PROOF 172
Prove.
The correlation coefficient lies between −1 and 1
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 sequences a₁, …, aₙ and b₁, …, bₙ, set A = Σaᵢ², B = Σbᵢ², P = Σaᵢbᵢ.
Proof.
(1)Σaᵢ denotes a₁ + ⋯ + aₙ, u − v denotes u + (−v), and u ≤ v denotes u < v or u = v.convention
(2)A and B are sums of squares, hence non-negative.A13, A19
∎
Definition D2
For data pairs (xᵢ, yᵢ), let dᵢ = xᵢ − x̄, eᵢ = yᵢ − ȳ and set aᵢ = dᵢ, bᵢ = eᵢ. When 0 < A and 0 < B, the quantity r = P(√(AB))⁻¹ is called the correlation coefficient.
Proof.
(1)Since 0 < A and 0 < B, 0 < ABassumption, A21
(2)Therefore √(AB) exists with 0 < √(AB), (√(AB))² = AB, and (√(AB))⁻¹ exists as well.(1), A28, A9, A23
∎
Lemma L1
Setting q(t) = Σ(aᵢt + bᵢ)² gives q(t) = A t² + 2P t + B, and 0 ≤ q(t) for all t.
Proof.
(1)(aᵢt + bᵢ)² = aᵢ²t² + 2(aᵢbᵢ)t + bᵢ²A10, A6, A7
(2)Summing all terms: q(t) = Σ(aᵢ²t²) + Σ(2(aᵢbᵢ)t) + Σbᵢ²(1), A11
(3)Σ(aᵢ²t²) = t²Σaᵢ² = A t² and Σ(2(aᵢbᵢ)t) = 2t Σaᵢbᵢ = 2P tA12, A6, A7, D1
(4)Therefore q(t) = A t² + 2P t + B(2), (3), D1
(5)Each term (aᵢt + bᵢ)² is non-negative, and a sum of non-negative quantities is non-negative, so 0 ≤ q(t)A13, A19
∎
Theorem T1
If 0 < A, then P² ≤ AB
Proof.
(1)Set D = (2P)² − 4AB.convention
(2)Suppose 0 < D.assumption
(3)Since 0 < A and 0 < D, there exist α, β with α < β such that q(t) = A(t − α)(t − β)assumption, (2), L1, A16
(4)By density there is a t₁ with α < t₁ < β.(3), A25
(5)Adding −α and −β to both sides respectively gives 0 < t₁ − α and t₁ − β < 0(4), A18, A1, A2, A3, A4
(6)Since 0 < A and 0 < t₁ − α, 0 < A(t₁ − α)assumption, (5), A21
(7)Since t₁ − β < 0, q(t₁) = A(t₁ − α)(t₁ − β) < 0(3), (5), (6), A22, A6
(8)But by L1, 0 ≤ q(t₁), contradicting trichotomy.(7), L1, A23
(9)Therefore 0 < D fails, and D ≤ 0(8), A23
(10)Since (2P)² = 4P², we have 4P² − 4AB ≤ 0, and adding 4AB to both sides gives 4P² ≤ 4AB(1), (9), A10, A6, A7, A17, A1, A2, A3, A4, N
(11)Adding 1 to 0 < 1 repeatedly gives 0 < 4, so cancelling 4 yields P² ≤ AB(10), A26, A18, A24, A20
∎
Theorem T2
Suppose 0 < A. Then P² = AB holds if and only if there is a k with bᵢ = k aᵢ for all i.
Proof.
(1)First suppose P² = AB.assumption
(2)Since 0 < A, A ≠ 0 and A⁻¹ exists. Set t₀ = −(P A⁻¹).assumption, A23, A9
(3)q(t₀) = A t₀² + 2P t₀ + B = P²A⁻¹ − 2P²A⁻¹ + B = B − P²A⁻¹L1, (2), A6, A7, A8, A9, A30, A32, A1, A2, A3, A4
(4)Since P² = AB, P²A⁻¹ = (AB)A⁻¹ = B, and hence q(t₀) = B − B = 0(1), (3), A6, A7, A9, A8, A4
(5)q(t₀) = Σ(aᵢt₀ + bᵢ)², and each term is non-negative.L1, A13
(6)A sum of non-negative terms equal to 0 forces every term to be 0.(4), (5), A15
(7)The square of a nonzero quantity is positive, so by trichotomy aᵢt₀ + bᵢ = 0(6), A14, A23
(8)Adding −(aᵢt₀) to both sides gives bᵢ = (−t₀)aᵢ, so setting k = −t₀ = P A⁻¹ suffices.(7), (2), A1, A2, A3, A4, A32, A7
(9)Conversely, suppose bᵢ = k aᵢ for all i.assumption
(10)P = Σaᵢ(k aᵢ) = k Σaᵢ² = kA(9), D1, A6, A7, A12
(11)B = Σ(k aᵢ)² = k² Σaᵢ² = k²A(9), D1, A6, A7, A12
(12)P² = (kA)² = k²A² = A(k²A) = AB(10), (11), A6, A7
(13)Therefore the two conditions are equivalent.(8), (12)
∎
Theorem T3
If 0 < A and 0 < B, then −1 ≤ r ≤ 1
Proof.
