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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 $[0, + \infty]$ . In this finite product of expectations, a zero factor makes the product zero even when another factor is $+ \infty$ ; otherwise multiplication has its usual extended nonnegative meaning.
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)

[L7]

A nonnegative measurable function has zero integral exactly when it vanishes almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[L8]

Increasing nonnegative measurable functions pass to the limit under the integral (Monotone convergence for the integral).

Proof

technique · direct

1.1L1L6

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.

2.1step 1.1L7

Assume each $g_{i}$ is nonnegative. If some $E [g_{j} (X_{j})] = 0$ , [L7] gives $Y_{j} = 0$ almost surely. The finite product $\prod_{i} Y_{i}$ is then zero almost surely, so [L7] gives expectation zero, agreeing with the Statement's explicit zero-factor convention even if another expectation is infinite.

2.2step 1.1L2L3L4L8

Otherwise every $E [g_{i} (X_{i})]$ is positive (possibly infinite). 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 $μ$ . For each integer $k \geq 1$ , put $Y_{i}^{(k)} = Y_{i} \land k$ . The finite product $\prod_{i} Y_{i}^{(k)}$ is bounded, so [L3] and repeated [L4] give $E [\prod_{i} Y_{i}^{(k)}] = \prod_{i} E [Y_{i}^{(k)}]$ with only finite factors. As $k ↑ \infty$ , the products increase pointwise to $\prod_{i} Y_{i}$ ; [L8] passes the limit through the left integral and each factor on the right. Since all limiting factors are strictly positive, the finite product of their increasing approximants tends to their ordinary extended product without an indeterminate zero-times-infinity expression.

3.1step 2.1step 2.2

Now assume every $g_{i} (X_{i})$ is integrable. Applying steps 2.1–2.2 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.

4.1step 1.1step 3.1L2L3L5

Let $μ_{i}$ be the law of $Y_{i} = g_{i} (X_{i})$ and $μ = ⨂_{i < n} μ_{i}$ as in step 2.2. Step 3.1 shows that the product map $m$ is $μ$ -integrable. By [L2], [L3], and repeated use of [L5], $E [\prod_{i < n} g_{i} (X_{i})] = \int_{R^{n}} m d μ = \prod_{i < n} \int_{R} y d μ_{i} = \prod_{i < n} E [g_{i} (X_{i})]$ .

5.1step 2.1step 2.2step 4.1∎

Steps 2.1–2.2 prove the nonnegative case, and step 4.1 proves the integrable case.

Expectations factor over finite products of independent random variables

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources