Choosing the Delta Scale for Continuous Outcomes

Why the Delta Scale Matters

A delta adjustment describes how imputed values for missing outcomes depart from a reference imputation under a missing-at-random (MAR) assumption. The numerical value of delta has no useful interpretation without its scale. A delta of 3 might mean an increase of 3 points, an increase of 3 standard deviations, or multiplication by \(e^3\), depending on the specification. These represent very different departures.

The choice should connect the sensitivity analysis to a meaningful statement about unobserved outcomes. For a continuous outcome, a fixed shift in its measurement units may be easiest to discuss. A shift expressed in standard deviations can make departures comparable across outcomes or settings. For a positive outcome, a shift on the log scale represents a fixed proportional change. None is universally preferable: the scale should match the plausible mechanism and the question being examined.

Definition: Let \(Y_{\mathrm{mis}}^{\mathrm{MAR}}\) denote a draw for a missing outcome from the MAR reference imputation model. A delta adjustment changes that draw according to a specified scale. On the raw scale, \(Y_{\mathrm{mis}}^\delta=Y_{\mathrm{mis}}^{\mathrm{MAR}}+\delta_{\mathrm{raw}}\). On a standardized scale, \(Y_{\mathrm{mis}}^\delta=Y_{\mathrm{mis}}^{\mathrm{MAR}}+\delta_{\mathrm{std}}s_{\mathrm{ref}}\), where \(s_{\mathrm{ref}}\) is a prespecified reference standard deviation. For a positive outcome on the natural-log scale, \(\log(Y_{\mathrm{mis}}^\delta)=\log(Y_{\mathrm{mis}}^{\mathrm{MAR}})+\delta_{\log}\), or equivalently \(Y_{\mathrm{mis}}^\delta=Y_{\mathrm{mis}}^{\mathrm{MAR}}e^{\delta_{\log}}\).

In each case, the adjustment applies to missing outcomes, not to observed outcomes. Delta is a sensitivity parameter: the observed data do not determine its value. The reference imputation model, its conditioning variables, and its MAR-based interpretation are taken as given here. The question is how to express and interpret departures from that reference.

Three Scales, Three Interpretations

Scale	Adjustment to a MAR draw	Meaning of a positive delta	Key requirement
Raw units	Add \(\delta_{\mathrm{raw}}\)	Missing outcomes are higher by a fixed number of outcome units	Outcome units and plausible shift must be clear
Standardized	Add \(\delta_{\mathrm{std}}s_{\mathrm{ref}}\)	Missing outcomes are higher by a specified number of reference SDs	Choose and report a fixed reference SD
Natural-log	Multiply by \(e^{\delta_{\log}}\)	Missing outcomes are higher by a fixed proportion	Outcome values must be positive for the log transformation

The raw and standardized adjustments are additive on the outcome scale. They preserve pairwise differences between adjusted and unadjusted draws, but their relative size depends on the outcome’s location and spread. The log adjustment is additive after taking logarithms and multiplicative after returning to the original scale. For example, \(\delta_{\log}=0.20\) corresponds to a multiplier of \(e^{0.20}\), approximately 1.221, or about a 22.1% increase in each missing draw.

The reference standard deviation \(s_{\mathrm{ref}}\) must be stated. A common choice is the observed-outcome standard deviation in the target analysis population. A model-based conditional residual standard deviation is another possible reference when the intended interpretation is in units of conditional residual variation. These choices answer different questions. Pick one before comparing delta values and hold it fixed across scenarios; recalculating it for each delta would change the meaning of the scale.

Worked Examples

Worked Example: A Raw-Unit Shift in Blood Pressure

Suppose a hypothetical study measures systolic blood pressure in mmHg. Three missing outcomes have MAR imputation draws of 118, 126, and 122 mmHg. Consider \(\delta_{\mathrm{raw}}=4\) mmHg, a scenario in which each missing outcome is assumed to be 4 mmHg higher than its MAR draw.

The adjustment is \(Y_{\mathrm{mis}}^\delta=Y_{\mathrm{mis}}^{\mathrm{MAR}}+4\). Thus the adjusted draws are \(118+4=122\), \(126+4=130\), and \(122+4=126\) mmHg. Their MAR mean is \((118+126+122)/3=122\) mmHg; their adjusted mean is \((122+130+126)/3=126\) mmHg.

