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Sensitivity of Conclusions to Imputation Assumptions · Tutorial 418 of 497

Multiplicative Delta Adjustments for Positive Outcomes

Apply a positive multiplier to MAR imputations, interpret its ratio-scale meaning, and assess how that choice changes estimates relative to an additive adjustment.

Advanced 12 min read

What You'll Learn

  • Define a multiplicative delta adjustment for a positive missing outcome.
  • Interpret delta as the ratio of an adjusted imputation to its MAR reference.
  • Derive how multiplication on the outcome scale becomes addition on the log scale.
  • Calculate changes in means and group contrasts under a multiplicative adjustment.
  • Compare multiplicative adjustments with additive shifts and identify key diagnostics.

Why Multiply a Positive Outcome?

For some outcomes, a departure from the missing at random (MAR) reference is more naturally described as a proportional change than as a fixed number of units. For example, a sensitivity assumption might be that the unobserved values of a positive laboratory measurement are, on average, 20% lower than their MAR reference values. Multiplication expresses that assumption directly: it lowers a reference value of 100 by more units than a reference value of 10, while applying the same proportion to both.

A multiplicative delta adjustment changes imputations for missing values, not the observed values. It is a sensitivity assumption: the data generally do not identify the departure from MAR, so an analyst examines how conclusions change across specified multipliers. The MAR imputation model supplies a reference; the multiplier describes a departure from that reference.

Definition: Let \(Y_{\mathrm{MAR}}>0\) be a draw for a missing outcome from its MAR imputation distribution. A multiplicative delta adjustment with multiplier \(\delta>0\) replaces that draw by \(Y_{\delta}=\delta Y_{\mathrm{MAR}}\). The multiplier is the ratio \(Y_{\delta}/Y_{\mathrm{MAR}}\). Thus, \(\delta=1\) gives the MAR reference; \(0<\delta<1\) lowers every adjusted draw by the same proportion; and \(\delta>1\) raises every adjusted draw by the same proportion.

The ratio interpretation is individual to each reference draw: \(Y_{\delta}/Y_{\mathrm{MAR}}=\delta\). For instance, \(\delta=0.8\) means each adjusted imputation is 80% of its MAR reference, or 20% lower. It does not mean “0.8 outcome units lower.” The adjusted value retains the outcome's units, while \(\delta\) is dimensionless.

The adjustment is applied to missing values only. Observed outcomes remain as recorded, and a value imputed under one sensitivity scenario should not be treated as observed data. In multiple imputation, the adjustment is made to the relevant missing-value draws in each completed data set, and the analysis is repeated for each chosen \(\delta\).

Assumptions and a Log-Scale Justification

Before applying the adjustment, make its assumptions explicit. The outcome must be positive on the scale being multiplied. The MAR imputation model must provide a suitable reference distribution conditional on the variables used in that model. The sensitivity assumption is that, for the missing cases being adjusted, the departure from that reference is proportional and represented by the specified multiplier. In the simplest version, the same \(\delta\) applies to all those cases.

This assumption does not follow from the observed data. It is a chosen scenario, not an estimated correction. It also does not claim that the observed cases should be rescaled: their outcomes are observed and are left unchanged. If zero or negative outcomes are possible, multiplication may not express the intended departure, and it may not preserve the required support.

The connection with a log transformation explains why multiplication is convenient for positive outcomes. Taking logarithms of the adjustment gives an additive shift on the log scale:

$$ \log(Y_{\delta}) =\log(\delta Y_{\mathrm{MAR}}) =\log(Y_{\mathrm{MAR}})+\log(\delta). $$

So a fixed multiplier corresponds to the same additive shift, \(\log(\delta)\), for every missing value on the log scale. This is a mathematical representation of the adjustment, not a requirement that the original imputation model was fitted on the log scale. If \(Y_{\mathrm{MAR}}\) is positive, multiplication also preserves positivity. For any positive quantile \(q\), the adjusted quantile is \(\delta q\); in particular, the conditional mean is multiplied by \(\delta\) when it exists.

The log identity provides a useful check on interpretation. A multiplier of 0.8 gives a log-scale shift of \(\log(0.8)\), which is negative. A multiplier of 1.25 gives a log-scale shift of \(\log(1.25)\), which is positive. These shifts are not equal in magnitude: multiplying by 0.8 and then by 1.25 returns to the original value, but the percentages “20% lower” and “25% higher” are not symmetric descriptions around the original value.

