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MSE of a sum of error components includes covariance terms

Statement

Let $E_{1}, \dots, E_{m}$ be square-integrable real random variables on one probability space, where $m \geq 1$ . Write $b_{j} = E E_{j}$ . Then $E [{(\sum_{j = 1}^{m} E_{j})}^{2}] = {(\sum_{j = 1}^{m} b_{j})}^{2} + \sum_{j = 1}^{m} Var (E_{j}) + 2 \sum_{1 \leq j < k \leq m} Cov (E_{j}, E_{k}) .$ The correction to the sum of component variances is exactly $(\sum_{j} b_{j})^{2} + 2 \sum_{j < k} Cov (E_{j}, E_{k})$ . Thus the mean square equals the sum of component variances exactly when this combined correction is zero. In particular, if the total bias is zero, pairwise independence suffices.

Facts & Assumptions

Given: A probability space and a finite family of square-integrable real random variables $E_{1}, \dots, E_{m}$ , with $m \geq 1$ . Prove the formula and its pairwise-independent special case.

[F1]

For a square-integrable random variable $X$ , $Var (X) = E [(X - E X)^{2}]$ ; for square-integrable $X, Y$ , $Cov (X, Y) = E [(X - E X) (Y - E Y)]$ . (Moments, variance, and covariance on a probability space)

[F2]

The Lebesgue integral is linear on $L^{1}$ : finite sums of integrable functions are integrable and their integrals add. (The Lebesgue integral is linear on $L^{1} (μ)$ )

[F3]

For independent real random variables and integrable coordinate functions, the expectation of their product factors into the product of expectations. (Expectations factor over finite products of independent random variables)

Proof

1.1givenF2algebra

Since $∣ E_{j} ∣ \leq 1 + E_{j}^{2}$ and $E E_{j}^{2} < \infty$ , each $E_{j}$ is integrable. For $S = \sum_{j = 1}^{m} E_{j}$ , [F2] gives $E S = \sum_{j = 1}^{m} b_{j}$ .

2.1step 1.1algebra

Put $C_{j} = E_{j} - b_{j}$ . By step 1.1, $S - E S = \sum_{j = 1}^{m} C_{j}$ .

3.1step 2.1algebra

Expanding the finite square gives $(S - E S)^{2} = \sum_{j} C_{j}^{2} + 2 \sum_{j < k} C_{j} C_{k}$ . Each cross product is integrable because $2 ∣ C_{j} C_{k} ∣ \leq C_{j}^{2} + C_{k}^{2}$ .

4.1step 3.1F1F2

Taking expectations in step 3.1, using [F2] and [F1], yields $Var (S) = \sum_{j} Var (E_{j}) + 2 \sum_{j < k} Cov (E_{j}, E_{k})$ .

5.1step 1.1step 4.1F1algebra

Since $E (S - E S) = 0$ , expansion of $S^{2} = ((S - E S) + E S)^{2}$ gives $E S^{2} = Var (S) + (E S)^{2}$ ; substitute step 4.1 and step 1.1 to obtain the stated formula. Subtracting $\sum_{j} Var (E_{j})$ shows that equality with the variance sum holds exactly when $(\sum_{j} b_{j})^{2} + 2 \sum_{j < k} Cov (E_{j}, E_{k}) = 0$ .

6.1givenF3step 5.1algebra∎

In the pairwise-independent case, apply [F3] to the coordinate functions $x \mapsto x - b_{j}$ and $y \mapsto y - b_{k}$ ; their means are zero, so every covariance in step 5.1 is zero. If all components are also centered, the squared-total-bias term is zero as well.

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Sources

Anthony Davison and David Hinkley, Bootstrap Methods and Their Application, Chapter 2 §2.5.1–§2.5.2 (standard reference, not scraped)
Art B. Owen, Monte Carlo Book: The Quasi-Monte Carlo Parts, Chapter 17 §§17.1 and 17.4 and end notes (standard reference, not scraped)