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Actuarial Statistics · Expectations and conditional expectations

Properties of Expectation, Variance and Moments for Actuarial Statistics

Updated 11 October 2026 · Fact-checked

Expectation is linear: E[aX + bY + c] = aE[X] + bE[Y] + c, always, even for dependent variables. Variance is not linear: Var(aX + bY) = a²Var(X) + b²Var(Y) + 2ab Cov(X, Y). For independent variables the covariance term is zero. Moments summarise a distribution's shape.

Understand Properties of Expectation, Variance and Moments

The expectation E[X] is the long-run average value of X. It is the balance point of the distribution. For a discrete X, E[X] = Σ x P(X = x). For a continuous X, it is the integral of x f(x) over the range of X.

The key property is linearity. E[aX + bY + c] = aE[X] + bE[Y] + c. This holds whether or not X and Y are independent. It is the most useful rule in the topic, because you never need the joint distribution to find the mean of a sum.

Variance measures spread: Var(X) = E[(X − μ)²] = E[X²] − (E[X])². It is not linear. Adding a constant does not change it. Multiplying by a constant a multiplies it by a². For a sum, you must add a covariance term: Cov(X, Y) = E[XY] − E[X]E[Y]. Covariance is positive when X and Y tend to move together and negative when they move opposite ways.

If X and Y are independent, then E[XY] = E[X]E[Y] and Cov(X, Y) = 0. The reverse is not true. Zero covariance does not prove independence.

Moments go further. The k-th raw moment is E[Xᵏ]. The k-th central moment is E[(X − μ)ᵏ]. The second central moment is the variance. The third gives skewness and the fourth gives kurtosis. Moments are used to describe shape and to build the moment generating function.

Key rules to remember

Linearity of expectation
E[aX + bY + c] = aE[X] + bE[Y] + c
Always true. No independence needed. Extends to any number of variables.
Variance (computational form)
Var(X) = E[X²] − (E[X])²
Usually the fastest route. Variance is never negative.
Variance of a linear function
Var(aX + c) = a²Var(X)
The constant c drops out. The multiplier a is squared, so a sign change does not matter.
Covariance
Cov(X, Y) = E[XY] − E[X]E[Y]
Also Cov(X, Y) = E[(X − μX)(Y − μY)]. Cov(X, X) = Var(X).
Variance of a linear combination
Var(aX + bY) = a²Var(X) + b²Var(Y) + 2ab Cov(X, Y)
For independent variables, Cov = 0 and the last term vanishes.
Covariance with constants
Cov(aX + b, cY + d) = ac Cov(X, Y)
Bilinear in both arguments. Constants added do not matter.
Correlation
ρ(X, Y) = Cov(X, Y) ÷ (σX σY)
Always between −1 and 1, when both standard deviations are non-zero.
Variance of a sum of n variables
Var(ΣXᵢ) = ΣVar(Xᵢ) + 2 Σ(i<j) Cov(Xᵢ, Xⱼ)
For independent variables, Var(ΣXᵢ) = ΣVar(Xᵢ).
Moments
k-th raw moment = E[Xᵏ]; k-th central moment = E[(X − μ)ᵏ]
Skewness = E[(X − μ)³] ÷ σ³. Kurtosis = E[(X − μ)⁴] ÷ σ⁴.

How to solve Properties of Expectation, Variance and Moments questions

Use this routine for any question asking for a mean, variance or covariance of a combination of random variables.

  1. 1Write the quantity you need in symbols, for example Var(2X − 3Y + 5).
  2. 2Find E[X], Var(X) and the other basic values you are given or can compute.
  3. 3For the mean, apply linearity directly. Do not worry about dependence.
  4. 4For the variance, check whether the variables are independent. If not, you need Cov(X, Y) or the correlation ρ.
  5. 5If you are given ρ, convert: Cov(X, Y) = ρ σX σY.
  6. 6Expand using Var(aX + bY) = a²Var(X) + b²Var(Y) + 2ab Cov(X, Y). Keep the sign of ab.
  7. 7If you need E[XY], compute it from the joint distribution, or use E[X]E[Y] only when independent.
  8. 8Check that your variance is not negative, and state units and any assumptions.

Quickest way: Constants out, then expand with covariance

When to use it: Use when the question gives means, variances and a covariance or correlation, and asks for a linear combination.

