What Does a Delta Shift?
A delta adjustment describes a departure from a reference prediction for outcomes that are missing. The key question here is whether the adjustment specifies only a new conditional mean or specifies how the entire conditional distribution changes. These are not equivalent claims. A mean is one feature of a distribution; it does not determine its spread, tail probabilities, or shape.
Let \(Y\) be a continuous outcome, \(X=x\) the covariates for a case, and \(R=0\) indicate that the outcome is missing. Write \(F_0(y\mid x)\) for the reference conditional predictive distribution used for missing outcomes, and \(m_0(x)\) for its mean. An additive delta \(\delta(x)\) is defined here as the change in the missing-outcome mean relative to that reference. Thus a negative delta says that the adjusted mean is lower than the reference mean.
The reference distribution and the delta are sensitivity-analysis inputs: the observed outcomes do not, by themselves, reveal the distribution of outcomes that are missing. A mean-only statement is deliberately limited, but that limitation has a practical consequence. To generate predictive draws or calculate a probability such as \(P(Y\geq c\mid X=x,R=0)\), one must specify more than the mean.
A Mean Constraint Is Not a Distribution
Many different distributions can share the same mean. For example, two normal distributions can have the same mean but different variances. A normal distribution and a two-component mixture can also have the same mean while differing in shape. Consequently, the mean-only condition permits many predictive distributions unless additional assumptions are imposed.
One common choice is to assume that the adjusted distribution is normal with the shifted mean and the reference variance. That is a complete predictive model, but it combines the mean constraint with an extra assumption about variance and shape. Another choice is a full location translation of the reference distribution. That makes a stronger distributional claim: every location feature moves by the same amount.
A full location shift does change the mean by \(\delta(x)\), so it satisfies the mean-only constraint. The reverse implication does not hold: satisfying the constraint does not require a full location shift. The two descriptions coincide only when the distributional construction makes them coincide, such as when the chosen model is \(Y_0+\delta(x)\).
Why a Full Location Shift Moves More Than the Mean
Suppose \(Q_0(p\mid x)\) is the \(p\)th quantile of the reference distribution. Under a full location shift, the event \(Y_0+\delta(x)\leq y\) is the same event as \(Y_0\leq y-\delta(x)\). Therefore the shifted distribution function is \(F_0(y-\delta(x)\mid x)\). Its \(p\)th quantile is \(Q_0(p\mid x)+\delta(x)\). This establishes that the entire set of quantiles moves, rather than just the mean.
The variance equality follows because adding a constant does not change deviations from the mean: $$ (Y_0+\delta)-(E(Y_0\mid x)+\delta)=Y_0-E(Y_0\mid x). $$ Thus a full location shift moves the distribution without stretching or compressing it. A mean-only adjustment does not imply this invariance.
Worked Example: Same Shifted Mean, Different Lower-Tail Probabilities
Suppose, hypothetically, the reference distribution for a missing continuous outcome at a fixed covariate value is \(N(70,100)\), where the second argument is the variance. Set \(\delta=-8\), so the target mean is \(70-8=62\). Compare two possible distributions satisfying that mean constraint: \(N(62,100)\) and \(N(62,400)\).
For the first distribution, the standard deviation is \(\sqrt{100}=10\). The probability of an outcome below 50 is $$ P(Y<50)=\Phi\left(\frac{50-62}{10}\right)=\Phi(-1.2)\approx 0.1151. $$
For the second distribution, the standard deviation is \(\sqrt{400}=20\), and $$ P(Y<50)=\Phi\left(\frac{50-62}{20}\right)=\Phi(-0.6)\approx 0.2743. $$
Both distributions have the same adjusted mean, 62 outcome units, but their probabilities of falling below 50 differ by about \(0.1592\), or 15.92 percentage points. The second distribution is not a full location shift of the stated reference \(N(70,100)\), because it changes the variance. The mean-only condition alone cannot rule it out; a fixed-variance assumption would.
Worked Predictive Comparisons
A predictive interval or threshold probability depends on the whole distribution, not only its mean. The next examples make clear what is gained by adopting a particular distributional assumption, and what remains undetermined if the assumption is not made.
Worked Example: A Normal Mean Model and a Full Location Shift
Continue with a hypothetical reference distribution \(Y_0\sim N(70,100)\) and delta \(\delta=-8\). Consider the additional assumption that the adjusted distribution is normal with variance fixed at 100. Under this assumption, the adjusted distribution is \(Y\sim N(62,100)\). This is also exactly the full location shift \(Y=Y_0-8\); it is not a consequence of the mean constraint alone.
The central 90% predictive interval for a normal distribution uses the standard-normal multiplier 1.645. Its endpoints are $$ 62\pm 1.645\sqrt{100} =62\pm16.45 =[45.55,\ 78.45]. $$
Under the model, this interval contains 90% of the predictive distribution for a missing outcome at the specified covariate value. It is not an interval expressing uncertainty about the unknown mean parameter. For comparison, the reference distribution’s central 90% interval is \(70\pm16.45=[53.55,\ 86.45]\); translating the distribution by \(-8\) translates both endpoints by \(-8\).
