Why Feasibility Needs More Than a Plausible Mean
A delta adjustment expresses a sensitivity assumption about outcomes that are missing: relative to a reference distribution, how different might those outcomes be? When the outcome is bounded—for example, a score from 0 to 100—the adjustment must also respect the possible values. A shifted mean inside the permitted range is not enough: a shifted distribution can still assign positive probability to impossible values.
This tutorial focuses on the mathematical feasibility of delta values, not on choosing which values are clinically plausible. The key distinction is between shifting the outcome directly and applying a shift on a transformed scale. A direct shift has a constant change in outcome units, but may cross a boundary. A suitable transformation can keep values within the limits, but its delta has a different interpretation.
The reference distribution and the range \([L,U]\) are taken as specified. If the adjustment is conditional on covariates or a missingness pattern, feasibility can differ across those conditions: the same delta must be checked against the relevant reference distributions. A feasible delta also does not establish that the reference model is suitable or that the sensitivity assumption is credible.
Deriving the Direct-Shift Range
For \(Y_\delta=Y_0+\delta\) to be bounded almost surely, every value with positive probability under the reference distribution must remain within the permitted limits after adding \(\delta\). Let \(\operatorname{ess\,inf}(Y_0)\) and \(\operatorname{ess\,sup}(Y_0)\) denote the lower and upper essential bounds of that distribution: values cannot fall below the first or above the second, except on events of probability zero. The direct-shift constraint is therefore
The equivalence follows by subtracting \(Y_0\) from each inequality. To satisfy the lower bound for all possible draws, \(\delta\) must be at least \(L-Y_0\) for every such draw; the most restrictive value is \(L-\operatorname{ess\,inf}(Y_0)\). The upper bound similarly requires \(\delta\leq U-\operatorname{ess\,sup}(Y_0)\). These two conditions together are necessary and sufficient. If the lower endpoint of the resulting interval exceeds the upper endpoint, there is no feasible direct shift.
If the reference distribution can approach both outcome limits, so its essential bounds are \(L\) and \(U\), the only direct shift that keeps its entire distribution inside the limits is \(\delta=0\). By contrast, a model whose reference draws lie well inside the outcome range may allow nonzero shifts. The relevant bounds come from the reference distribution's support—not just its mean.
Worked Example: A Shift Feasible for the Distribution
Suppose a hypothetical reference model for a score bounded from 0 to 100 gives \(Y_0\sim\operatorname{Uniform}(20,80)\). Find the range of direct shifts that keeps every adjusted score within the bounds, and check \(\delta=15\) and \(\delta=25\).
The reference distribution has essential lower bound 20 and essential upper bound 80. Thus
For \(\delta=15\), the adjusted support is \([20+15,80+15]=[35,95]\), which lies within \([0,100]\). The adjusted mean is \(50+15=65\) score points. For \(\delta=25\), the adjusted support is \([45,105]\), so the shift is not distributionally feasible even though its adjusted mean, \(50+25=75\), is within the score range. Feasibility is about all possible draws, not just the mean.
For comparison, if the only requirement were that the adjusted mean stay between 0 and 100, the reference mean of 50 would give \(0\leq50+\delta\leq100\), or \(-50\leq\delta\leq50\). That mean-feasible range is wider than the distributionally feasible range \([-20,20]\). It would include shifts that put part of the reference distribution outside the allowable score range.
Transforming the Outcome Before Shifting
A transformation can encode the bounded range directly. For an outcome strictly inside \((L,U)\), define its scaled logit as
A transformed-scale delta \(\delta\) adds to \(g(Y_0)\), not to \(Y_0\). Inverting the transformation gives
For any finite transformed value, \(0<\operatorname{expit}(z)<1\). Therefore, if \(Y_0\) is strictly inside \((L,U)\) and \(\delta\) is finite, the adjusted value is also strictly inside \((L,U)\). This guarantees the bounds draw by draw. It does not keep the adjusted distribution unchanged in any other respect, nor does it make the transformed-scale delta a constant change in the original outcome units.
Worked Example: A Transformed-Scale Shift for a Score
Consider a hypothetical score on the scale 0 to 100. A reference draw is 60, and the transformed-scale shift is \(\delta=\log(1.5)\), approximately 0.4055. Calculate the adjusted score.
The scaled score is \(60/100=0.60\), so its starting odds are \(0.60/(1-0.60)=0.60/0.40=1.50\). The shift adds \(\log(1.5)\) to the log odds, which multiplies the odds by 1.5. The adjusted odds are \(1.50\times1.5=2.25\), and the adjusted scaled score is
Thus \(Y_\delta=100(0.6923)=69.23\) score points, which is within the allowed range. The increase is about 9.23 points, not 0.4055 points: delta is on the logit scale. The same transformed shift need not produce the same raw-scale increase for other starting scores.
For instance, starting at 20 gives scaled odds \(0.20/0.80=0.25\). After multiplying by 1.5, the odds are \(0.375\), giving a scaled score of \(0.375/(1+0.375)=0.2727\), or 27.27 points. That is an increase of about 7.27 points. The raw-scale effect depends on the starting value.
