Why the Scale of a Tipping Threshold Matters
A tipping point describes where a specified sensitivity analysis changes its conclusion. Its numerical value is expressed on some scale: for example, a shift of \(-6\) points, a shift of \(-0.5\) standard deviations, or a log-scale shift of \(\log(0.8)\). Those numbers can refer to the same missing-outcome scenario, or they can encode different scenarios. The distinction matters because a threshold without its scale and parameterization is difficult to interpret.
The basic setting is a pattern-mixture sensitivity analysis in which missing outcomes are imputed under a reference assumption and then adjusted by a sensitivity parameter. In a simple continuous-outcome model, let \(Y_{\mathrm{mis}}(0)\) denote an imputed missing outcome under the reference assumption. An additive raw-scale adjustment is \(Y_{\mathrm{mis}}(\delta)=Y_{\mathrm{mis}}(0)+\delta\). A tipping threshold is a value of \(\delta\) at which a stated decision rule changes. For example, the rule might concern whether an estimated mean difference is positive or whether a confidence interval excludes zero. The threshold depends on that rule as well as on the data and model.
Earlier material on delta adjustment establishes that the direction and meaning of \(\delta\) depend on how it is defined. Here the question is narrower: how should a threshold be interpreted when outcomes are expressed in different units or transformed? A useful principle is to first identify the outcomes that each scenario assigns to missing cases. Only then translate the threshold.
Raw Units and Standardized Units
For a raw-scale additive adjustment, \(\delta\) has the same units as the outcome. If \(Y\) is a symptom score measured in points, a threshold of \(-6\) means that the assumed missing outcomes are shifted downward by 6 points relative to their reference imputations. This is often the most direct clinical or operational interpretation.
A standardized shift divides the raw shift by a chosen reference standard deviation. Let \(s_{\mathrm{ref}}\) be a fixed, positive standard deviation, such as the standard deviation of the outcome in the observed reference group. Define the standardized outcome as \(Z=(Y-c)/s_{\mathrm{ref}}\), where \(c\) is a fixed centering constant. A raw shift \(\delta_Y\) then corresponds to a standardized shift \(\delta_Z=\delta_Y/s_{\mathrm{ref}}\). The centering constant cancels because the same \(c\) is used before and after adjustment.
This conversion is exact for a fixed linear rescaling. It is not enough to say “standardized units,” however: the reference standard deviation must be named or otherwise defined. If the standard deviation is recalculated separately for each sensitivity scenario, the denominator can also change, so the converted values need not represent a simple translation of one fixed raw-scale threshold.
Worked Example: Converting a Point Threshold to Standard Deviations
Suppose a hypothetical symptom score is measured in points. A sensitivity analysis finds that its conclusion changes when the missing outcomes are shifted downward by 6 points relative to the reference imputations. The fixed reference standard deviation is 12 points. What is the threshold in standardized units, and how should it be interpreted?
The raw threshold is \(\delta_Y=-6\) points and the reference standard deviation is \(s_{\mathrm{ref}}=12\) points. Thus \(\delta_Z=\delta_Y/s_{\mathrm{ref}}\).
The units cancel, so the standardized threshold is \(-0.500\) standard deviations. For example, a missing outcome with a reference imputation of 70 points would be assigned \(70-6=64\) points under this scenario. If the score is standardized using the stated reference standard deviation, the same change is \(64/12-70/12=-0.500\) standardized units (with centering canceling in the difference).
The threshold is not a claim that every participant’s observed score fell by half a standard deviation. It describes the assumed adjustment to missing outcomes under this sensitivity model. Its interpretation also depends on the particular decision rule whose conclusion changed.
Transformed Outcomes: Translate the Scenario, Not Just the Number
A nonlinear transformation requires more care. Let \(g\) be a one-to-one transformation over the outcome values under consideration. If a reference-imputed value \(y_0\) is changed to \(y_1\), the corresponding change on the transformed scale is \(g(y_1)-g(y_0)\). In general, that difference depends on \(y_0\). Therefore, one raw additive shift need not correspond to a constant additive shift on the transformed scale for every missing case.
For a positive outcome, the logarithm is a common example. An additive shift \(\delta_{\log}\) to the log outcome gives \(\log(Y_{\mathrm{mis}})=\log(Y_{\mathrm{mis}}(0))+\delta_{\log}\). Exponentiating both sides shows that the original-scale outcome is multiplied by \(\exp(\delta_{\log})\). Thus the log-scale shift has a multiplicative interpretation. A negative log shift is a reduction; it does not generally mean a fixed reduction in the outcome’s original units.
