Definition D1
A random variable X is called finite-valued when the values it takes are only the distinct x₁, …, x_k.
Proof.
(1)The values are distinct, so the events {X = xᵢ} are mutually disjoint.convention
(2)Every sample point takes some value, so their union is Ω.convention
(3)Therefore {X = x₁}, …, {X = x_k} form a partition of Ω.(1), (2)
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Definition D2
Let pᵢ = P(X = xᵢ), and define the expectation of X as E(X) = x₁p₁ + ⋯ + x_k p_k.
Proof.
The values are finitely many, so this sum has finitely many terms.convention, D1
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Lemma L1
p₁ + ⋯ + p_k = 1
Proof.
(1){X = x₁}, …, {X = x_k} are mutually disjoint and their union is Ω.D1
(2)P(Ω) = P(X=x₁) + ⋯ + P(X=x_k)(1), A1
(3)P(Ω) = 1A2
(4)p₁ + ⋯ + p_k = 1(2), (3), D2, E
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Theorem T1
Let m be the smallest and M the largest of the values of X. Then m ≤ E(X) ≤ M
Proof.
(1)For every i, pᵢ = P(X = xᵢ) ≥ 0A3, D2
(2)For every i, xᵢ ≤ Massumption
(3)xᵢ pᵢ ≤ M pᵢ(1), (2), A4
(4)Σ xᵢ pᵢ ≤ Σ M pᵢ(3), A5
(5)Σ M pᵢ = M Σ pᵢA6
(6)Since Σ pᵢ = 1, Σ M pᵢ = M(5), L1, N
(7)E(X) ≤ M(4), (6), D2
(8)For every i, m ≤ xᵢ, so m pᵢ ≤ xᵢ pᵢ(1), assumption, A4
(9)Σ m pᵢ ≤ Σ xᵢ pᵢ and Σ m pᵢ = m Σ pᵢ = m(8), A5, A6, L1, N
(10)m ≤ E(X) ≤ M(7), (9), D2
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Theorem T2
If there are at least two values and every pᵢ is positive, then m < E(X) < M
Proof.
(1)E(X) − m = Σ xᵢpᵢ − m Σ pᵢ = Σ (xᵢ − m) pᵢD2, L1, A6
(2)For every i, xᵢ − m ≥ 0 and pᵢ > 0, so (xᵢ − m)pᵢ ≥ 0assumption, A4
(3)There are at least two values, so some j has x_j > m.assumption
(4)x_j − m > 0 and p_j > 0, so (x_j − m)p_j > 0(3), assumption, A7
(5)Every term is non-negative and one term is positive, so Σ (xᵢ − m)pᵢ > 0(2), (4), A8
(6)m < E(X)(1), (5)
(7)Applying the same argument to M − E(X) = Σ (M − xᵢ)pᵢ gives E(X) < M(1), (2), (4), (5), A6, A8
(8)Since m < E(X) < M, E(X) is neither the minimum nor the maximum.(6), (7)
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Corollary C1
If there are only two values, both with positive probability, then E(X) is neither of them. The E(X) of a die is 3.5, which is no face at all.
Proof.
(1)Suppose the only values are x₁ < x₂ with p₁ > 0, p₂ > 0.assumption, D1
(2)x₁ < E(X) < x₂, so E(X) is neither x₁ nor x₂.(1), T2
(3)Suppose the values of a die are 1, 2, 3, 4, 5, 6, each with probability 1/6.assumption, D1
(4)The six terms sum to 6 × (1/6) = 1, consistent with L1.(3), L1, N
(5)E(X) = (1+2+3+4+5+6) × (1/6) = 21/6 = 3.5(3), D2, N
(6)All values are integers and 3.5 is not, so no i has 3.5 = xᵢ.(3), (5), N
(7)1 < 3.5 < 6, consistent with T2.(5), T2, N
(8)Therefore the expectation is a point between the values, and there is no guarantee that this point is a value of X.(2), (6), (7)
∎