(1)Since r = P(√(AB))⁻¹ and (√(AB))² = AB, r² = P²(AB)⁻¹D2, A6, A7, A9, A28
(2)By T1, P² ≤ AB, and since 0 < AB, 0 < (AB)⁻¹T1, D2, A27
(3)Multiplying both sides by (AB)⁻¹ gives r² = P²(AB)⁻¹ ≤ (AB)(AB)⁻¹ = 1(1), (2), A20, A9
(4)Adding −r² to both sides gives 0 ≤ 1 − r², and since (1 − r)(1 + r) = 1 − r², we get 0 ≤ (1 − r)(1 + r)(3), A17, A5, A7, A10, A1, A2, A3, A4
(5)Suppose 1 < r. Adding −r to both sides gives 1 − r < 0assumption, A18, A1, A2, A3, A4
(6)Since 0 < 1 and 1 < r, we have 0 < r, and adding 1 to both sides gives 1 < 1 + r, so 0 < 1 + r(5), A26, A24, A18, A2, A3
(7)Then (1 − r)(1 + r) < 0, contradicting (4).(4), (5), (6), A22, A7, A23
(8)Suppose r < −1. Adding 1 to both sides gives 1 + r < 0assumption, A18, A2, A3, A4
(9)Adding −r to both sides also gives 0 < −1 − r, and since 0 < 2, 2 < 2 + (−1 − r) = 1 − r, hence 0 < 1 − r(8), A18, A26, A24, A1, A2, A3, A4
(10)Then (1 − r)(1 + r) < 0, again contradicting (4).(4), (8), (9), A22, A7, A23
(11)By trichotomy neither 1 < r nor r < −1 holds, so −1 ≤ r ≤ 1(7), (10), A23
∎
Corollary C
r = 1 or r = −1 holds if and only if all the points lie on a single line.
Proof.
(1)Since r² = P²(AB)⁻¹, r² = 1 is equivalent to P² = AB.D2, A28, A9, A6, A7, E
(2)If r² = 1, then (r − 1)(r + 1) = r² − 1 = 0, so r = 1 or r = −1(1), A5, A10, A1, A2, A3, A4, A31
(3)Conversely, if r = 1 or r = −1, then r² = 1.A30, A8, N
(4)Therefore r = ±1 is equivalent to P² = AB, and by T2 this is equivalent to the existence of a k with eᵢ = k dᵢ.(1), (2), (3), T2
(5)eᵢ = k dᵢ means yᵢ − ȳ = k(xᵢ − x̄), that is, yᵢ = k xᵢ + (ȳ − k x̄), so all the points lie on one line.(4), D2, A5, A32, A1, A2, A3, A4
(6)B = k²A and 0 < B, so k ≠ 0. That is, this line is not horizontal.T2, D2, A23
(7)If 0 < k, then 0 < kA and AB = (kA)², so √(AB) = kA and r = kA(kA)⁻¹ = 1(6), T2, A21, A29, A9, A6, A7
(8)If k < 0, then kA < 0 and (kA)² = (−(kA))², so √(AB) = −(kA) and r = kA(−(kA))⁻¹ = −1(6), T2, A22, A30, A29, A9, A32, A8
(9)Consider the data x = 1, 2, 3, 4, 5 and y = 7, 5, 3, 1, −1. Since x̄ = 3, ȳ = 3, we get dᵢ = −2, −1, 0, 1, 2 and eᵢ = 4, 2, 0, −2, −4.D2, N
(10)A = 10, B = 40, P = −20, so P² = 400 = AB, and √400 = 20, so r = −20 × 20⁻¹ = −1(9), D1, D2, A28, A9, N
(11)Indeed eᵢ = −2dᵢ and yᵢ = −2xᵢ + 9, so the five points lie on one line.(9), N
(12)Changing the data to y = 1, 2, 3, 5, 4 gives eᵢ = −2, −1, 0, 2, 1 and A = 10, B = 10, P = 9, so P² = 81 < 100 = ABD1, D2, N
(13)Here the fourth point demands k = 2 and the fifth point demands k = 0.5, so there is no k with eᵢ = k dᵢ and equality does not hold.(12), T2, N
(14)So −1 ≤ r ≤ 1 is given by the inequality, and the endpoints ±1 are given by the equality condition. |r| = 1 means not that the relationship is strongest, but that the data lie exactly on one line.T3, (5), (10), (13)
∎
Remark
Can A15 be removed from this proof?
The validity of the inequality and the condition for equality do not live on the same level. Check which block rests on A15.