Solution
Sample Variance Under Serial Correlation
Show the problem again
Three observations X₁, X₂, X₃ each have mean μ and variance 1, with Corr(Xᵢ, Xⱼ) = (1/2)^|i − j| (a first-order autoregressive pattern). Let S² = (1/2) Σ (Xᵢ − X̄)². Find E[S²] as a reduced fraction.
Worked solution
Use E[Σ(Xᵢ − X̄)²] = Σ Var(Xᵢ) − n Var(X̄). Here ΣVar(Xᵢ) = 3. Next, n² Var(X̄) = Var(ΣXᵢ) = Σᵢⱼ Cov(Xᵢ, Xⱼ) = 3 + 2(1/2 + 1/2 + 1/4) = 3 + 5/2 = 11/2, so n Var(X̄) = (11/2)/3 = 11/6. Therefore E[Σ(Xᵢ − X̄)²] = 3 − 11/6 = 7/6 and E[S²] = (7/6)/2 = 7/12. The sample variance underestimates the true variance 1 because positively correlated observations cluster, so the sample mean is unusually close to them. The answer is 7/12.
Source: Original problem written for AxiomIQ; expectation of the sample variance for dependent data.