Worked Example: Adjusting MAR Draws by a Ratio

Suppose a hypothetical positive outcome is measured in milligrams. In one completed data set, the observed values are 12, 18, 20, 25, 30, and 35 mg, while two missing outcomes have MAR draws of 10 and 20 mg. Use \(\delta=0.8\) for the missing outcomes and compare the completed-data mean with the MAR-reference mean.

The MAR-reference total is \(12+18+20+25+30+35+10+20=170\) mg, so the reference mean is \(170/8=21.25\) mg. The adjusted missing draws are \(0.8(10)=8\) mg and \(0.8(20)=16\) mg. The observed values are unchanged, so the adjusted total is \(12+18+20+25+30+35+8+16=164\) mg.

The adjusted completed-data mean is \(164/8=20.50\) mg, which is \(0.75\) mg lower than the MAR-reference mean. The ratio statement applies to the missing draws: each is 80% of its own MAR reference. It does not imply that the completed-data mean is 80% of the MAR-reference mean, because the observed outcomes were not adjusted.

Consequences for Means and Contrasts

The adjustment's effect on a completed-data mean depends on how many values are missing and on the sizes of their MAR draws. Suppose a data set has \(n\) individuals, \(r\) observed outcomes with total \(S_{\mathrm{obs}}\), and \(n-r\) missing outcomes with MAR draws \(y_1,\ldots,y_{n-r}\). The MAR-reference mean and the adjusted mean are:

$$ \bar{Y}_{\mathrm{MAR}} =\frac{S_{\mathrm{obs}}+\sum_{j=1}^{n-r}y_j}{n}, \qquad \bar{Y}_{\delta} =\frac{S_{\mathrm{obs}}+\delta\sum_{j=1}^{n-r}y_j}{n}. $$

Subtracting the first mean from the second shows exactly what determines the change:

$$ \bar{Y}_{\delta}-\bar{Y}_{\mathrm{MAR}} =\frac{(\delta-1)\sum_{j=1}^{n-r}y_j}{n}. $$

Because the MAR draws are positive, their sum is positive. Therefore, \(\delta<1\) lowers this completed-data mean, \(\delta>1\) raises it, and \(\delta=1\) leaves it unchanged. The change grows with the total of the missing draws and shrinks as the total sample size grows. This calculation concerns one completed data set; with multiple imputations, the analysis and combination are repeated across the adjusted completed data sets.

Worked Example: A Treatment-Group Difference

Consider a hypothetical study with positive outcomes measured in units. In one completed data set, Group A has four observed outcomes totaling 80 and two missing outcomes with MAR draws 20 and 20. Group B has four observed outcomes totaling 72 and two missing outcomes with MAR draws 18 and 18. Apply \(\delta=0.75\) to the missing outcomes in both groups. Calculate the mean difference, defined as Group A minus Group B, under the MAR reference and after adjustment.

Under the MAR reference, Group A's mean is \((80+20+20)/6=120/6=20\) units. Group B's mean is \((72+18+18)/6=108/6=18\) units. The MAR-reference difference is \(20-18=2\) units.

After adjustment, Group A's two missing draws are \(0.75(20)=15\) each, so its mean is \((80+15+15)/6=110/6=18.33\) units. Group B's adjusted draws are \(0.75(18)=13.5\) each, so its mean is \((72+13.5+13.5)/6=99/6=16.50\) units. The adjusted difference is \(18.33-16.50=1.83\) units (more precisely, \(11/6\) units).

Both group means decrease, but the difference does not decrease by the same absolute amount as either mean. The example illustrates why the effect on a contrast depends on the missing-value patterns and reference values in both groups. These are completed-data calculations, not a claim about a population treatment effect or statistical significance.

For an inferential analysis, each chosen multiplier defines a separate sensitivity scenario. For each scenario, generate the adjusted imputations, calculate the analysis estimate and its complete-data variance in each imputed data set, and combine the results using the usual multiple-imputation rules. The resulting interval and estimate describe uncertainty under that particular scenario and its imputation model. Conventional within-scenario uncertainty does not account for uncertainty about which multiplier is appropriate; that uncertainty is explored by comparing scenarios.

Multiplicative and Additive Adjustments

An additive delta adjustment instead adds a fixed quantity to each missing MAR draw: \(Y_{\mathrm{add}}=Y_{\mathrm{MAR}}+a\), where \(a\) is measured in the outcome's units. Multiplication and addition encode different assumptions. A fixed additive shift changes every value by the same number of units; a fixed multiplier changes every value by the same proportion. Consequently, the two adjustments generally produce different ordering of changes across small and large reference values.