Worked Example: Interpreting a Log-Scale Threshold
Consider a hypothetical positive biomarker measured in units per litre. A log-scale sensitivity analysis has a tipping threshold of \(\delta_{\log}=\log(0.80)\), approximately \(-0.2231\). Interpret this threshold for reference-imputed missing values of 40 and 80 units per litre.
Because \(\exp(\delta_{\log})=\exp(\log(0.80))=0.80\), each reference-imputed value is multiplied by 0.80. For a value of 40, the adjusted value is \(40(0.80)=32\) units per litre, a raw-scale decrease of \(32-40=-8\) units per litre. For a value of 80, it is \(80(0.80)=64\) units per litre, a decrease of \(64-80=-16\) units per litre.
The same log-scale threshold means a 20% reduction relative to each reference-imputed value, not a reduction of 20 units per litre and not a fixed raw-unit decrease. This multiplicative interpretation follows from the model, while the tipping conclusion still refers to the decision rule evaluated at this scenario.
Why a Nonlinear Scale Can Define a Different Analysis
A nonlinear transformation can do more than change the units used to display a threshold. Imposing a constant additive adjustment on the transformed scale generally creates outcome assignments different from those made by a constant raw-scale adjustment. This is not an error if the transformed-scale assumption is intended, but it should not be described as a mere conversion of the raw-scale threshold.
Worked Example: A Raw Shift Is Not a Constant Log Shift
Suppose two hypothetical missing cases have reference imputations of 20 and 80 units. Consider first a raw-scale adjustment of \(-10\) units. Then consider a log-scale adjustment of \(\log(0.75)\). Compare the scenarios and determine the log shifts that would reproduce the raw-scale assignments case by case.
Under the raw-scale adjustment, the assigned values are \(20-10=10\) and \(80-10=70\) units. To reproduce the first assignment on the log scale, the required shift is \(\log(10)-\log(20)=\log(10/20)=\log(0.50)\), approximately \(-0.6931\). To reproduce the second, it is \(\log(70)-\log(80)=\log(70/80)=\log(0.875)\), approximately \(-0.1335\). The two required log shifts differ, so one constant additive log shift cannot reproduce both raw-scale assignments.
Conversely, \(\log(0.75)\) multiplies both reference values by 0.75, assigning 15 and 60 units. Those are not the 10 and 70 units assigned by the raw shift. The two models make different assumptions about how missing outcomes depart from their reference imputations. Their tipping results may consequently differ; one threshold should not be reported as a unit conversion of the other.
Decision Rules, Comparisons, and Diagnostics
A threshold is interpretable only alongside the decision rule that defines it. For an effect-direction rule, it may mark where an estimated effect changes sign. For an inferential rule, it may mark where a confidence-interval conclusion changes. Those are different tipping definitions, even if the sensitivity parameter and outcome scale are identical. Changing an outcome scale can also change the estimand or the fitted model, so a conclusion about the mean on one scale cannot automatically be carried over to a mean on another.
For a fixed scenario and a decision rule defined on the same underlying quantity, translating units does not change the scenario itself. For example, converting a score from points to a linearly rescaled score changes the numerical threshold but not the adjusted outcomes represented by it. With a nonlinear transformation, it is safer to reconstruct the missing outcomes and state what is being summarized than to assume that a test or mean-based conclusion is invariant.
| Threshold expression | What it says | Key qualification |
|---|---|---|
| Raw additive shift, such as \(-6\) points | Missing outcomes are shifted by a fixed number of outcome units. | State the outcome units and reference imputation. |
| Standardized shift, such as \(-0.5\) SD | Missing outcomes are shifted by a fraction of a stated reference standard deviation. | Define the fixed standard deviation used for scaling. |
| Log additive shift, such as \(\log(0.8)\) | Missing outcomes are multiplied by 0.8 on the original positive scale. | This is a proportional change, not a fixed raw-unit decrement. |
| Other nonlinear-scale shift | Missing outcomes follow an additive shift after applying the specified transformation. | Translate through the transformation to establish the original-scale scenario and domain. |
Several diagnostics help reveal a scale mismatch before a threshold is reported:
- Check the parameter units. A raw-scale delta has outcome units; a standardized delta is dimensionless; a log-scale delta is also dimensionless but has a multiplicative interpretation after exponentiation.
