# Covariance

> Covariance for JEE Mathematics: E[XY] − E[X]E[Y], variance of sums, correlation coefficient, independence counterexamples and joint distribution work.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/covariance-jee
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Covariance", PrepElephant, https://prepelephant.com/topics/jee/mathematics/covariance-jee

## Direct answer

How two random variables move together gets its own number: Cov(X, Y) = E[XY] − E[X]E[Y], the expected product of paired deviations — positive when the variables rise together, negative when they oppose, zero under independence. The converse fails: zero covariance does not force independence, the standard counterexample being X uniform on {−1, 0, 1} with Y = X², where E[XY] = 0 yet Y is completely determined by X. The operational identities: Cov(X, X) = Var(X); Var(aX + bY) = a²Var X + b²Var Y + 2ab Cov(X, Y); and the correlation coefficient ρ = Cov(X, Y)/(σXσY) is trapped in [−1, 1], hitting ±1 only under perfect linear dependence.

## What you must remember

- **Definition:** Cov(X, Y) = E[XY] − E[X]E[Y] — computing E[XY] from the joint distribution is the whole of most problems.
- **Independence one-way street:** independence forces zero covariance; zero covariance does not force independence, with the X versus X² counterexample the quotable witness.
- **Self-covariance:** Cov(X, X) = Var(X), so variance is a special case — the formulas below then specialise to familiar ones.
- **Variance of a sum:** Var(X + Y) = Var X + Var Y + 2Cov(X, Y); Var(X − Y) = Var X + Var Y − 2Cov(X, Y); independence deletes the cross term, not the others.
- **Bilinearity:** Cov(aX + b, Y) = a Cov(X, Y) — constants drop out, coefficients scale through.
- **Correlation coefficient:** ρ = Cov/(σXσY) ∈ [−1, 1], sign matching the covariance's, magnitude 1 exactly on a perfect straight line.
- **Computational route:** from a joint distribution, first the marginals (E[X], E[Y]), then E[XY] as Σ x·y·P(X = x, Y = y), then one subtraction.

## One joint table, three numbers

Take a joint distribution on {0, 1} × {0, 1}: P(0, 0) = 0.3, P(0, 1) = 0.2, P(1, 0) = 0.1 and P(1, 1) = 0.4. The marginals first: P(X = 1) = 0.1 + 0.4 = 0.5, so E[X] = 0.5; P(Y = 1) = 0.2 + 0.4 = 0.6, so E[Y] = 0.6. The product expectation is E[XY] = 1 × 1 × 0.4 = 0.4 — the only cell where both variables equal 1 contributes. Then Cov(X, Y) = 0.4 − 0.5 × 0.6 = 0.4 − 0.30 = 0.1, positive: high X travels with high Y in this table. Now the variance identity as a self-check: Var X = 0.5 − 0.25 = 0.25, Var Y = 0.6 − 0.36 = 0.24, so Var(X + Y) = 0.25 + 0.24 + 2(0.1) = 0.69. Verify directly: X + Y takes 0 with probability 0.3, 1 with 0.3 and 2 with 0.4, giving E = 1.1, E[(X + Y)²] = 0.3 + 1.6 = 1.9, and 1.9 − 1.21 = 0.69. Two routes agreeing is the arithmetic audit worth performing on every joint-distribution answer.

## Covariance is not independence

JEE Main asks the joint-table computation and the variance-of-sum identity, and the recurring loss is the forgotten cross term — Var(X + Y) computed as the plain sum 0.49 instead of 0.69 whenever positive covariance binds the variables. Data-based variants give paired observations and ask for the sample covariance: Σ(xi − x̄)(yj − ȳ)/n, the same machinery with frequencies in disguise. JEE Advanced leans on the conceptual boundary: construct or identify a pair with zero covariance but dependence (the X, X² example, or the uniform angle pair cos Θ, sin Θ, both zero-mean with zero product expectation yet locked together by cos² + sin² = 1), and interpret ρ — knowing |ρ| = 1 certifies a linear relation, while ρ = 0 certifies nothing about nonlinear structure. The unit trap deserves respect: covariance carries the units of X times Y, which is exactly why the correlation coefficient divides by both standard deviations before any comparison. Guard the order of computation — marginals, product expectation, subtraction — because E[XY] computed from marginals instead of the joint table is the single most common and least detectable error in the chapter.

## Frequently asked questions

### What is the formula for covariance?

Cov(X, Y) = E[XY] − E[X]E[Y], with E[XY] computed from the joint distribution of the pair.

### Does zero covariance imply independence?

No — independence implies zero covariance, but X with Y = X² (X uniform on {−1, 0, 1}) has zero covariance with full dependence.

### How does covariance enter Var(X + Y)?

Var(X + Y) = Var X + Var Y + 2Cov(X, Y); independence removes the cross term, giving the familiar additive form.

### What is the correlation coefficient?

ρ = Cov(X, Y)/(σX σY), a unit-free measure in [−1, 1] that reaches ±1 only under exact linear dependence.

### How do you compute E[XY] from a joint table?

Sum x·y·P(X = x, Y = y) over all cells — only cells where both variables are nonzero contribute.