The model also gives \(P(Y<50)=\Phi(-1.2)\approx0.1151\), as in the first distribution of the preceding example. This probability is available because the normal, fixed-variance assumption supplies a full distribution, not because the mean shift by itself determines it.
Worked Example: Same Mean, Different Probability of Crossing a Threshold
Consider two hypothetical predictive models for a missing outcome, measured in clinical units. Model A is \(N(62,100)\). Model B is a two-component mixture: with probability 0.5 the outcome is \(N(52,36)\), and with probability 0.5 it is \(N(72,36)\). The mixture mean is $$ 0.5(52)+0.5(72)=62. $$ Thus both models have the same mean.
The mixture variance is the within-component variance plus the variance of the component means. The component means are 10 units below and above 62, so $$ \operatorname{Var}(Y)=0.5(36)+0.5(36)+0.5(52-62)^2+0.5(72-62)^2 =36+100=136. $$
Suppose the clinical question concerns the probability that the outcome is at least 70. Under Model A, $$ P(Y\geq70)=1-\Phi\left(\frac{70-62}{10}\right) =1-\Phi(0.8)\approx0.2119. $$ Under Model B, the component-specific probabilities give $$ P(Y\geq70)=0.5\left[1-\Phi\left(\frac{70-52}{6}\right)\right] +0.5\left[1-\Phi\left(\frac{70-72}{6}\right)\right] $$ $$ =0.5[1-\Phi(3)]+0.5[1-\Phi(-0.3333)] \approx0.5(0.00135)+0.5(0.6306)=0.3160. $$
The mixture assigns about 31.60% probability to crossing the threshold, compared with about 21.19% under Model A, despite the identical mean. The calculation depends on the mixture specification. The mean-only delta does not select between these models or justify either one on its own.
Assumptions, Diagnostics, and Alternatives
Every predictive comparison above is conditional on the stated covariates and on the specified missing-outcome model. Interpreting a mean-only delta requires a reference mean \(m_0(x)\) that is suitable for the target cases, a clearly defined scale and sign for \(\delta(x)\), and a decision about what distributional features—if any—are held fixed. A full location shift additionally assumes that the reference distribution’s shape and spread are appropriate for the missing outcomes after translation.
The observed outcomes cannot directly verify the distribution of the unobserved outcomes. Diagnostics can still reveal problems with the reference model or with the predictions it generates. Compare predicted and observed outcome distributions in covariate regions with data; inspect residual spread and tail behavior; and check whether adjusted predictions violate the outcome’s possible range or produce implausible threshold probabilities. If an assumed normal model predicts impossible values for a bounded outcome, for example, the chosen predictive family is inadequate even if its mean is plausible. These checks can expose a poor construction, but they cannot identify the correct missing-outcome distribution from observed outcomes alone.
The main choices have different implications:
| Approach | What is specified | What it implies |
|---|---|---|
| Mean-only delta | The adjusted conditional mean | Other distributional features remain unspecified without extra assumptions |
| Full location shift | The entire reference distribution translated by \(\delta(x)\) | All quantiles shift equally; variance and shape are preserved |
| Specified distributional model | A full predictive family, such as a normal model with stated variance | Predictive probabilities and intervals follow from that family and its assumptions |
These approaches answer different questions. A mean-only delta is appropriate when the sensitivity claim concerns the average missing outcome and the analyst avoids claiming more than that. A full location shift is appropriate when translating the reference distribution is a defensible sensitivity assumption. A different full distribution may be needed when clinical knowledge suggests different tail behavior or variability. Such a choice should be stated and examined as a sensitivity assumption, not presented as something learned from the mean delta.
A common mistake is to apply \(\delta\) to the mean and then treat every predictive quantity as though the full distribution had necessarily shifted by \(\delta\). That reasoning is valid only if the chosen model defines a full location shift. The correction is to state the distributional assumption explicitly, then calculate predictions from that model. If only the mean is specified, report mean-based conclusions and acknowledge that tail probabilities and predictive intervals are not determined.
Check Your Understanding
Answer each question using the distinction between a mean constraint and a full distributional assumption. Brief solutions follow the questions.
- If the reference mean is 40 units and \(\delta=3\), what adjusted mean is specified? Does this determine the adjusted variance?
- A full location shift adds 5 units to every outcome. What happens to each quantile and to the variance?
- Two candidate predictive distributions both have mean 10 units. Can their probabilities of exceeding 15 units differ? Explain.
- A model specifies \(Y\sim N(25,49)\) after adjustment. What are its standard deviation and central 90% predictive interval?
- Why can observed-data diagnostics fail to determine which missing-outcome distribution is correct?
Solutions. (1) The adjusted mean is \(40+3=43\) units; the mean constraint alone does not determine variance. (2) Every quantile increases by 5 units, while variance is unchanged. (3) Yes. A mean does not determine tail probabilities; different spreads or shapes can yield different probabilities above 15. (4) The standard deviation is \(\sqrt{49}=7\) units, so the interval is \(25\pm1.645(7)=25\pm11.515=[13.49,\ 36.52]\) units, conditional on the stated normal model. (5) The outcomes with \(R=0\) are unobserved, so diagnostics can assess the reference model against observed data but cannot directly reveal the distribution of those missing outcomes.