- Reconstruct adjusted outcomes. Apply the proposed shift to representative reference imputations. If the implied changes are not the intended ones, the proposed “conversion” changed the scenario.
- Check the transformation’s domain. For example, a logarithm requires positive values. A proposed transformed-scale adjustment that produces invalid values cannot be used as stated.
- Hold the reference scale fixed when standardizing. Re-estimating the denominator across scenarios can make apparent changes in standardized thresholds reflect both a changing shift and a changing scale.
- Recheck the estimand and decision rule. A transformation can change what is being averaged or tested. Confirm that a reported change in conclusion is the one the analysis set out to assess.
A practical reporting comparison is between the raw-unit additive model, which is easy to interpret as a fixed absolute change, and a transformed-scale model, which can express changes such as a fixed proportion. Neither is universally preferable. The appropriate choice depends on the outcome, the intended assumption about departures from reference imputations, and the target estimand. A standardized summary can help compare the magnitude of shifts across contexts, but it does not replace the clinically meaningful raw-unit description.
Deriving a Valid Scale Translation
The conversion rule can be justified directly. For standardization, both the reference and adjusted outcome are transformed with the same fixed center and scale. Subtracting them eliminates the center, leaving the raw difference divided by the reference standard deviation. For a general one-to-one transformation, the same adjusted outcome \(y_1\) must be obtained after transforming back. The corresponding transformed shift is therefore determined by the transformed endpoints, not by a universal conversion factor.
Write the reference-imputed value \(y_0\) and the value \(y_1\) assigned under the tipping scenario.
For a one-to-one transformation \(g\), the shift that represents these endpoints is \(\delta_g=g(y_1)-g(y_0)\).
If \(\delta_g\) changes with \(y_0\), then a single additive shift on the transformed scale cannot represent the original scenario for all cases.
Report either the converted description of the same scenario or the distinct scale-specific assumption being evaluated.
For standardization, take \(g(y)=(y-c)/s_{\mathrm{ref}}\). Then \(\delta_g=(y_1-c)/s_{\mathrm{ref}}-(y_0-c)/s_{\mathrm{ref}}=(y_1-y_0)/s_{\mathrm{ref}}\), which is constant for a fixed raw difference. For the logarithm, \(g(y)=\log(y)\), so \(\delta_g=\log(y_1/y_0)\). This depends on the ratio rather than the raw difference, explaining why a fixed raw shift usually produces different log shifts at different reference values.
Common Mistake and Practical Limits
Wrong reasoning: “The tipping threshold is \(-0.5\) on both scales, so the missing outcomes have the same adjustment.” This ignores what one unit means on each scale. A half-unit standardized shift, a half-unit log shift, and a half-unit raw shift are not interchangeable. The correction is to specify the transformation and reference scale, then translate the implied missing outcomes. For a log shift, exponentiate the parameter to obtain the multiplier; for a standardized shift, multiply by the stated reference standard deviation.
Scale translation also has limits. Standardization depends on a defensible reference standard deviation, and a nonlinear transformation may be difficult to interpret or may be undefined for some outcomes. More fundamentally, expressing a threshold in a different scale does not make the sensitivity assumption empirically verified. It makes the assumption easier to communicate or compare; conclusions still depend on the chosen missing-data model, the observed data, and the decision rule.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A raw-scale tipping threshold is \(+4\) points and the fixed reference standard deviation is 8 points. What is the standardized threshold?
- A log-scale shift is \(\log(1.10)\). What multiplier does it imply, and how does it change a reference-imputed value of 50 units?
- A fixed raw-scale decrease of 5 units is applied to reference values of 10 and 100 units. Is the corresponding log-scale decrease the same for both? Calculate each.
- Why must a report identify the reference standard deviation used to express a tipping threshold in standardized units?
Solutions. (1) \(\delta_Z=4/8=+0.500\) standard deviations. (2) The multiplier is \(\exp(\log(1.10))=1.10\), so the adjusted value is \(50(1.10)=55\) units, an increase of 5 units. (3) No. For reference value 10, the adjusted value is 5, giving \(\log(5/10)=\log(0.5)\approx-0.6931\). For reference value 100, the adjusted value is 95, giving \(\log(95/100)=\log(0.95)\approx-0.05129\). (4) The numerical standardized shift is the raw shift divided by that standard deviation; using a different reference standard deviation changes the standardized threshold and its interpretation.