Probability-Scale Delta Adjustments for Binary Outcomes

Why Shift Binary Probabilities Directly?

A binary outcome has only two possible values, but its conditional event probability can take any value between zero and one. In a sensitivity analysis, an analyst may ask how conclusions would change if, among cases with missing outcomes, the event risk were a fixed amount higher or lower than the risk implied by the reference imputation model. A probability-scale delta expresses that difference directly in probability units.

The adjustment is made to the model-implied probability for each missing case, before drawing its binary imputed value. It is not an arithmetic change to an imputed zero or one: adding 0.10 to a binary outcome would produce a non-binary value, not a valid imputation.

Definition: Let $Y_i$ be a binary outcome, with $Y_i=1$ denoting the event, and let $p_i^{\text{MAR}}$ be the event probability for a case with a missing outcome under the MAR reference imputation model, conditional on the information used by that model. A probability-scale delta $\delta$ defines the adjusted probability $$ p_i^{(\delta)}=p_i^{\text{MAR}}+\delta. $$ If this value is valid, an adjusted imputation is drawn as $Y_i^{(\delta)}\sim\operatorname{Bernoulli}(p_i^{(\delta)})$. The delta is measured in probability units: for example, $\delta=0.10$ represents a ten-percentage-point increase in event probability.

Here, the MAR model supplies a reference distribution for missing outcomes; the delta specifies a departure from that reference. In the usual sensitivity-analysis interpretation, the shift is applied to missing outcomes, while observed outcomes remain as recorded. A positive delta raises the event probability and a negative delta lowers it. The value of delta is a sensitivity assumption, not a quantity that the observed data alone identify.

The adjustment also presumes that a common absolute change is a useful way to describe the departure. It need not imply that every case has the same relative change in risk. For instance, raising probabilities from 0.10 to 0.20 and from 0.70 to 0.80 adds ten percentage points in both cases, but doubles the first probability and raises the second by much less proportionally.

Validity and Expected Changes

A probability must lie between zero and one, inclusive. For a set of missing cases with MAR probabilities $p_i^{\text{MAR}}$, an unbounded additive shift is valid for every case only when it satisfies the intersection of the individual restrictions:

-p_i^{\text{MAR}}\leq\delta\leq 1-p_i^{\text{MAR}}\quad\text{for every missing case }i.

Equivalently, a single delta is valid for all those cases when it lies between the negative of the smallest reference probability and one minus the largest reference probability. The permissible range therefore depends on the probabilities in the target set. A delta that is valid for one analysis or subgroup need not be valid for another.

Key result: If $p_i^{(\delta)}=p_i^{\text{MAR}}+\delta$ is valid, then the adjusted conditional expected outcome increases by exactly $\delta$ for each missing case: $$ E\!\left[Y_i^{(\delta)}\mid\text{model information}\right] =p_i^{\text{MAR}}+\delta. $$ For $m$ missing outcomes all shifted by the same valid delta, the conditional expected number of events increases by $m\delta$ relative to the MAR reference. This is an expected-count difference, not a guarantee about the particular random imputed values.

This result follows from the mean of a Bernoulli random variable. If $Y_i^{(\delta)}\sim\operatorname{Bernoulli}(p_i^{(\delta)})$, then $E(Y_i^{(\delta)}\mid\text{model information})=p_i^{(\delta)}$. Subtracting the reference expectation $p_i^{\text{MAR}}$ gives $\delta$. Summing these differences across $m$ cases gives $m\delta$, by linearity of expectation; this does not require the imputed outcomes to be independent.

Worked Example: A Ten-Percentage-Point Increase

Suppose three hypothetical missing cases have MAR event probabilities 0.12, 0.35, and 0.78. The analyst considers $\delta=0.10$. Calculate the adjusted probabilities and the change in expected events.

Check validity. The lower bound for a common shift is $-0.12$, and the upper bound is $1-0.78=0.22$. Since $-0.12\leq0.10\leq0.22$, the shift is valid for all three cases.

Adjust each probability. The values are $0.12+0.10=0.22$, $0.35+0.10=0.45$, and $0.78+0.10=0.88$. Each probability rises by ten percentage points.

Compare expected counts. The MAR expected event count is $0.12+0.35+0.78=1.25$. The adjusted expected count is $0.22+0.45+0.88=1.55$. The difference is $1.55-1.25=0.30$ expected events, also given by $m\delta=3(0.10)=0.30$. Neither expected count has to be an integer, and a particular set of Bernoulli draws can differ from either expectation.

Worked Examples: Heterogeneous Risks and Boundaries

Worked Example: The Same Absolute Shift at Different Risks

For four hypothetical missing cases, suppose the MAR probabilities are 0.05, 0.20, 0.60, and 0.90. Apply $\delta=0.10$, then summarize the mean probability before and after adjustment.

The largest baseline probability is 0.90, so the upper bound on a common valid delta is $1-0.90=0.10$. The smallest is 0.05, giving a lower bound of $-0.05$. Thus $\delta=0.10$ is valid, with the last adjusted probability exactly one.

The adjusted probabilities are $0.15, 0.30, 0.70,$ and $1.00$. The mean reference probability is $(0.05+0.20+0.60+0.90)/4=1.75/4=0.4375$. The adjusted mean is $(0.15+0.30+0.70+1.00)/4=2.15/4=0.5375$, an increase of $0.1000$.

The absolute increase is the same for each case, but its relative size is not: for the first case, $0.10/0.05=2$, whereas for the third, $0.10/0.60\approx0.167$. Thus a probability-scale delta encodes a constant absolute risk difference, not a constant relative risk change.

