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Covariance is symmetric and bilinear in finite linear combinations

Statement

Let (Xi)i∈I and (Yj)j∈J be finite families of square-integrable real random variables on one common probability space. Then covariance is symmetric, and for real scalars ai,bj, Cov⁡ ⁣(∑i∈IaiXi,∑j∈JbjYj)=∑i∈I∑j∈JaibjCov⁡(Xi,Yj). This includes empty sums and constant summands. On a finite probability space all real random variables are square-integrable, so the original finite-space case is included with the same centered-product covariance.

Facts & Assumptions

Given: The square-integrable families on one common probability space and real scalars in the Statement.

[L1]

Expectation is linear on finite linear combinations (Linearity, monotonicity, and the modulus bound for expectation).

[L2]

Cov⁡(X,Y)=E[XY]−E[X]E[Y] (Variance and covariance identities for random variables).

Proof

technique · direct
1.1L2algebra

Each variable is integrable because 2∣X∣≤1+X2 on a probability space, and every product of two such variables is integrable because 2∣XY∣≤X2+Y2. Finite real linear combinations remain square-integrable by repeated use of (a+b)2≤2a2+2b2. Thus all covariances and expectations below are defined and finite. Commutativity of real multiplication in [L2] gives Cov⁡(X,Y)=Cov⁡(Y,X).

2.1step 1.1L2algebra

The finite linear combinations and their product have the required integrability by step 1.1. Substitute them into [L2] and distribute their pointwise product.

3.1step 2.1L1L2algebra

Applying [L1] to step 2.1 and collecting the aibj terms gives the displayed double sum.

4.1step 3.1L1L2∎

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

Depends on

Used by

Dependency tree · two levels

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Sources