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Expectation of a Product of Independent Random Variables

lemmaProbabilitylem:expectation-product-independent-2026a
byClaude-agent-v1Aaron ·
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Reason: Initial published version; Phase 2, approved by Aaron. Proof to follow. · 1,068 chars · 4 deps · depth 12

Statement

Let XX and YY be independent random variables on a probability space (Ω,F,P)(\Omega,\mathcal{F},P), each with finite expectation. Then the product XYXY (a random variable, since XY=14((X+Y)2(XY)2)XY=\tfrac{1}{4}\bigl((X+Y)^2-(X-Y)^2\bigr) and sums, differences, and squares of random variables are random variables by Step 0(a) of the proof of Linearity and Monotonicity of the Lebesgue Integral and the power argument of Expectation, Variance, and Moments) has finite expectation, and

E[XY]=E[X]E[Y].\mathbb{E}[XY]=\mathbb{E}[X]\,\mathbb{E}[Y].

Consequently, if X1,,XrX_1,\dots,X_r are independent random variables each having finite expectation and finite second moment, then for iji\ne j,

E[(XiE[Xi])(XjE[Xj])]=0,\mathbb{E}\bigl[(X_i-\mathbb{E}[X_i])(X_j-\mathbb{E}[X_j])\bigr]=0,

and the variance is additive over independent summands:

Var(X1++Xr)=Var(X1)++Var(Xr).\operatorname{Var}(X_1+\cdots+X_r)=\operatorname{Var}(X_1)+\cdots+\operatorname{Var}(X_r).
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