If the observed-outcome standard deviation is 15 mmHg, the 4 mmHg shift is equivalent in size to \(4/15=0.267\) observed standard deviations. That conversion can help describe the shift, but the primary meaning remains “4 mmHg higher.” The scenario is defensible only if a constant 4 mmHg departure is a plausible sensitivity question; the arithmetic alone cannot establish that plausibility.

Worked Example: Expressing a Shift in Standard Deviations

Suppose two hypothetical continuous outcomes have different units and spreads. A symptom score has \(s_{\mathrm{ref}}=8\) points, while a laboratory measurement has \(s_{\mathrm{ref}}=40\) units. Consider a common standardized delta of \(\delta_{\mathrm{std}}=0.25\).

For the symptom score, the raw-scale adjustment is \(0.25(8)=2\) points. A MAR draw of 30 points becomes \(30+2=32\) points. For the laboratory measurement, the adjustment is \(0.25(40)=10\) units. A MAR draw of 100 units becomes \(100+10=110\) units.

Both scenarios posit a shift of one quarter of the chosen reference SD, but they do not posit equal raw-unit changes. This can be useful for comparing departures across outcomes with different scales. It is less useful when a quarter of an SD has no clear scientific interpretation or when the chosen standard deviation does not represent the variation relevant to the sensitivity question.

Worked Example: A Log-Scale Shift in a Positive Biomarker

Suppose a hypothetical biomarker is strictly positive and measured in units for which proportional differences are scientifically meaningful. Three MAR draws are 20, 30, and 40 units. Choose \(\delta_{\log}=0.20\), using natural logarithms.

The multiplier is \(e^{0.20}\approx1.2214\). Therefore, the adjusted draws are \(20(1.2214)\approx24.43\), \(30(1.2214)\approx36.64\), and \(40(1.2214)\approx48.86\) units. The MAR arithmetic mean is \((20+30+40)/3=30\) units; the adjusted arithmetic mean is approximately \((24.43+36.64+48.86)/3=36.64\) units, also equal to \(30(1.2214)\).

Each missing draw is assumed to be about 22.1% higher, not higher by a common number of units. The increase is approximately 4.43 units for the draw of 20 and 8.86 units for the draw of 40. Positivity is preserved because each positive draw is multiplied by a positive factor. This scale would not be suitable for outcomes that can be zero or negative without an explicitly justified alternative transformation.

Why the Conversions Hold

The scale conversions follow directly from the adjustment definitions. For a standardized shift, the raw-unit change is the standardized delta multiplied by the fixed reference standard deviation. For a log shift, exponentiating both sides of the adjusted log equation gives the corresponding multiplicative change.

\begin{aligned} Y_{\mathrm{mis}}^\delta-Y_{\mathrm{mis}}^{\mathrm{MAR}} &=\delta_{\mathrm{std}}s_{\mathrm{ref}},\\ \log(Y_{\mathrm{mis}}^\delta)-\log(Y_{\mathrm{mis}}^{\mathrm{MAR}}) &=\delta_{\log},\\ \frac{Y_{\mathrm{mis}}^\delta}{Y_{\mathrm{mis}}^{\mathrm{MAR}}} &=e^{\delta_{\log}}. \end{aligned}

The final equality requires positive values so that the logarithms and ratio are defined. It shows why a log-scale delta is a log ratio: a change of \(\delta_{\log}\) in log units corresponds to multiplying the original-scale value by \(e^{\delta_{\log}}\). For a positive delta, that factor exceeds 1; for a negative delta, it is below 1.

There is also a useful implication for an overall arithmetic mean when the adjustment is a common raw shift. Suppose a completed data set has \(n\) outcomes, of which \(n_{\mathrm{mis}}\) are missing before imputation, and all other values stay fixed. The complete-data mean increases by \(n_{\mathrm{mis}}\delta_{\mathrm{raw}}/n\), or \(q\delta_{\mathrm{raw}}\), where \(q=n_{\mathrm{mis}}/n\). This is an arithmetic identity, not a model for why values are missing. For a log adjustment, the missing values are multiplied, while observed values are unchanged; consequently the overall arithmetic mean generally does not simply multiply by \(e^{\delta_{\log}}\).

Choosing a Scale for the Scientific Question

A practical choice begins with the statement the sensitivity analysis should represent. A raw-unit delta is often clearest when clinicians or decision-makers can assess a plausible difference directly in units such as mmHg, days, or points. A standardized delta is useful when the aim is to describe a departure relative to outcome variability, especially when comparing outcomes measured on different scales. A log-scale delta is natural when proportional differences are meaningful and the outcome is positive.

