that the conditions have different probabilities.

    Fix: Weight each conditional mean by P(Y = y).

Worked examples

Example 1

The joint distribution of X ∈ {0, 1} and Y ∈ {1, 2} is: P(0,1) = 0.10, P(0,2) = 0.30, P(1,1) = 0.20, P(1,2) = 0.40. Find E(X | Y = 2), and state whether X and Y are independent.

Show the solution
  1. Marginal of Y = 2: 0.30 + 0.40 = 0.70. Marginal of Y = 1: 0.10 + 0.20 = 0.30.
  2. Conditional on Y = 2: P(X = 0 | Y = 2) = 0.30 ÷ 0.70 = 3/7; P(X = 1 | Y = 2) = 0.40 ÷ 0.70 = 4/7.
  3. E(X | Y = 2) = 0 × 3/7 + 1 × 4/7 = 4/7 ≈ 0.5714.
  4. Independence: P(X = 1) = 0.20 + 0.40 = 0.60, and P(X = 1 | Y = 1) = 0.20 ÷ 0.30 = 2/3.
  5. Check: 2/3 ≠ 0.60, so f(x | y) ≠ f(x). Check the product rule: P(1,1) = 0.20, but P(X=1) × P(Y=1) = 0.60 × 0.30 = 0.18. Not equal.

Answer: E(X | Y = 2) = 4/7 ≈ 0.571. X and Y are not independent.

Example 2

Use the same joint distribution as above, but now X and Y take values X ∈ {0, 1}, Y ∈ {1, 2} with P(0,1) = 0.12, P(0,2) = 0.28, P(1,1) = 0.18, P(1,2) = 0.42. Verify independence, then use the law of iterated expectations to find E(X).

Show the solution
  1. Marginals of Y: P(Y = 1) = 0.12 + 0.18 = 0.30; P(Y = 2) = 0.28 + 0.42 = 0.70.
  2. Marginals of X: P(X = 0) = 0.12 + 0.28 = 0.40; P(X = 1) = 0.18 + 0.42 = 0.60.
  3. Product rule: 0.40 × 0.30 = 0.12 ✓; 0.40 × 0.70 = 0.28 ✓; 0.60 × 0.30 = 0.18 ✓; 0.60 × 0.70 = 0.42 ✓. All four cells match, so X and Y are independent.
  4. Conditional means: E(X | Y = 1) = 0.18 ÷ 0.30 = 0.60; E(X | Y = 2) = 0.42 ÷ 0.70 = 0.60.
  5. Iterated expectations: E(X) = 0.30 × 0.60 + 0.70 × 0.60 = 0.18 + 0.42 = 0.60.
  6. Check directly: E(X) = 1 × P(X = 1) = 0.60 ✓.

Answer: X and Y are independent, and E(X) = 0.60.

Exam tips

  • Questions often hand you a joint table. Compute marginals first, then the conditional you need.
  • For independence, find one cell that fails. It is faster than checking them all.
  • Expect a distractor that says uncorrelated variables are independent. It is true only for the bivariate normal.
  • Watch the conditioning direction: P(X | Y) and P(Y | X) use different denominators.
  • If a conditional probability you compute exceeds 1, you divided by the wrong marginal.

Practice questions from Multivariate Random Variables

Conditional Distributions and Independence: frequently asked questions

How do I check whether two random variables are independent?

Compute both marginals and test whether f(x, y) = f(x) × f(y) in every cell. If any cell fails, the variables are dependent. A zero joint cell with positive marginals is an instant failure.

What is the difference between independent and uncorrelated?

Independent means the joint distribution factors into the marginals. Uncorrelated means only that covariance is zero, which rules out linear dependence. Independence implies uncorrelated, but not the other way, except for the bivariate normal.

What is the law of iterated expectations?

It says E(X) = E[E(X | Y)]. You compute the mean of X for each value of Y, then average those means using the probabilities of Y. It is a quick way to find an unconditional mean from conditional ones.

How do I calculate a conditional distribution from a joint table?

Take the row or column for the given value, then divide each entry by that row or column total. The results are the conditional probabilities and must sum to 1.