Definition D1
When the sample space Ω is finite, an assignment of a real number X(ω) to each sample point ω is called a random variable.
Proof.
(1)For two random variables X, Y, define (X+Y)(ω) = X(ω) + Y(ω).convention
(2)Define (XY)(ω) = X(ω) · Y(ω).convention
(3)Both definitions act pointwise, one sample point at a time. No relation between X and Y enters here.(1), (2)
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Definition D2
The expectation of a random variable X is defined as E(X) = Σ_{ω∈Ω} X(ω) P(ω).
Proof.
Since Ω is finite, this sum has finitely many terms.convention, D1
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Definition D3
X and Y are called independent when P(X = x, Y = y) = P(X = x) P(Y = y) for all x, y.
Proof.
Independence is a condition on the values the two variables take together; it is not determined by their individual distributions.convention
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Theorem T1
For two random variables X, Y, E(X + Y) = E(X) + E(Y)
Proof.
(1)E(X+Y) = Σ_ω (X+Y)(ω) P(ω)D2
(2)(X+Y)(ω) = X(ω) + Y(ω)D1
(3)E(X+Y) = Σ_ω [X(ω) + Y(ω)] P(ω)(1), (2)
(4)[X(ω) + Y(ω)] P(ω) = X(ω)P(ω) + Y(ω)P(ω)A1
(5)E(X+Y) = Σ_ω [X(ω)P(ω) + Y(ω)P(ω)](3), (4)
(6)The terms are finitely many, so they may be reordered, gathering the X terms together and the Y terms together.(5), A2, D2
(7)E(X+Y) = Σ_ω X(ω)P(ω) + Σ_ω Y(ω)P(ω)(6)
(8)E(X+Y) = E(X) + E(Y)(7), D2
(9)No line from (1) to (8) used any relation between X and Y. All that was used is pointwise addition and its rearrangement.(2), (6)
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Theorem T2
E(XY) = E(X) E(Y) does not hold in general.
Proof.
(1)Let Ω = {ω₁, ω₂} with P(ω₁) = P(ω₂) = 1/2.assumption
(2)Both values are non-negative, the two sample points are disjoint, and their union is Ω, so P(Ω) = 1/2 + 1/2 = 1. The three axioms are satisfied.(1), A3, A4, A5, N
(3)Let X(ω₁) = 1, X(ω₂) = 0, and let Y = X.assumption, D1
(4)E(X) = 1·(1/2) + 0·(1/2) = 1/2 and E(Y) = 1/2(1), (3), D2, N
(5)(XY)(ω₁) = 1·1 = 1 and (XY)(ω₂) = 0·0 = 0(3), D1, N
(6)E(XY) = 1·(1/2) + 0·(1/2) = 1/2(1), (5), D2, N
(7)E(X) E(Y) = (1/2)(1/2) = 1/4(4), N
(8)Since 1/2 ≠ 1/4, E(XY) ≠ E(X)E(Y)(6), (7), N
(9)P(X=1, Y=1) = 1/2 while P(X=1)P(Y=1) = 1/4, so this X and Y are not independent.(1), (3), D3, N
(10)For the same X, Y: (X+Y)(ω₁) = 2 and (X+Y)(ω₂) = 0, so E(X+Y) = 2·(1/2) + 0·(1/2) = 1, and E(X) + E(Y) = 1/2 + 1/2 = 1.(1), (3), D1, D2, N
(11)For one and the same non-independent pair, the product identity fails while the addition identity holds.(8), (10), (9), T1
(12)Therefore the product identity requires an additional condition such as independence, and the addition identity does not.(8), (9), (11)
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