Why Worst-Case Bounds Require More Than a Weight Range
A marginal sensitivity model does not specify one hidden-confounding adjustment. Instead, it limits how far the treatment propensity conditional on unobserved factors may depart from the propensity given measured covariates. Each allowed departure implies a possible set of inverse-probability weights, and each possible set gives a weighted outcome mean. The goal is to find the smallest and largest means compatible with the model.
The endpoints are not obtained by assigning the largest weight to every outcome. Such an assignment can violate the model’s calibration requirement: averaging over people in an observed treatment arm, the sensitivity multiplier must have mean one within measured-covariate strata. Sharp bounds respect both the multiplier limits and this requirement.
From Odds Distortion to Weight-Multiplier Limits
Under inverse-probability weighting, the nominal treated weight is \(1/e(X)\), and the nominal control weight is \(1/[1-e(X)]\). Replacing the nominal propensity by the propensity conditional on \(U\) multiplies these weights by different factors. Define \(q_1=e(X)/e(X,U)\) for a treated observation and \(q_0=[1-e(X)]/[1-e(X,U)]\) for a control observation. Thus the corresponding weights are \(q_1/e(X)\) and \(q_0/[1-e(X)]\).
To find the limits, let \(r\) be the odds ratio in the sensitivity model, so \(e(X,U)=r e(X)/[1-e(X)+r e(X)]\). Substitution into the treated multiplier gives \(q_1=e(X)+[1-e(X)]/r\). Substitution into the control multiplier gives \(q_0=1-e(X)+r e(X)\). Since \(r\) ranges from \(1/\Gamma\) to \(\Gamma\), these expressions imply the following intervals.
The intervals depend on the nominal propensity. In general, a single common multiplier interval is not the implication of an odds-distortion bound across all covariate values.
Worked Example: Finding the Multiplier Limits
Suppose the nominal propensity is \(e(X)=0.4000\) in a covariate stratum and the sensitivity parameter is \(\Gamma=2\). Find the treated and control multiplier intervals.
For treated observations, $$ q_1\in\left[0.4000+\frac{0.6000}{2},\ 0.4000+2(0.6000)\right]=[0.7000,\ 1.600]. $$ For controls, $$ q_0\in\left[0.6000+\frac{0.4000}{2},\ 0.6000+2(0.4000)\right]=[0.8000,\ 1.400]. $$
The limits differ by treatment arm because changing the conditional propensity affects \(1/e(X,U)\) and \(1/[1-e(X,U)]\) differently. These are limits on multipliers of the nominal weights, not the inverse-probability weights themselves.
Calibration and the Weighted Arm Means
The propensity \(e(X)\) is marginal over \(U\), so the conditional propensity values cannot vary arbitrarily. This imposes a calibration condition on the multipliers. For treated observations, their distribution of \(U\) is weighted by \(e(X,U)\). Consequently, $$ E(q_1\mid A=1,X) =\int \frac{e(X)}{e(X,u)} \frac{e(X,u)f(u\mid X)}{e(X)}\,du =1. $$
For controls, their distribution of \(U\) is weighted by \(1-e(X,U)\), giving \(E(q_0\mid A=0,X)=1\) by the same cancellation. In a finite-sample calculation within a covariate stratum, these conditions are represented by requiring the average multiplier in each arm to be one. With equal base weights, this is \(\sum_i q_i=n\). With unequal base weights \(w_i\), it is \(\sum_i w_iq_i=\sum_i w_i\).
Calibration fixes the denominator within a stratum, so finding an endpoint reduces to optimizing a weighted sum of outcomes. For the lower bound, begin with every multiplier at its lower limit. The remaining calibration budget is \(\sum_i w_i(1-L_i)\). Allocate that budget to the smallest outcomes first, raising each multiplier toward its upper limit before moving to the next outcome. For the upper bound, allocate it to the largest outcomes first. If the budget runs out partway through an observation’s available increase, that multiplier takes an intermediate value. This rule follows directly from an exchange argument: moving a small amount of allowed weighted multiplier from a larger outcome to a smaller one cannot increase the weighted sum.
With multiple covariate strata, apply the calibration condition and the allocation rule within each stratum, then combine the resulting weighted sums using the nominal stratum weights. This preserves the conditional calibration implied by the model rather than imposing only a pooled constraint.
Worked Example: Sharp Bounds for a Treated-Arm Mean
In a hypothetical treated arm, three observations have outcomes \(2\), \(5\), and \(9\) outcome units. Their nominal weights are equal, and the multiplier interval is \([0.7000,1.600]\), as in the preceding example. Calibration requires the three multipliers to sum to \(3\).
For the lower bound, allocate the available multiplier increase to the smallest outcome first: \(q=(1.600,0.7000,0.7000)\). The multipliers sum to \(3.000\), and the mean is $$ \frac{1.600(2)+0.7000(5)+0.7000(9)}{1.600+0.7000+0.7000} =\frac{3.200+3.500+6.300}{3.000} =\frac{13.00}{3.000} \approx 4.333\text{ outcome units}. $$
For the upper bound, allocate the increase to the largest outcome first: \(q=(0.7000,0.7000,1.600)\). Again the multipliers sum to \(3.000\), and $$ \frac{0.7000(2)+0.7000(5)+1.600(9)}{3.000} =\frac{1.400+3.500+14.40}{3.000} =\frac{19.30}{3.000} \approx 6.433\text{ outcome units}. $$
Both assignments satisfy the multiplier limits and calibration condition. The sharp treated-arm mean bounds for this stratum are therefore approximately \(4.333\) to \(6.433\) outcome units.