The transformation above is undefined at \(L\) and \(U\): an outcome at either endpoint has an infinite logit. If the reference distribution assigns positive probability to an endpoint, this method cannot be applied to those draws as written. A separate endpoint model or another explicitly defined approach is required; silently replacing endpoints by nearby interior values changes the model.
Worked Example: A Valid Odds Shift for a Binary Outcome
Suppose a hypothetical reference model gives probability \(p_0=0.40\) of a binary outcome being 1 among missing cases. Apply a log-odds delta \(\delta=\log(1.5)\), and calculate the adjusted probability.
The starting odds are \(p_0/(1-p_0)=0.40/0.60=2/3\), approximately 0.6667. Adding \(\log(1.5)\) to the log odds multiplies the odds by 1.5, giving \((2/3)(1.5)=1\). Converting odds back to a probability,
The adjusted distribution assigns probability 0.50 to 1 and 0.50 to 0, so it remains a valid distribution for an outcome bounded by 0 and 1. Here delta is a log odds ratio: \(\exp(\delta)=1.5\) is the multiplier on the odds, not a 0.4055-unit increase in probability. Any finite delta maps a reference probability strictly between 0 and 1 to another probability strictly between 0 and 1. If \(p_0\) is exactly 0 or 1, its logit is not finite and this formula does not move it away from that endpoint.
Choosing Among Ways to Respect the Bounds
A transformed-scale shift is not the only response to an infeasible direct shift. The options differ in what they assume about the adjusted distribution. They should not be treated as interchangeable ways of implementing the same delta.
| Approach | How it handles bounds | Important consequence |
|---|---|---|
| Direct additive shift | Use only deltas satisfying the support-based range. | Delta is a constant change in outcome units; a nonzero feasible shift may be impossible if the reference support reaches both boundaries. |
| Shift on a transformed scale | Transform interior values, shift, and invert to the bounded scale. | Bounds are respected, but the raw-scale change varies with the starting outcome; delta is in transformed-scale units. |
| Clipping at the limits | Replace values below \(L\) by \(L\) and values above \(U\) by \(U\). | Creates probability mass at the boundaries and is not the same distribution as an unmodified additive shift. |
| Truncation to the limits | Restrict to values in range and renormalize the retained distribution. | Changes the distribution by conditioning on being in range; it is not merely a shift with impossible values deleted. |
Clipping and truncation can produce valid bounded distributions, but they introduce additional modeling choices. Clipping accumulates probability at an endpoint. Truncation changes the relative probabilities of values that remain by renormalizing the distribution. If either is used, its effect should be part of the stated sensitivity model rather than an unreported repair after simulation.
Practical Checks and Common Pitfalls
Before using a delta with bounded outcomes, state whether it is a direct-scale or transformed-scale adjustment, identify the permitted range, and check the resulting distribution—not only its mean. For a direct shift, calculate the support-based interval from the reference distribution. For a transformed shift, verify that the transformation and its inverse are defined for the reference values and that the inverse maps into the permitted range.
- Confusing a mean constraint with distributional feasibility. An adjusted mean inside \([L,U]\) does not rule out draws below \(L\) or above \(U\).
- Assuming a bounded reference variable guarantees a feasible direct shift. If the reference distribution reaches both limits, any nonzero constant shift crosses one of them.
- Interpreting transformed delta in raw outcome units. A logit-scale shift changes raw values by amounts that depend on their starting values.
- Ignoring endpoint values. The logit transformation is defined only for values strictly between the limits; endpoint mass needs explicit treatment.
- Repairing invalid draws without naming the change. Clipping or truncation changes the distribution and must be recognized as part of the model.
The practical implication is that feasibility is a constraint on model specification. It can rule out a proposed direct delta for a particular reference distribution, or motivate a transformed-scale adjustment when preserving the outcome limits is essential. It does not select a clinically credible delta: that question requires substantive justification, and the scale on which delta is defined must remain explicit.
Check Your Understanding
Use these questions to check the distinction between support constraints and transformed-scale adjustments. Brief solutions follow.
- A bounded outcome has limits 0 and 10, and a reference distribution with support \([2,8]\). What direct delta range is distributionally feasible?
- The reference mean in Question 1 is 5. What is the mean-feasible delta range, and why is it not the answer to Question 1?
- A reference distribution has support \([0,10]\). Which direct shifts are distributionally feasible?
- For a 0-to-1 outcome, a reference probability is 0.25. What probability results from adding \(\log(2)\) to its log odds?
- Why is clipping an out-of-range shifted value not equivalent to a direct shift that preserves the original shifted distribution?
Solutions. (1) The support rule gives \(0-2\leq\delta\leq10-8\), so \([-2,2]\). (2) The mean condition is \(0\leq5+\delta\leq10\), so \([-5,5]\). It is weaker because it checks only the mean, not every possible draw. (3) The support rule gives \(0-0\leq\delta\leq10-10\), so only \(\delta=0\). (4) Starting odds are \(0.25/0.75=1/3\); multiplying by \(\exp\{\log(2)\}=2\) gives odds \(2/3\), and the probability is \((2/3)/(1+2/3)=0.40\). (5) Clipping moves any out-of-range probability to an endpoint, creating boundary mass and altering the shifted distribution.