Why Combine the Two Restrictions?
When treatment is not randomly assigned, the observed outcome difference between treated and untreated units need not equal the average treatment effect. Bounds describe a set of causal effects compatible with the observed data and stated assumptions. Monotone treatment response restricts how treatment can affect each unit; monotone treatment selection restricts how the potential outcomes of treated and untreated units compare. Used together, these restrictions can tighten the bounds, but they can also be incompatible with the observed group means.
Consider a binary treatment \(D\), with \(D=1\) for treated units, and a binary outcome \(Y\). Let \(Y_1\) and \(Y_0\) be the potential outcomes under treatment and no treatment. The observed outcome is \(Y=DY_1+(1-D)Y_0\). Assume consistency and no interference, so the observed outcome for each unit equals its potential outcome under the treatment actually received, and one unit's treatment does not alter another unit's outcome. The target is the population average treatment effect, \(E[Y_1-Y_0]\).
For binary outcomes, MTR says treatment cannot change an individual outcome from 1 to 0. MTS is a pair of comparisons of conditional means; it does not say that every treated unit has higher potential outcomes than every untreated unit. The two restrictions concern different comparisons: MTR compares potential outcomes within a unit, while MTS compares groups under the same potential treatment state.
Observed and Missing Potential-Outcome Means
Let \(q=P(D=1)\), and suppose \(0<q<1\), so both treatment groups occur in the population. Write \(a=E[Y\mid D=1]\) and \(b=E[Y\mid D=0]\). By consistency, \(a=E[Y_1\mid D=1]\) and \(b=E[Y_0\mid D=0]\) are observed. Define the unobserved group means \(x=E[Y_0\mid D=1]\) and \(y=E[Y_1\mid D=0]\). Because the outcome is binary, all four means lie between 0 and 1.
Partitioning the population into treated and untreated groups gives an expression for the target in terms of those observed and missing means:
Here \(a-x\) is the average effect among treated units, and \(y-b\) is the average effect among untreated units. The population effect is their treatment-prevalence-weighted average. The key task is to restrict \(x\) and \(y\) using the assumptions.
How Each Assumption Restricts the Endpoints
Under MTR, treated units have \(Y_1\geq Y_0\), so their group means satisfy \(x\leq a\). Untreated units also have \(Y_1\geq Y_0\), so \(y\geq b\). Binary outcomes add \(x\geq0\) and \(y\leq1\). Thus MTR alone permits \(0\leq x\leq a\) and \(b\leq y\leq1\). Substituting the largest and smallest allowable values into the expression for \(\tau\) gives the MTR bounds.
Under MTS, the restriction for \(Y_0\) is \(x\geq b\), and the restriction for \(Y_1\) is \(a\geq y\). With the binary-outcome limits, this gives \(b\leq x\leq1\) and \(0\leq y\leq a\). MTS alone therefore gives:
The lower endpoint uses the largest allowed \(x\) and smallest allowed \(y\); the upper endpoint uses the smallest allowed \(x\) and largest allowed \(y\). Without MTR, MTS alone does not require either group’s treatment effect to be nonnegative.
When both assumptions hold, their restrictions intersect: \(b\leq x\leq a\) and \(b\leq y\leq a\). In particular, compatibility requires \(a\geq b\). The smallest possible effect sets \(x=a\) and \(y=b\); the largest sets \(x=b\) and \(y=a\). This yields the combined bounds:
Why These Bounds Are Sharp
An interval is sharp when every value in it is compatible with the observed data and assumptions, and no value outside it is compatible. To see why the combined endpoints are attainable, first take \(x=a\) and \(y=b\). In both groups, the two potential-outcome means are then equal. Binary potential outcomes can be coupled so that \(Y_1=Y_0\) for each unit in each group. MTR holds, MTS holds because both group means for each treatment state are ordered equally when \(a\geq b\), and the resulting average effect is 0.
For the upper endpoint, take \(x=b\) and \(y=a\). In each group, the mean under treatment is \(a\) and the mean without treatment is \(b\). A valid binary construction assigns potential outcomes \((Y_0,Y_1)=(1,1)\) to a fraction \(b\), \((0,1)\) to a fraction \(a-b\), and \((0,0)\) to a fraction \(1-a\). These nonnegative fractions sum to 1 when \(0\leq b\leq a\leq1\). This construction satisfies MTR and gives group means \(b\) and \(a\). MTS holds with equal group means under each treatment state, and the average effect is \(a-b\).
Mixing these endpoint constructions in any proportion preserves the observed means and assumptions and produces every intermediate average effect. Thus the combined interval is not merely a conservative envelope: it is the full set of effects allowed by these restrictions.
Worked Example: Computing the Combined Bounds
Suppose a hypothetical population has treatment prevalence \(q=0.30\), an observed treated outcome mean \(a=0.70\), and an observed untreated outcome mean \(b=0.40\). The outcome is binary. Assume consistency, no interference, MTR, and MTS.
First check compatibility: \(a=0.70\geq b=0.40\), so the assumptions do not conflict with the observed means. The missing means must satisfy \(0.40\leq x\leq0.70\) and \(0.40\leq y\leq0.70\). The average effect is \(\tau=0.30(0.70-x)+0.70(y-0.40)\).
