Expectation and Variance Properties
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Direct answer
Expectation is linear; variance is not. For any random variables, E(aX + b) = aE(X) + b and E(X + Y) = E(X) + E(Y) hold with no independence requirement whatsoever — linearity is unconditional. Variance transforms as Var(aX + b) = a^2 Var(X): scaling stretches spread by the square of the multiplier, while shifting does nothing to it. Independence upgrades variance arithmetic: for independent X and Y, Var(X + Y) = Var(X) + Var(Y) (and equally for X - Y), with E(XY) = E(X)E(Y). The general addition rule runs through covariance: Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y), where Cov(X, Y) = E(XY) - E(X)E(Y), and the computational keystone Var(X) = E(X^2) - (E(X))^2 converts between raw and central moments.
What you must remember
- Linearity of expectation: E(aX + b) = aE(X) + b and E(X + Y) = E(X) + E(Y) always — no independence needed; this is the property behind expected-value shortcuts in sums of dice.
- Variance under transformation: Var(aX + b) = a^2 Var(X); adding a constant never changes spread, and SD(aX + b) = |a| SD(X).
- Working formula: Var(X) = E(X^2) - (E(X))^2; conversely E(X^2) = Var(X) + (E(X))^2 — most numerical questions shuttle between these two.
- Variance of a sum: Var(X ± Y) = Var(X) + Var(Y) ± 2Cov(X, Y); independence forces Cov = 0, but Cov = 0 does not force independence.
- Product rule: E(XY) = E(X)E(Y) when X and Y are independent; without independence use the definition E(XY) from the joint distribution.
- Die numbers: a fair die has E(X) = 3.5 and Var(X) = 35/12; two independent dice sum to mean 7 with variance 35/6 — reusable anchors for checking arithmetic.
- Non-negativity: Var(X) ≥ 0 always, and Var(X) = 0 exactly when X is constant; a negative computed variance certifies an arithmetic slip.
An anchor computation with dice
Let S be the sum on two fair dice. Write S = X + Y with X, Y the individual faces. For one die, E(X) = (1 + 2 + ... + 6)/6 = 21/6 = 3.5, and E(X^2) = (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6, so Var(X) = 91/6 - (7/2)^2 = 91/6 - 49/4 = (182 - 147)/12 = 35/12. By linearity, E(S) = 7 with no distribution table. By independence, Var(S) = 35/12 + 35/12 = 35/6, hence SD(S) = √(35/6) ≈ 2.42. Now invert the keystone formula: E(S^2) = Var(S) + (E S)^2 = 35/6 + 49 = 329/6 ≈ 54.83 — a quantity painful to compute from the 36-cell table but free here. This is the pattern JEE exploits: hand over E and Var, demand E(S^2) or SD of a linear combination like 3X - 2, whose variance is 9 Var(X) (the -2 shifts nothing) — Var(3X - 2) = 9 × 35/12 = 105/4.
Where students slip
JEE Main frames these as two-minute transformations: Var(2X + 3), E(5X - 4), or the variance of a sum of independent variables given in a small distribution — the die anchor 35/12 appears recurrently in numerical answers. Advanced adds covariance arithmetic (Var(X + Y) given Var, Var of each and correlation-like data), conditional expectations feeding E(XY), and the distinction between uncorrelated and independent. The chronic errors: writing Var(aX + b) = a Var(X) + b or a^2 Var(X) + b — the constant must vanish; assuming Var(X - Y) = Var(X) - Var(Y), which is false even for independent variables (variances always add); and concluding independence from E(XY) = EX · EY, valid only in the reverse direction. One more quiet trap: variance of a sum needs independence, but expectation of a sum never does — statements mixing these two requirements are planted as true/false options in both papers. The topic belongs to probability and statistics, examined in both Main and Advanced.
Frequently asked questions
Does E(X + Y) = E(X) + E(Y) require independence?
No — expectation is linear unconditionally; even for dependent variables the means simply add.
What is Var(aX + b) equal to?
a^2 Var(X); the additive constant b contributes nothing, and the standard deviation becomes |a| SD(X).
How is E(X^2) found from the mean and variance?
E(X^2) = Var(X) + (E(X))^2, the rearranged working formula — the standard route when a distribution's second moment is asked.
What is the variance of the sum of two independent random variables?
Var(X + Y) = Var(X) + Var(Y), and equally Var(X - Y) = Var(X) + Var(Y) — variances never subtract under independence.
Does Cov(X, Y) = 0 imply X and Y are independent?
No — zero covariance (uncorrelatedness) is necessary but not sufficient for independence; independent variables always have zero covariance, not vice versa.