Definition D1
For paired data (x₁,y₁), …, (xₙ,yₙ), let dᵢ = xᵢ − x̄, eᵢ = yᵢ − ȳ, and set A = Σdᵢ², B = Σeᵢ², P = Σdᵢeᵢ. When 0 < A and 0 < B, r = P·(√(AB))⁻¹ is called the correlation coefficient.
Proof.
(1)Σaᵢ means a₁ + ⋯ + aₙ; u − v means u + (−v); u ≤ v means u < v or u = v.convention
(2)Since 0 < A and 0 < B, 0 < ABassumption, A8
(3)Therefore √(AB) exists and 0 < √(AB), (√(AB))² = AB(2), A15
(4)0 < A means the data x is not constant. For constant data the correlation coefficient is undefined.convention
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Definition D2
Changing only the slot that determines x in the rule that generated the data to a different value is called an intervention on x. If, with everything else held fixed, intervening on x changes the value of y, then x is called a cause of y.
Proof.
An intervention is not an observation but a change to the rule. The same data can arise from different rules.convention
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(1)For every t, set q(t) = Σ(dᵢt + eᵢ)². Each term is a square, so 0 ≤ q(t)A2, A3
(2)(dᵢt + eᵢ)² = dᵢ²t² + 2dᵢeᵢt + eᵢ²A12, A17, A18
(3)q(t) = A t² + 2P t + B(2), A13, A14
(4)Since 0 < A, the minimum of q is B − (2P)²(4A)⁻¹ = B − P²A⁻¹.D1, A1, A17, A18, A16, A19, E
(5)Since 0 ≤ q(t) holds for every t, the minimum is also at least 0. That is, 0 ≤ B − P²A⁻¹(1), (4)
(6)Adding P²A⁻¹ to both sides gives P²A⁻¹ ≤ B(5), A4, A22, A23, A24, A25
(7)Since 0 < A, multiplying both sides by A gives P² = A(P²A⁻¹) ≤ AB(6), D1, A6, A17, A18, A16, A19
∎
Theorem T1
−1 ≤ r ≤ 1
Proof.
(1)r² = P²·((√(AB))⁻¹)² = P²(AB)⁻¹D1, A15, A16, A17, A18, E
(2)P² ≤ AB and 0 < AB, so multiplying both sides by (AB)⁻¹ gives r² ≤ 1L1, D1, A6, A16, A17, A18, A19
(3)Suppose 1 < r or r < −1.assumption
(4)If r < −1, adding −r to both sides gives 0 < −r − 1, and adding 1 again gives 1 < −r. Set s = −r; then s² = r².(3), A5, A22, A23, A24, A25, A20
(5)If 1 < r, set s = r. In either case 1 < s and s² = r².(3), (4)
(6)Since 0 < 1 < s, we have 0 < s, and multiplying both sides of 1 < s by s gives s = s×1 < s×s = s²(5), A11, A9, A7, A18, A19
(7)By transitivity, 1 < s² = r²(5), (6), A9
(8)But by (2), r² ≤ 1, contradicting trichotomy.(2), (7), A10
(9)Therefore neither 1 < r nor r < −1 holds, and −1 ≤ r ≤ 1(8), A10
∎
Theorem T2
There exists data with r = 1 in which x is not a cause of y.
Proof.
(1)Consider the rule in which z takes the values 1, 2, 3, 4, 5 and x = z, y = 2z.assumption, D2
(2)The observed data are (1,2), (2,4), (3,6), (4,8), (5,10).(1), N
(3)x̄ = 3, ȳ = 6, and dᵢ = −2, −1, 0, 1, 2, eᵢ = −4, −2, 0, 2, 4(2), D1, N
(4)P = 8 + 2 + 0 + 2 + 8 = 20, A = 10, B = 40(3), D1, N
(5)AB = 400 and √400 = 20, so r = 20 × 20⁻¹ = 1(4), D1, A15, A16, N
(6)Now fix z = 3. By the rule, y = 2 × 3 = 6(1), N
(7)Changing the slot that determines x from 3 to 7 leaves y at 2z = 6.(1), (6), D2, N
(8)Intervening on x with everything else fixed does not change y, so x is not a cause of y.(7), D2
(9)There is data with r = 1 in which x is not a cause of y.(5), (8)
(10)The proposition "if r is close to 1, then x is a cause of y" is falsified by this single case.(9), A21
∎
Corollary C
The same data and the same r arise from different causal structures.
Proof.
(1)Under the rule of T2, r = 1 and x is not a cause of y.T2
(2)Consider another rule. Set x directly to 1, 2, 3, 4, 5 and define y = 2x.D2
(3)The data this rule produces are (1,2), (2,4), (3,6), (4,8), (5,10), the same as in T2, so r = 1.(2), T2, N
(4)But changing x from 3 to 7 changes y to 14, so x is a cause of y.(2), D2, N
(5)The two rules give the same data and the same r, yet their causal structures differ.(1), (3), (4)
(6)Data does not determine the rule. Therefore data and r alone cannot settle causation.(5), D2
∎