Testing Mediation: Principal Strata and Observable Bounds
Suppose treatment changes both an intermediate variable and the final outcome. Could the entire outcome change be explained by movement in that intermediate variable? Randomizing treatment does not randomize the mediator, but it identifies distributions that any such explanation must reproduce. (Kwon and Roth 2024) uses those distributions to test the sharp claim of full mediation.
1. Principal strata and the null
Let \(D\) be randomized, \(M(d)\) the potential mediator, and \(Y(d,m)\) the potential outcome. The sharp null is \(Y(1,m)=Y(0,m)\) almost surely for every \(m\). For a finite mediator support \(\{m_0,\ldots,m_{K-1}\}\), define
The observed mediator distributions give the row and column sums of \(\theta\). Additional restrictions, such as monotonicity, define the feasible set \(\Theta_I\). For an ordered mediator, monotonicity sets \(\theta_{lk}=0\) when \(l>k\).
2. The observable inequality
Define the difference between two outcome subdistributions,
Let \(\nu_k\) be the fraction of \(k\)-stayers—those with \(M(0)=M(1)=m_k\)—whose outcome changes under treatment.
Proposition 1. Random assignment implies
Hence the sharp null requires a single \(\theta\in\Theta_I\) for which \(\sup_A\Delta_k(A)\le\sum_{l\ne k}\theta_{lk}\) simultaneously for every \(k\).
3. Account for the possible sources of change
An increase in the mass of \((Y\in A,M=m_k)\) can come from people who enter mediator state \(k\) under treatment. Their total mass is \(\sum_{l\ne k}\theta_{lk}\). It can also come from stayers whose outcomes change. Their mass is at most \(\theta_{kk}\nu_k\). People leaving state \(k\) can only subtract from this upper bound.
Taking the supremum over outcome sets gives the inequality. The argument only counts how much probability mass entrants and affected stayers can contribute; it does not require random assignment of the mediator.
For a binary monotone mediator, the two observed mediator probabilities determine the stratum shares. With several mediator values, the same marginal change can result from different movements between states. Compatibility with full mediation therefore asks whether at least one feasible allocation explains all the outcome differences.
4. How much can be learned
Under ordered monotonicity, the smallest feasible stayer share has the closed form
The allocation below supplies the remaining step: a feasible matrix attaining these minima jointly. Separate minimizations would not suffice, because their solutions might assign the same people to incompatible strata.
The restrictions are most informative when a substantial stayer mass is unavoidable. Binning a continuous mediator can create apparent stayers who actually change mediator values within a bin, so it needs an additional restriction on within-bin outcome effects. The sharpness construction and mean-effect bounds explain what these inequalities exhaust and what remains unidentified.
5. The mass-accounting proof in indicators
Let \(S_{lk}=\{M(0)=m_l,M(1)=m_k\}\), and abbreviate \(Y_d=Y(d,M(d))\). Random assignment lets us write the observed difference as
The last sum is nonnegative and is subtracted. Each entrant term is at most \(\theta_{lk}\). For stayers,
The first term is therefore at most \(\theta_{kk}\nu_k\). Taking the supremum over \(A\) proves the inequality. When \(\theta_{kk}=0\), the unconditional affected mass is zero and the conditional fraction \(\nu_k\) is not defined; one should retain the inequality in terms of mass rather than divide by zero.
The supremum permits \(A\) to contain disconnected parts of the outcome support. Threshold sets \((-\infty,y]\) form a smaller class and can miss a violation, as the three-outcome example below shows. For a discrete outcome, the full restriction ranges over all subsets of cells.
6. A binary-mediator example
Suppose \(M(1)\geq M(0)\), \(\Pr(M=1\mid D=0)=0.3\), and \(\Pr(M=1\mid D=1)=0.6\). The feasible strata are then
There are no entrants to mediator state zero. Full mediation thus requires \(\Delta_0(A)\leq0\) for every \(A\): the treated state-zero subdistribution must be a submeasure of the control state-zero subdistribution.
For a binary outcome, imagine state-zero subdistribution masses \((0.30,0.10)\) under treatment and \((0.25,0.45)\) under control, listed at outcomes zero and one. Their totals are \(0.4\) and \(0.7\), as required. Nevertheless,
Treatment has increased the mass at outcome zero within a mediator state nobody enters. At least \(0.05\) of the population must therefore be state-zero stayers whose outcomes change. Among the \(0.4\) stayers, the affected fraction is at least \(0.05/0.4=0.125\).
