TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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Covariance is symmetric and bilinear in finite linear combinations

Statement

Covariance is symmetric, and for finite families $(X_{i})_{i \in I}$ and $(Y_{j})_{j \in J}$ and real scalars $a_{i}, b_{j}$ , $Cov ⁣ (\sum_{i \in I} a_{i} X_{i}, \sum_{j \in J} b_{j} Y_{j}) = \sum_{i \in I} \sum_{j \in J} a_{i} b_{j} Cov (X_{i}, Y_{j}) .$ This includes empty sums and constant summands.

Facts & Assumptions

Given: Finite families and scalars as in the Statement.

[L1]

Expectation is linear on finite linear combinations (Expectation is linear for every finite family of random variables, without any independence hypothesis).

[L2]

$Cov (X, Y) = E [X Y] - E [X] E [Y]$ ( $Var (X) = E [X^{2}] - E [X]^{2}$ and $Cov (X, Y) = E [X Y] - E [X] E [Y]$ ).

Proof

technique · direct

1.1

Commutativity of real multiplication in [L2] gives $Cov (X, Y) = Cov (Y, X)$ .

L2algebra

1.2

Substitute the two finite linear combinations into [L2] and distribute their pointwise product.

L2algebra

2.1

Applying [L1] to step 1.2 and collecting the $a_{i} b_{j}$ terms gives the displayed double sum.

step 1.2L1L2algebra

3.1

If either index set is empty, both sides are zero; covariance with a constant is zero by [L1] and [L2].

step 2.1L1L2∎

Depends on

Used by

Variance of a finite sum as the sum of all variances and covariances Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 24 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

H. Pishro-Nik, Introduction to Probability, Statistics, and Random Processes, Section 5.3.1 (standard reference, not scraped)