Efficiency for evaluation of disease etiologic heterogeneity in case-case and case-control studies

ARE = 1 for categorical exposure with the constrained model

Consider (K + 1)-level categorical exposure, represented by a K-dimension binary indicator X _i = (X_i1, …, X _iK ). Let the (K + 1)th category be the reference category. X _i = (X_i1, …, X _iK ) = 1 _k denotes a K-dimension vector with the kth element = 1 and all other elements = 0. The components of S β 1 o = 0 include

and,

Subtract (10) from (9), we have ∑ i = 1 n I ( X i = 0 ) { I ( Y i = 0 ) − p ( Y i = 0 | U i ) } = 0 , where 0 is a K-dimension zero vector.

Let a ̃ j k ( 1 ) and a ̃ j ( 0 ) denote a j i − with X _i = 1 _k and with X _i = 0, respectively. We have

Also,

It follows that S α j o = S α j A .

ARE = 1 for α _x = α _w = 0

Suppose J = 2. Since α _xj = α _wj = 0, a_2i = a₂₁ for all i (i = 1, …, n) and thus, I α β 1 o = a 21 ∑ i = 1 n q i U i ′ U i under the unconstrained model. From (6), it follows that Δ I = I α o − a 21 I α β 1 o = 0 , that is, I α o = I α A . This proof can be straightforwardly extended to situations in which J > 2 and/or under the constrained model.

ARE derivation for continuous exposure and no confounder

When there is continuous exposure and no confounders, we have U i ′ U i = 1 X i b X i X i 2 and ∑ i = 1 n q i U i ′ U i − 1 = 1 / e ∑ i = 1 n q i X i 2 − ∑ i = 1 n q i X i − ∑ i = 1 n q i X i ∑ i = 1 n q i , where e = ∑ i = 1 n q i X i 2 { ∑ i = 1 n q i } − ∑ i = 1 n q i X i 2 . The (1,1)th element of the (d₁, d₂)-block of ΔI (6) is

It follows that e δ d 1 d 2 11 can be written as:

This can be simplified by rewriting the terms according to the orders of i₁, i₂, i₃ as follows:

where s i 1 i 2 i 3 , d = a d i 1 ( X i 2 − X i 3 ) + a d i 3 ( X i 1 − X i 2 ) − a d i 2 ( X i 1 − X i 3 ) and t i 1 i 2 i 3 , d = a d i 1 X i 1 ( X i 2 − X i 3 ) + a d i 3 X i 3 ( X i 1 − X i 2 ) − a d i 2 X i 2 ( X i 1 − X i 3 ) . Using similar algebra for the other elements of the (d₁, d₂)-block matrix of ΔI, we obtain

Now, assume J = 2. It follows that ΔI, I ^A (α) and I ^o (α) are all 2 × 2 matrices. This gives

and,

We denote the set of {i₁, i₂, i₃ : 1 ≤ i₁ < i₂ < i₃ ≤ n} as {m : m = 1, …, M}, q i 1 q i 2 q i 3 as Q _m , s i 1 i 2 i 3 , 1 as s _m , and t i 1 i 2 i 3 , 1 as t _m . It follows that

where ζ 1 = ∑ i = 1 n 1 a 2 i ( 1 − a 2 i ) X i 2 { ∑ i = 1 n 1 a 2 i ( 1 − a 2 i ) } − ∑ i = 1 n 1 a 2 i ( 1 − a 2 i ) X i 2 , and

where ζ₂ = ζ₁ + ζ₃/e + ζ₄/e², ζ 3 = ∑ i = 1 n 1 a 2 i ( 1 − a 2 i ) x i 2 ∑ m = 1 M Q m s m 2 + ∑ i = 1 n 1 a 2 i ( 1 − a 2 i ) ∑ m = 1 M Q m t m 2 − 2 ∑ i = 1 n 1 a 2 i ( 1 − a 2 i ) x i ∑ m = 1 M Q m s m t m , and ζ 4 = ∑ m = 1 M Q m s m 2 ∑ m = 1 M Q m t m 2 − ∑ m = 1 M Q m s m t m 2 . Then,

ARE ≤ 1 for a continuous exposure and no confounder

We can rewrite ζ₄ as

Similarly, e can be rewritten as ∑ i 1 < i 2 q i 1 q i 2 ( X i 1 − X i 2 ) 2 ≥ 0 . Denote the numerator and denominator of ARE in (11) as A and B, respectively. Therefore,

where b_2i = a_2i(1 − a_2i). Because ζ₄ ≥ 0, we have ∑ m = 1 M Q m s m 2 ∑ m = 1 M Q m t m 2 1 / 2 ≥ ∑ m = 1 M Q m s m t m . It follows that

Note that e, ζ₄, and b_2i are all non-negative. Therefore B− A ≥ 0, that is, ARE ≤ 1.