FeatureMultiplicative adjustmentAdditive adjustment
Definition\(Y_{\delta}=\delta Y_{\mathrm{MAR}}\)\(Y_{\mathrm{add}}=Y_{\mathrm{MAR}}+a\)
Parameter meaningRatio; dimensionlessDifference; has outcome units
Change for a reference value \(y\)\((\delta-1)y\)\(a\)
PositivityPreserved if \(y>0\) and \(\delta>0\)Not necessarily preserved

Worked Example: Same Reference Values, Different Shifts

Suppose two hypothetical missing outcomes have positive MAR reference draws of 5 and 20 mg. Compare a multiplier of \(\delta=0.7\) with an additive shift of \(a=-3\) mg.

With multiplication, the adjusted values are \(0.7(5)=3.5\) mg and \(0.7(20)=14\) mg. Each is 70% of its reference, a 30% proportional decrease. The absolute decreases are \(5-3.5=1.5\) mg and \(20-14=6\) mg.

With the additive adjustment, the adjusted values are \(5-3=2\) mg and \(20-3=17\) mg. Each decreases by 3 mg, but the relative decreases are \(3/5=60\%\) and \(3/20=15\%\). The additive shift leaves both values positive in this example, but a sufficiently large negative shift could produce a nonpositive value. The two scenarios therefore represent distinct assumptions even when both lower the imputations.

Diagnostics, Failure Modes, and Limitations

A multiplicative adjustment is simple to calculate, but its simplicity makes checking the implications important. Useful checks include:

  • Verify the scale and support. Confirm that the outcome values being multiplied are positive and that the multiplier is positive. If zero or negative values are meaningful possibilities, this adjustment may not be suitable as written.
  • Inspect adjusted values and summaries. Compare the MAR-reference and adjusted distributions of missing values, including their means, quantiles, and upper tails. Confirm that the intended ratio is applied and that the resulting values remain scientifically plausible.
  • Check completed-data analyses across scenarios. Track the estimate and its uncertainty for each multiplier. A rapid change, a change in the substantive conclusion, or an unexpected direction can reveal that conclusions depend strongly on the unobserved outcomes.
  • Check the model reference. A multiplier cannot repair a poor MAR imputation model. If the reference draws are implausible or the model omits important predictors, the adjusted results inherit those problems.

A common mistake is to say that \(\delta=0.8\) means “subtract 0.8 units from each imputation.” That is an additive interpretation and is wrong: the adjustment is multiplication, so a reference draw of 10 becomes 8, while a draw of 100 becomes 80. Another mistake is to multiply observed values as well as missing imputations. That changes the observed data and no longer represents the stated sensitivity departure for missing outcomes.

The constant-ratio assumption is also a limitation. It imposes the same proportional departure for all adjusted missing cases, even though the actual departure could differ across people or parts of the outcome distribution. A chosen multiplier is not validated simply because the adjusted values are positive or plausible. Nor does one multiplier capture every possible departure from MAR. Results should therefore be reported as a sensitivity analysis across clearly described scenarios, not as a data-estimated correction.

Finally, do not confuse the ratio for each missing draw with a ratio between group means, or between an adjusted estimate and the MAR estimate. Observed values are unchanged, and the proportion of missing outcomes may differ across groups. Those features make the effect on a full-sample mean or contrast depend on the data, as the worked calculations demonstrate.

Key takeaway: A multiplicative delta adjustment replaces each positive MAR draw for a missing outcome by \(\delta\) times that draw. Interpret \(\delta\) as a ratio, apply it only to the missing values in the stated scenario, and compare completed-data analyses across multipliers to assess sensitivity.

Check Your Understanding

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

  1. A missing positive outcome has a MAR draw of 40 units. What adjusted value corresponds to \(\delta=1.15\), and how should the multiplier be interpreted?
  2. A completed data set has \(n=10\), observed outcomes totaling 60 units, and missing MAR draws totaling 30 units. Find the change in its mean under \(\delta=0.8\).
  3. For a reference draw of 12 mg, compare the adjusted value under \(\delta=0.5\) with the adjusted value under an additive shift of \(-3\) mg.
  4. Why can an adjusted Group A mean decrease while the Group A minus Group B mean difference increases?

Solutions. (1) The adjusted value is \(1.15(40)=46\) units; it is 115% of the MAR draw, or 15% higher. (2) The mean change is \((\delta-1)(30)/10=(-0.2)(30)/10=-0.6\) units, so the adjusted mean is 0.6 units lower. (3) Multiplication gives \(0.5(12)=6\) mg; addition gives \(12-3=9\) mg. The adjustments differ because one halves the reference value and the other subtracts a fixed 3 mg. (4) The difference depends on the changes in both group means. If Group B decreases more than Group A, then Group A minus Group B can increase even though Group A's mean decreases.

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