Worked Example: When a Shift Exceeds the Probability Range

Suppose three hypothetical missing cases have MAR probabilities 0.04, 0.35, and 0.92. Consider a proposed $\delta=0.10$. What happens if the analyst caps any result above one at one?

The uncapped adjusted probabilities would be $0.14, 0.45,$ and $1.02$. The third is not a probability. If the analyst clips at one, the values become $0.14, 0.45,$ and $1.00$. The resulting increases are $0.10, 0.10,$ and $0.08$, so clipping does not preserve a ten-percentage-point increase for every case.

The reference mean is $(0.04+0.35+0.92)/3=1.31/3\approx0.4367$. The clipped mean is $(0.14+0.45+1.00)/3=1.59/3=0.5300$, a difference of about $0.0933$, not $0.10$. Alternatively, the largest common shift that is valid without clipping is $1-0.92=0.08$. At $\delta=0.08$, the adjusted probabilities are $0.12, 0.43,$ and $1.00$, and their mean increases by exactly $0.08$.

Contrast With Log-Odds Adjustments

A log-odds adjustment adds a delta to the logit of the reference probability and then transforms back to the probability scale. The earlier material on shifting binary imputations on the log-odds scale defines that approach; the key contrast here is the scale on which the change is constant. A probability-scale delta specifies a constant absolute difference $p_i^{(\delta)}-p_i^{\text{MAR}}$, when valid. A log-odds shift specifies a constant difference in log odds, so the resulting probability difference varies with the baseline probability.

For a numerical comparison, let the reference probability be 0.20. A probability-scale shift of $-0.10$ gives $0.20-0.10=0.10$. A log-odds shift of $-0.5$ gives:

\operatorname{logit}(0.20)=\log(0.25)\approx-1.3863,\qquad \operatorname{expit}(-1.3863-0.5) =\operatorname{expit}(-1.8863) \approx0.1317.

The log-odds adjustment therefore reduces the probability by about $0.20-0.1317=0.0683$, while the probability-scale adjustment reduces it by 0.10. The two delta values are on different scales and should not be treated as numerically equivalent. Log-odds shifts automatically map finite shifted logits to probabilities strictly between zero and one; additive probability shifts can reach or cross the boundaries.

Diagnostics, Pitfalls, and Limitations

For each proposed delta, inspect the reference and adjusted probabilities across the missing cases. Check the minimum and maximum after adjustment, identify cases at or near zero and one, and compare the observed change in average probability with the intended shift. If clipping is used, report both the nominal delta and the effective changes after clipping. A sensitivity result can be misleading if the analyst describes a constant shift when boundary adjustments have made the changes case-specific.

Applying delta to the completed binary value. Adding 0.10 to an imputed zero or one does not create a valid binary outcome. Adjust the event probability first, then draw a binary value.
Calling 0.10 a ten-percent relative increase. It is an increase of ten percentage points. For a baseline risk of 0.20, the adjusted risk is 0.30, a relative increase of 50%.
Ignoring invalid probabilities. If $p_i^{\text{MAR}}+\delta>1$ or $<0$, it cannot be used as a Bernoulli probability. Restrict delta to the common feasible range or explicitly define a boundary rule and acknowledge that it changes the shift.
Assuming the delta is learned from the observed data. The missing outcomes are unobserved, so this departure from the reference model is a sensitivity assumption. A single chosen delta cannot resolve uncertainty about the missingness mechanism.

The probability scale is often attractive when subject-matter experts can reason directly about absolute risk differences. Its limitations are equally important: a common shift can be infeasible when reference probabilities approach a boundary, and equal absolute changes can have very different relative meanings. Clipping maintains valid probabilities but alters the assumed shift; restricting delta maintains its interpretation but limits the range that can be explored. The log-odds scale avoids invalid probabilities for finite shifts, but expresses a different assumption. Neither scale is universally preferable; the choice should match the scientific interpretation, and conclusions should be examined across plausible values.

Check Your Understanding

Work each question, then compare with the solutions below.

A missing case has reference event probability 0.25. What probability results from $\delta=0.07$, and what does the delta mean?
Three missing cases have reference probabilities 0.12, 0.48, and 0.93. Find the range of common deltas that keeps all three adjusted probabilities valid.
Two reference probabilities are 0.03 and 0.40. Apply $\delta=-0.08$ using clipping at zero. What are the adjusted probabilities, and what is the change in their mean?
For a reference probability of 0.20, compare a probability-scale delta of $-0.10$ with a log-odds shift of $-0.5$. Why are the resulting probabilities different?

Solutions. (1) The adjusted probability is $0.25+0.07=0.32$. The delta is an increase of seven percentage points, not a seven-percent relative increase. (2) The lower bound is $-\min(0.12,0.48,0.93)=-0.12$, and the upper bound is $1-\max(0.12,0.48,0.93)=0.07$. Thus $-0.12\leq\delta\leq0.07$. (3) The uncapped values are $-0.05$ and $0.32$; clipping gives 0 and 0.32. The reference mean is $(0.03+0.40)/2=0.215$, and the adjusted mean is $(0+0.32)/2=0.160$, a change of $-0.055$. (4) The probability-scale result is $0.20-0.10=0.10$. The log-odds result is $\operatorname{expit}(\operatorname{logit}(0.20)-0.5)=\operatorname{expit}(-1.8863)\approx0.1317$. The probability-scale delta fixes an absolute probability difference, whereas the log-odds shift fixes a log-odds difference.