Bounding a Weighted Average Treatment Effect
For each allowed assignment of treated and control multipliers, the weighted average treatment effect is the treated weighted mean minus the control weighted mean. When the arm-specific multiplier choices are unrestricted by any additional cross-arm condition, the smallest difference combines the smallest treated mean with the largest control mean. The largest difference combines the largest treated mean with the smallest control mean. These combinations are attainable because the two arms’ multiplier assignments can be chosen separately under this model.
Worked Example: Combining Both Arms
Continue with the hypothetical treated outcomes \(2,5,9\) and treated multiplier limits \([0.7000,1.600]\). Suppose the three control outcomes are \(3,4,7\), with equal nominal weights and multiplier limits \([0.8000,1.400]\). In each arm, calibration requires the three multipliers to sum to \(3\).
The treated mean bounds from the previous calculation are \(4.333\) and \(6.433\) outcome units. For the smallest control mean, assign the largest multiplier to the smallest outcome: \(q=(1.400,0.8000,0.8000)\). This assignment sums to \(3.000\), and its weighted mean is $$ \frac{1.400(3)+0.8000(4)+0.8000(7)}{3.000} =\frac{4.200+3.200+5.600}{3.000} =\frac{13.00}{3.000} \approx 4.333\text{ outcome units}. $$
For the largest control mean, assign the largest multiplier to the largest outcome: \(q=(0.8000,0.8000,1.400)\). Then $$ \frac{0.8000(3)+0.8000(4)+1.400(7)}{3.000} =\frac{2.400+3.200+9.800}{3.000} =\frac{15.40}{3.000} \approx 5.133\text{ outcome units}. $$
The lower treatment-effect endpoint is the smallest treated mean minus the largest control mean, \(4.333-5.133=-0.8000\) outcome units. The upper endpoint is the largest treated mean minus the smallest control mean, \(6.433-4.333=2.100\) outcome units. Thus the sharp sensitivity interval in this hypothetical example is approximately \([-0.8000,\ 2.100]\) outcome units. It describes the range compatible with the specified sensitivity model and these data; it is not a confidence interval for sampling uncertainty.
Why the Allocation Rule Is Sharp
The lower-bound allocation rule can be justified by considering any feasible assignment that gives a larger outcome more multiplier than a smaller outcome when the smaller one could still receive more. Suppose \(Y_i<Y_j\), \(q_i<U_i\), and \(q_j>L_j\). Increase \(q_i\) by \(\delta/w_i\) and decrease \(q_j\) by \(\delta/w_j\), where \(\delta>0\) is small enough to keep both multipliers within their limits. The calibration sum is unchanged, while the weighted numerator changes by $$ \delta Y_i-\delta Y_j=\delta(Y_i-Y_j)<0. $$ So the original assignment could not have minimized the numerator. At a minimum, multiplier increases must therefore be exhausted on lower outcomes before being assigned to higher ones. Reversing the inequality proves the corresponding ordering for the maximum. Since the feasible set is closed and bounded, a minimum and maximum exist; the allocation rule constructs feasible assignments attaining them.
If there are ties, any allocation among tied outcomes gives the same objective value. If multiplier limits vary by observation, the same exchange logic applies, but each observation has its own available increase \(w_i(U_i-L_i)\). The optimizer allocates that capacity in outcome order, using a partial increase where the remaining calibration budget is insufficient to fill the next capacity.
Common Mistakes and Scope
- Giving every observation the largest multiplier. This generally violates calibration. The multipliers must average to one within covariate strata, so an increase assigned to some observations must be offset elsewhere.
- Using the treated limits for controls. The odds-distortion model yields distinct formulas for \(q_1\) and \(q_0\). Calculate each from the nominal propensity and \(\Gamma\).
- Calling the sensitivity interval a confidence interval. These endpoints describe worst-case values allowed by a specified model. They do not by themselves quantify sampling variability.
- Interpreting the endpoints as proof of causation. The bounds rely on the causal design assumptions needed for weighting, including consistency and positivity, as well as the stated sensitivity model. They do not establish those assumptions.
The sensitivity parameter controls the range of departures considered: at \(\Gamma=1\), the odds cannot depart from the nominal odds and the multipliers equal one. Larger values allow more distortion and can widen the identified interval. A separate uncertainty analysis is needed to account for sampling error, and the conclusions remain conditional on the measured-covariate adjustment and model specification.
Check Your Understanding
Use the definitions and allocation rule above to answer these questions.
- For \(e(X)=0.2500\) and \(\Gamma=2\), calculate the treated and control multiplier intervals.
- Why must the average sensitivity multiplier equal one within a covariate stratum?
- With equal base weights, outcomes \(1,4,8\), multiplier limits \([0.5,1.5]\), and calibration \(\sum_i q_i=3\), which outcomes receive the multiplier increases for the lower-bound mean?
- In a treatment-effect bound, which control-arm endpoint is paired with the smallest treated-arm mean to form the lower endpoint?
- Explain why a sensitivity interval is not, on its own, a confidence interval.