At the lower endpoint, set \(x=0.70\) and \(y=0.40\). Then \(\tau=0.30(0.70-0.70)+0.70(0.40-0.40)=0\). At the upper endpoint, set \(x=0.40\) and \(y=0.70\). Then \(\tau=0.30(0.70-0.40)+0.70(0.70-0.40)=0.09+0.21=0.30\). The sharp combined bounds are therefore \([0,0.30]\), or from 0 to 30 percentage points on the binary-outcome scale.
For comparison, MTR alone would give an upper endpoint of \(0.30(0.70)+0.70(1-0.40)=0.21+0.42=0.63\). MTS alone would give a lower endpoint of \(0.30(0.70-1)-0.70(0.40)=-0.09-0.28=-0.37\), and an upper endpoint of \(0.70-0.40=0.30\). The joint restrictions raise the lower endpoint from \(-0.37\) to 0 and retain the tighter upper endpoint.
Worked Example: Detecting Incompatible Assumptions
Suppose hypothetical population means are \(a=0.55\) among treated units and \(b=0.62\) among untreated units. MTS requires \(E[Y_1\mid D=0]=y\leq a=0.55\), while MTR among untreated units requires \(y\geq b=0.62\). No value of \(y\) can meet both inequalities, so the assumptions and these population means cannot all be true.
The contradiction can also be seen from the implication \(a\geq y\geq b\), which would require \(0.55\geq0.62\). This is a compatibility failure, not an instruction to report reversed bounds. If these are sample estimates, sampling variation may produce this ordering even when population restrictions are plausible; the sample discrepancy alone does not establish a population-level logical contradiction.
Worked Example: Comparing Bounds Under Different Treatment Prevalence
Consider another hypothetical population with \(q=0.40\), \(a=0.65\), and \(b=0.25\). Compatibility holds because \(0.65\geq0.25\). The combined bounds are \([0,0.65-0.25]=[0,0.40]\).
The separate restrictions show what each assumption contributes. Under MTR alone, the upper endpoint is \(0.40(0.65)+0.60(1-0.25)=0.26+0.45=0.71\), giving \([0,0.71]\). Under MTS alone, the lower endpoint is \(0.40(0.65-1)-0.60(0.25)=-0.14-0.15=-0.29\), and the upper endpoint is \(0.65-0.25=0.40\), giving \([-0.29,0.40]\). Together, the assumptions raise the lower endpoint to zero and reduce the MTR-only upper endpoint from 0.71 to 0.40.
In the combined formula, \(q\) cancels from the upper endpoint because both groups’ maximum average effects are \(a-b\). This does not mean treatment prevalence is irrelevant in general: it affects the separate-assumption bounds and the decomposition of the average effect.
Diagnostics, Interpretation, and Limitations
A practical check is to calculate the observed group means before reporting any combined interval. If \(a<b\), inspect the outcome coding, treatment-group definitions, and data summaries. If those are correct, the population versions of MTR and MTS cannot both hold for these means. If \(a\geq b\), that necessary compatibility check passes, but it does not establish that either substantive assumption is credible.
The bounds are identification bounds, not a point estimate and not a confidence interval. An interval such as \([0,0.30]\) says that the stated population assumptions and observed means permit average effects from 0 to 0.30; it does not say that the effect is equally likely to be any value in that interval. In data analysis, estimated means are subject to sampling uncertainty, which must be handled separately.
The restrictions are also deliberately substantive. MTR rules out treatment harm for every unit; MTS asserts favorable selection on both potential outcomes. Either may be implausible in some applications. Violations can make the resulting interval uninformative or misleading, and passing the compatibility check cannot verify the assumptions. Without these restrictions, the worst-case bounds for a binary outcome are generally wider. With only one restriction, the bounds may also be wider, as the examples demonstrate. The central advantage of combining the assumptions is tighter endpoints when both are defensible and jointly compatible.
Check Your Understanding
Try each question, then check the solutions below.
- For binary outcomes, state the restrictions on \(x\) and \(y\) imposed by MTR alone.
- Suppose \(a=0.80\), \(b=0.30\), and \(q=0.20\). Find the combined bounds and the MTR-only upper endpoint.
- Why does \(a<b\) rule out the joint assumptions?
- For the values in question 2, explain why the upper combined endpoint does not depend on \(q\).
Solutions. (1) MTR gives \(0\leq x\leq a\) and \(b\leq y\leq1\), because treatment cannot lower either group’s mean outcome. (2) The combined bounds are \([0,a-b]=[0,0.50]\). The MTR-only upper endpoint is \(0.20(0.80)+0.80(1-0.30)=0.16+0.56=0.72\). (3) MTR among untreated units gives \(y\geq b\), while MTS for \(Y_1\) gives \(y\leq a\). If \(a<b\), these inequalities cannot both hold. (4) Under both assumptions, the upper endpoint sets \(x=b\) and \(y=a\), so \(\tau=q(a-b)+(1-q)(a-b)=a-b=0.50\); the weights sum to one.