Probability Distributions and Expectation
Expectation summarises a distribution by its centre; variance summarises its spread. The exam tests the algebra of expectation and variance under linear transformations and sums, and the use of the moment generating function to extract mean and variance.
Core rules
- E(X) = Σx p(x) for discrete X, or ∫x f(x)dx for continuous X.
- E(aX + b) = aE(X) + b, and E(X ± Y) = E(X) ± E(Y) always.
- E(XY) = E(X)E(Y) only when X and Y are independent.
- V(X) = E(X²) − [E(X)]²; V(aX + b) = a²V(X).
- V(X ± Y) = V(X) + V(Y) only for independent X and Y; otherwise the covariance term must be included.
- Moment generating function M_X(t) = E(e^{tX}); then M′(0) = E(X) and M″(0) = E(X²), so V(X) = M″(0) − [M′(0)]².
Worked example
Given E(X) = 3 and V(X) = 4, find E(2X + 5) = 2(3) + 5 = 11 and V(2X + 5) = 4 × 4 = 16, because the additive constant shifts the distribution without changing spread. Also E(X²) = V(X) + [E(X)]² = 4 + 9 = 13. If Y is independent of X with E(Y) = 1, then E(XY) = 3 and V(X + Y) = 4 + V(Y).
Traps that cost marks
Applying V(aX + b) = aV(X) + b; adding variances without independence; writing E(1/X) = 1/E(X); forgetting the square in the variance transformation; substituting t = 0 into M(t) rather than into its derivatives.