  1. Drop all added constants for the variance. They do not affect it.
  2. Square each multiplier for its variance term.
  3. Add 2ab × Cov for each pair, with the sign of a × b.
  4. Set Cov = 0 only if the question says independent.
  5. Use E[X²] = Var(X) + (E[X])² when you need a second moment quickly.

Common mistakes in Properties of Expectation, Variance and Moments

  • Writing Var(aX) = a Var(X).

    Students copy the linearity rule for expectation.

    Fix: Remember the multiplier is squared: Var(aX) = a²Var(X).

  • Writing Var(X − Y) = Var(X) − Var(Y).

    Subtraction feels like it should reduce spread.

    Fix: Use Var(X − Y) = Var(X) + Var(Y) − 2Cov(X, Y). For independent variables the variances add.

  • Assuming zero covariance means independence.

    The converse is true, so students assume both directions hold.

    Fix: Independence implies zero covariance, but not the other way. Only say independent if told so or shown from the joint distribution.

  • Using E[XY] = E[X]E[Y] for dependent variables.

    It is the easy shortcut.

    Fix: This needs independence, or at least zero covariance. Otherwise compute E[XY] from the joint distribution.

  • Forgetting the 2 in the covariance term.

    The term comes from the cross products appearing twice when you expand.

    Fix: Always write 2ab Cov(X, Y) in full before substituting.

  • Confusing the raw second moment E[X²] with the variance.

    Both are called second moments in some contexts.

    Fix: Variance is the second central moment. Subtract (E[X])² from E[X²].

Worked examples

Example 1

X and Y have E[X] = 4, E[Y] = 6, Var(X) = 9, Var(Y) = 16 and Cov(X, Y) = 6. Find E[2X − 3Y + 5] and Var(2X − 3Y + 5).

Show the solution
  1. Mean: E[2X − 3Y + 5] = 2(4) − 3(6) + 5 = 8 − 18 + 5 = −5.
  2. The constant 5 does not affect the variance.
  3. Var(2X − 3Y) = 2²(9) + (−3)²(16) + 2(2)(−3)(6).
  4. = 4 × 9 + 9 × 16 − 12 × 6.
  5. = 36 + 144 − 72 = 108.

Answer: E = −5 and Var = 108.

Example 2

X takes values 1, 2 and 3 with probabilities 0.2, 0.5 and 0.3. Find E[X], Var(X) and the third central moment E[(X − μ)³].

Show the solution
  1. E[X] = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1.0 + 0.9 = 2.1.
  2. E[X²] = 1(0.2) + 4(0.5) + 9(0.3) = 0.2 + 2.0 + 2.7 = 4.9.
  3. Var(X) = 4.9 − 2.1² = 4.9 − 4.41 = 0.49.
  4. Deviations from 2.1 are −1.1, −0.1 and 0.9.
  5. Cubes are −1.331, −0.001 and 0.729.
  6. E[(X − μ)³] = 0.2(−1.331) + 0.5(−0.001) + 0.3(0.729) = −0.2662 − 0.0005 + 0.2187 = −0.048.

Answer: E[X] = 2.1, Var(X) = 0.49, third central moment = −0.048.

Exam tips

  • In written answers, state which rule you use, for example linearity, and whether independence is needed.
  • For MCQs, check the variance sign trap: Var(X − Y) with independent X and Y adds the variances.
  • When a question gives ρ instead of covariance, convert first using Cov = ρ σX σY. Remember σ is the square root of the variance.
  • In computer-based papers, check results against sample versions, and note that R's var() uses divisor n − 1.
  • Show the expansion line in full. Method marks are awarded even if the arithmetic slips.

Practice questions from Expectations and conditional expectations

Properties of Expectation, Variance and Moments in other exams

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

Properties of Expectation, Variance and Moments: frequently asked questions

Does linearity of expectation need independence?

No. E[X + Y] = E[X] + E[Y] holds for any random variables with finite means. Independence is only needed for results like E[XY] = E[X]E[Y] or for dropping the covariance term in a variance.

How do I find the variance of a sum of dependent variables?

Add all the individual variances and then add 2 × covariance for every pair. For two variables this is Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y). If you have the correlation, use Cov = ρ σX σY.

Can covariance be negative?

Yes. A negative covariance means the variables tend to move in opposite directions. Then the variance of the sum is smaller than the sum of the variances.

What is the difference between raw and central moments?

Raw moments are E[Xᵏ], taken about zero. Central moments are E[(X − μ)ᵏ], taken about the mean. The variance is the second central moment, and skewness uses the third.