At mediator state one, up to \(0.3\) of the population enters. The corresponding restriction is \(\sup_A\Delta_1(A)\leq0.3\). A change there can be explained by these entrants before it forces any direct effect among stayers. This asymmetry is the information supplied by mediator monotonicity.
7. Why a single feasible stratum matrix matters
Let \(p_{dk}=\Pr(M=m_k\mid D=d)\). The matrix must satisfy
For ordered monotonicity, entries below the diagonal vanish. The net mass crossing the threshold \(m_k\) is
Every entrant to state \(k\) from below is part of this crossing mass. Thus
Nonnegativity now gives the stated positive-part bound. This establishes necessity. Joint attainment requires an allocation preserving both margins; the construction below supplies it.
For general restrictions on \(\Theta_I\), the corresponding test is a joint feasibility problem: find one nonnegative matrix with the observed margins satisfying every outcome inequality. Optimizing coordinates separately need not produce such a matrix.
8. What coarsening changes
For discrete outcomes with cells \(y_1,\ldots,y_J\), let \(\delta_{kj}\) be the treated-minus-control subdistribution difference in cell \(j\). Then
The maximizing set includes exactly the positive-difference cells. After merging two cells, their contribution becomes the positive part of their sum. Since \((a+b)_+\leq a_++b_+\), merging cannot strengthen the restriction.
The difference between threshold tests and the full set restriction can be substantial. Consider three ordered outcome values, with state-\(k\) subdistribution masses \((0.18,0.04,0.18)\) under treatment and \((0.10,0.20,0.10)\) under control. Both cell totals are \(0.40\), and the differences are \((0.08,-0.16,0.08)\).
Every lower-threshold difference has absolute value at most \(0.08\): the cumulative differences are \(0.08,-0.08,0\). But the set consisting of the first and third outcomes gives
If the feasible entrant mass is \(0.10\) and the stayer mass is \(0.30\), the full inequality requires at least \(0.06\) of population mass, or \(0.06/0.30=0.20\) of these stayers, to change outcome. A test examining only threshold differences would miss this violation.
For general subdensities, write \(p_{dk}\) for their total masses. The identity \(x_+=(|x|+x)/2\) gives
These measures have their original mediator-cell masses. Normalizing each to a probability distribution would discard the composition change that the entrant and stayer calculation uses.
Treatment randomization thus identifies the two margins and the outcome subdistributions. Restrictions on mediator movement bound how much of an outcome discrepancy those composition changes can explain. Any excess requires affected stayers.
Mediator coarsening changes the principal strata themselves. Someone moving from \(m\) to \(m'\) inside one bin becomes an apparent stayer, although full mediation permits \(Y(0,m)\ne Y(0,m')\). Treating that difference as evidence of a direct effect would be incorrect without an additional restriction on within-bin responses. The companion construction keeps the mediator states fixed and shows how the distributional discrepancy translates into sharp affected-mass and mean-effect bounds.
Constructing all minimum stayer shares at once
Write \(F_d(k)=\sum_{j\leq k}p_{dj}\) and set \(F_d(-1)=0\). Ordered monotonicity requires \(F_0(k)\geq F_1(k)\) at every cutoff. The crossing mass in the earlier calculation is \(c_k=F_0(k-1)-F_1(k-1)\). The following allocation proves joint attainment of the diagonal lower bounds.
Process destination columns \(k=0,\ldots,K-1\) in order. Before filling column \(k\), retain the unallocated mass from origin rows \(0,\ldots,k-1\) in a pool. Its total is \(c_k\): those rows supplied \(F_0(k-1)\) mass and the already filled columns used \(F_1(k-1)\). Fill as much of column \(k\) as possible from this pool, in any order. If its demand \(p_{1k}\) exceeds the pool, fill the deficit from row \(k\). This assigns
There is enough mass in row \(k\), because \(F_0(k)\geq F_1(k)\) is precisely \(p_{0k}+c_k\geq p_{1k}\). Add the unused part of row \(k\) to the pool. Its new total is \(c_k+p_{0k}-p_{1k}=c_{k+1}\). This proves the induction. At the last column the pool is exhausted because both margins total one. No mass ever moves to a lower destination, and every row and column margin is preserved. A single feasible matrix therefore attains all the minimum stayer shares simultaneously. This construction uses ordered monotonicity alone; additional restrictions on the stratum matrix can rule it out.