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PROOF 64

Prove.

The mean makes the deviations sum to zero

On finite data.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

Definition D1

For data x₁, …, xₙ (n ≥ 1), a value c satisfying n·c = x₁ + ⋯ + xₙ is called the mean of the data and is written x̄.

Proof.

(1)Σxᵢ means x₁ + ⋯ + xₙ; nc means c added n times; u − v means u + (−v).convention
(2)Since n is 1 added n times, n × c = (1 + ⋯ + 1)c = c + ⋯ + c. Therefore nc read as a product is the same.A5, A8, A6, A1
(3)Since n ≥ 1, n ≠ 0.A11
(4)This definition says nothing about whether such a c exists or is unique. Both are proved below.convention

∎

Definition D2

The difference xᵢ − c between a data point xᵢ and a reference value c is called the deviation with respect to c. When the reference value is the mean, it is simply called the deviation.

Proof.

A deviation is determined only once a reference value is fixed. Change the reference and the deviations change.convention

∎

Lemma L1

For every c, Σ(xᵢ − c) = Σxᵢ − nc

Proof.

(1)xᵢ − c = xᵢ + (−c)D2, convention
(2)Σ(xᵢ + (−c)) = Σxᵢ + n(−c)(1), A12
(3)n(−c) + nc is (−c) + c added n times, hence 0.A1, A2, A4, A3
(4)Adding −(nc) to both sides gives n(−c) = −(nc)(3), E, A1, A2, A3, A4
(5)Σ(xᵢ − c) = Σxᵢ + (−(nc)) = Σxᵢ − nc(1), (2), (4)

∎

Theorem T1

If c is the mean of the data, then Σ(xᵢ − c) = 0. That is, the deviations sum to 0.

Proof.

(1)Suppose c is the mean. That is, nc = Σxᵢ.assumption, D1
(2)Σ(xᵢ − c) = Σxᵢ − ncL1
(3)　　　　　　 = Σxᵢ + (−Σxᵢ) = 0(1), (2), A4

∎

Theorem T2

A c satisfying Σ(xᵢ − c) = 0 is the mean of the data, and there is at most one such c.

Proof.

(1)Suppose Σ(xᵢ − c) = 0.assumption
(2)Σxᵢ − nc = 0(1), L1
(3)Adding nc to both sides gives nc = Σxᵢ(2), E, A1, A2, A3, A4
(4)Therefore c is the mean.(3), D1
(5)Suppose c and c′ both satisfy the condition.assumption
(6)nc = Σxᵢ = nc′(3), (5)
(7)Since n ≠ 0, c = c′(6), A11, A10

∎

Corollary C

Such a c actually exists, and x̄ = (x₁ + ⋯ + xₙ)n⁻¹.

Proof.

(1)Since n ≠ 0, n⁻¹ exists and n·n⁻¹ = 1A11, A9
(2)Set c = (Σxᵢ)n⁻¹.(1)
(3)nc = n((Σxᵢ)n⁻¹) = (n·n⁻¹)Σxᵢ = 1 × Σxᵢ = Σxᵢ(2), A7, A8, (1), A6
(4)Therefore c is the mean, and by T2 it is the only mean.(3), D1, T2
(5)By T1 the deviations sum to 0 at this c, and by T2 this is the only value that makes the deviations sum to 0.(4), T1, T2

∎

Remark

Can A9 be removed from this proof?

That the mean is unique and that the mean exists are different claims. Check which block leans on division.