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Expectations factor over finite products of independent random variables

Statement

Let $n \geq 1$ , let $X_{0}, \dots, X_{n - 1}$ be independent real random variables on a common probability space, and let $g_{i} : R \to R$ be Borel measurable for each $i < n$ .

If every $g_{i}$ is nonnegative, then $E [\prod_{i < n} g_{i} (X_{i})] = \prod_{i < n} E [g_{i} (X_{i})]$ in $[0, + \infty]$ .
If every $g_{i} (X_{i})$ is integrable, then $\prod_{i < n} g_{i} (X_{i})$ is integrable and the same factorization holds in $R$ .

Facts & Assumptions

Given: Independent real random variables $X_{0}, \dots, X_{n - 1}$ and Borel measurable functions $g_{i} : R \to R$ .

[L1]

Measurable coordinatewise functions preserve independence. (Measurable coordinatewise functions preserve independence)

[L2]

Independent random elements have product joint law. (Independent random elements have product joint law)

[L3]

Expectation is integration against the law after a measurable change of variables. (Change of variables for expectation)

[L4]

Tonelli evaluates nonnegative product-measurable integrands on a sigma-finite product space. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

[L5]

Fubini evaluates absolutely integrable product-measurable integrands on a sigma-finite product space. (Fubini's theorem for L^1 functions on a sigma-finite product)

[L6]

On a product measurable space, coordinate projections are measurable, and finite sums and products of measurable real-valued functions remain measurable. (The product sigma-algebra and its finite iterates, Arithmetic and lattice operations preserve measurability whenever they are defined)

Proof

technique · direct

1.1

Put $Y_{i} : = g_{i} (X_{i})$ . By [L1], the family $Y_{0}, \dots, Y_{n - 1}$ is independent. On $R^{n}$ with the finite product sigma-algebra, each coordinate projection $π_{i}$ is measurable because $π_{i}^{- 1} (B) = R^{i} \times B \times R^{n - 1 - i}$ is a measurable rectangle. Repeated use of [L6] therefore makes the product map $m (y_{0}, \dots, y_{n - 1}) = \prod_{i < n} y_{i}$ measurable.

L1L6

2.1

Assume each $g_{i}$ is nonnegative. Let $μ_{i}$ be the law of $Y_{i}$ and let $μ = ⨂_{i < n} μ_{i}$ . By [L2], the joint law of $Y = (Y_{0}, \dots, Y_{n - 1})$ is $μ$ . Using [L3] for the measurable map $m$ and then applying [L4] repeatedly on the product measure space yields $E [\prod_{i < n} g_{i} (X_{i})] = \int_{R^{n}} m d μ = \prod_{i < n} \int_{R} y d μ_{i} = \prod_{i < n} E [g_{i} (X_{i})] .$

step 1.1L2L3L4

3.1

Now assume every $g_{i} (X_{i})$ is integrable. Applying step 2.1 to the nonnegative functions $∣ g_{i} ∣$ gives $E [\prod_{i < n} ∣ g_{i} (X_{i}) ∣] = \prod_{i < n} E [∣ g_{i} (X_{i}) ∣] < \infty .$ So $\prod_{i < n} g_{i} (X_{i})$ is integrable.

step 2.1

4.1

Let $μ_{i}$ be the law of $Y_{i} = g_{i} (X_{i})$ and $μ = ⨂_{i < n} μ_{i}$ as in step 2.1. Step 3.1 shows that the product map $m$ is $μ$ -integrable. By [L2], [L3], and repeated use of [L5], $E [\prod_{i < n} g_{i} (X_{i})] = \int_{R^{n}} m d μ = \prod_{i < n} \int_{R} y d μ_{i} = \prod_{i < n} E [g_{i} (X_{i})] .$

step 1.1step 3.1L2L3L5

5.1

Step 2.1 proves the nonnegative case, and step 4.1 proves the integrable case.

step 2.1step 4.1∎

Depends on

Used by

Independence forces covariance to vanish Corollary

Dependency tree · two levels

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Sources

Rick Durrett, Probability: Theory and Examples, 5th ed., Theorem 2.1.12 (standard reference, not scraped)
S. R. S. Varadhan, Probability Theory, Section 1.6 (standard reference, not scraped)