Causal null hypotheses of sustained treatment strategies: What can be tested with an instrumental variable?

Sometimes instrumental variable methods are used to test whether a causal effect is null rather than to estimate the magnitude of a causal effect. However, when instrumental variable methods are applied to time-varying exposures, as in many Mendelian randomization studies, it is unclear what causal null hypothesis is tested. Here, we consider different versions of causal null hypotheses for time-varying exposures, show that the instrumental variable conditions alone are insufficient to test some of them, and describe additional assumptions that can be made to test a wider range of causal null hypotheses, including both sharp and average causal null hypotheses. Implications for interpretation and reporting of instrumental variable results are discussed. Electronic supplementary material The online version of this article (10.1007/s10654-018-0396-6) contains supplementary material, which is available to authorized users.


SA Swanson, JA Labrecque, MA Hernán
The below results rely on combinations of the following assumptions: 0 if a 0 < a 0 for all subjects and all a 0 , a 0 ∈ Supp(A 0 ); (A3) for any a k = a k ∈ Supp(A k ) = [0, 1] such that a j ≤ a j , j ∈ {1, ..., k}.
Theorem 1. Under (A1), suppose the sharp time-fixed causal null holds: for all subjects and all a 0 , a 0 ∈ Supp(A 0 ).
Then we have: Proof. For any z, z ∈ Supp(Z) and any a 0 ∈ Supp(A 0 ), we have: where the first equality in each expression follows from (A1) and the second follows from the sharp time-fixed causal null and consistency. It immediately follows that Note: the logic of this proof could similarly apply for Y k and a single treatment at time j < k if Z Y aj k and an analogous sharp time-fixed causal null regarding Y aj k = Y a j k . However, this particular condition Z Y aj k will not be reasonable if Z affects treatment at other times besides time j and treatment at other times besides time j can affect Y k .
Theorem 2. Under (A2), suppose the sharp joint causal null holds: for all subjects and all a k , a k ∈ Supp(A k ).
Then we have: Proof. For any z, z ∈ Supp(Z) and any a k ∈ Supp(A k ), we have: where the first equality in each expression follows from (A2) and the second follows from the sharp joint causal null and consistency. It immediately follows that Proof. Define a min = min(Supp(A 0 )) and a max = max(Supp(A 0 )). For any z, z ∈ Supp(Z), we have: where the first equality in each expression follows from (A1) and the inequality follows from (A3 Theorem 4. Under (A2) and (A4), suppose the average joint causal null holds: where Supp(A k ) = [0, 1]. Then we have: Proof. For any z, z ∈ Supp(Z), we have: where the equality in each expression follows from (A2) and the inequality fol- k ] under the average joint causal null, it immediately follows that Note: A corollary of Theorem 4 is that (A2) and (A4) imply that the average causal effect under continuous treatment vs. continuous no treatment must be greater than or equal to the intention-to-treat association when we have a binary instrument Z. It follows immediately from (A4) that the average causal effect is non-negative, so we only need to show that this average causal effect is not between zero and the intention-to-treat association. The latter can be proven readily by contradiction. (Of course if we reversed the direction of the monotonic treatment effect condition (A4) then under similar logic we would likewise conclude that the absolute value of the average causal effect under continuous treatment vs. continuous no treatment must be greater than or equal to the absolute value of the intention-to-treat association when we have a binary instrument Z.)