Definition D1
The mean x̄ of data x₁, …, xₙ (n ≥ 1) is the value satisfying n x̄ = x₁ + ⋯ + xₙ.
Proof.
(1)Σaᵢ means a₁ + ⋯ + aₙ; u − v means u + (−v); t² means t·t.convention
(2)That such an x̄ exists and is unique was obtained in Proof 64.convention
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Definition D2
dᵢ = xᵢ − x̄ is called the deviation.
Proof.
The deviation records, sign included, in which direction and by how much a data point departs from the mean.convention
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Definition D3
For a reference value c, S(c) = Σ(xᵢ − c)² is called the sum of squares with respect to c.
Proof.
If c = x̄, then S(x̄) = Σdᵢ².D2
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Lemma L1
Let e = x̄ − c. Then for every c, S(c) = S(x̄) + n e²
Proof.
(1)xᵢ − c = (xᵢ − x̄) + (x̄ − c) = dᵢ + eD2, A4, A5, A6, A7
(2)(dᵢ + e)² = dᵢ² + 2dᵢe + e²(1), A14
(3)S(c) = Σdᵢ² + Σ(2dᵢe) + Σe²(2), D3, A11
(4)2dᵢe = (2e)dᵢA2, A3
(5)Σ((2e)dᵢ) = (2e)Σdᵢ(4), A12
(6)Σdᵢ = 0D2, A15
(7)(2e) × 0 = 0(5), (6), A9
(8)Σe² = n e²A13
(9)S(c) = Σdᵢ² + n e² = S(x̄) + n e²(3), (7), (8), A6, D3
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Theorem T1
The deviations sum to 0 whatever the data. Therefore that sum cannot distinguish different data sets.
Proof.
(1)For any data whatsoever, Σdᵢ = 0D2, A15
(2)The data 5, 5 have mean 5 and deviations 0, 0.D1, D2, N
(3)The data 0, 10 have mean 5 and deviations −5, 5.D1, D2, N
(4)The two data sets differ, yet the deviations sum to 0 in both.(1), (2), (3)
(5)The sum of deviations does not vary with the data, so it cannot express the degree of spread. The signs cancel each other.(1), (4)
(6)Therefore, to measure spread, one must use a quantity with the signs removed. The sum of squares S(c) is one such quantity.(5), D3
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Theorem T2
The only c that minimizes S(c) is the mean x̄.
Proof.
(1)Let e = x̄ − c. Then S(c) = S(x̄) + n e²L1
(2)Suppose c ≠ x̄. If e = 0, adding c to both sides would give x̄ = c, so e ≠ 0.assumption, E, A4, A5, A6, A7
(3)0 < e²(2), A16
(4)Since 0 < n, 0 < n e²A19, (3), A17
(5)S(x̄) < S(x̄) + n e²(4), A18, A5, A6
(6)Therefore S(x̄) < S(c)(1), (5)
(7)This holds for every c ≠ x̄, so the minimum of S occurs only at c = x̄.(6)
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Corollary C
Defining the variance as σ² = S(x̄)n⁻¹, for every c we have S(c)n⁻¹ = σ² + e².
Proof.
(1)Since n ≠ 0, n⁻¹ exists and n·n⁻¹ = 1A19, A10
(2)S(c) = S(x̄) + n e²L1
(3)S(c)n⁻¹ = S(x̄)n⁻¹ + (n e²)n⁻¹(2), A1, A3, E
(4)(n e²)n⁻¹ = (n·n⁻¹)e² = 1 × e² = e²A2, A3, (1), A8
(5)S(c)n⁻¹ = σ² + e²(3), (4)
(6)Moving the reference away from the mean increases the mean square by exactly e². The variance is the value measured at the point where it is smallest.(5), T2
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