These scales can describe the same shift only after a reference value is supplied. For example, with \(s_{\mathrm{ref}}=15\) mmHg, a standardized delta of 0.267 corresponds to about \(0.267(15)=4.01\) mmHg. Without that reference SD, “0.267 SD” is incomplete. Likewise, a log delta should be translated into a multiplier or percentage for interpretation; “0.20 log units” alone may be difficult to assess.

The scales also imply different patterns across the outcome range. A raw shift of 4 units changes every missing value by 4 units, so the proportional change is larger for smaller starting values. A log shift of 0.20 gives every positive missing value the same proportional increase, so the raw-unit increase grows with the starting value. A standardized shift is a raw shift once \(s_{\mathrm{ref}}\) is fixed; its interpretation depends on how that reference SD was chosen.

Diagnostics, Comparisons, and Limitations

A sensitivity analysis should be checked on the original outcome scale, even when delta is specified in standardized or log units. Inspect the adjusted missing draws alongside observed values and MAR draws, and compare ranges, quantiles, and summaries relevant to the analysis. Ask whether the implied values are credible in context. Also verify the unit conversion, the fixed reference SD, and—for log adjustments—the exponentiation and positivity of all values.

Implausible shifts: If a selected delta produces values outside a meaningful measurement range, reconsider the scenario or the scale. A mathematical transformation does not make an implausible assumption scientifically plausible.
Changing reference scales: If \(s_{\mathrm{ref}}\) is recomputed across scenarios, the same standardized delta no longer represents the same raw shift. Prespecify and report the reference.
Unclear log interpretation: Report \(e^{\delta_{\log}}\) and its percentage implication. Do not describe a log delta as a fixed number of outcome units.
Wrong outcome support: A log adjustment requires positive values. It cannot be applied directly to zero or negative values using the stated formula.
Overly simple departure: A single constant delta shifts all missing outcomes on the chosen scale in the same way. It does not represent departures that vary by observed predictors, outcome level, or missingness pattern.

The main alternative is not merely another scale: it is a different sensitivity assumption. Raw and standardized deltas posit fixed additive departures, while a log delta posits a fixed proportional departure. A single scale may be inadequate when the scientific concern is that departures vary across groups or outcome levels. Sensitivity values are not estimated from the observed data, and diagnostics cannot identify the true departure for outcomes that were never observed. Results should therefore be presented as conclusions across specified, interpretable scenarios rather than as proof that one delta is correct.

A useful justification for reporting more than one scale is that no scale is invariant to every scientific interpretation. If both a plausible fixed-unit shift and a plausible proportional shift are relevant, each can be examined and translated into its implications for the imputed values. The purpose is not to search for the scale that gives a preferred result, but to expose how conclusions respond to defensible assumptions.

Common Mistake

A common error is to call \(\delta_{\mathrm{std}}=0.25\) a 0.25-unit increase, or to interpret \(\delta_{\log}=0.20\) as a 0.20-unit increase. Neither is correct. The standardized delta must be multiplied by its reference SD to obtain outcome units; the log delta must be exponentiated to obtain the multiplier. Another error is to shift observed outcomes as well as imputed ones. That changes the data rather than expressing a departure assumption for missing outcomes.

Key takeaway: Choose the delta scale by matching it to the scientific meaning of a plausible departure: fixed outcome units, a fixed fraction of a prespecified reference SD, or a fixed proportional change on the positive outcome scale. State the reference clearly and inspect the resulting values in the original units.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

A missing outcome has MAR draw 50 units. What value results from a raw delta of \(-3\) units?
If \(s_{\mathrm{ref}}=12\) points, what raw-unit adjustment corresponds to \(\delta_{\mathrm{std}}=0.5\)?
For a positive outcome, what multiplier corresponds to \(\delta_{\log}=-0.10\), approximately?
Why must the reference SD be held fixed when comparing standardized-delta scenarios?
Which stated scale directly represents the same proportional change for MAR draws of 10 and 100 units?

Solutions. (1) \(50+(-3)=47\) units. (2) \(0.5(12)=6\) points. (3) \(e^{-0.10}\approx0.9048\), so adjusted values are about 9.52% lower than their MAR draws. (4) The raw shift equals the standardized delta times the reference SD; changing that SD changes the raw shift and therefore changes what the same standardized delta means. (5) The log scale: multiplying both values by \(e^{\delta_{\log}}\) applies the same proportional change, although the resulting raw-unit increases differ.