Location-Scale Delta Models for Residual Variability

Why Model Location and Residual Variability Separately?

A sensitivity analysis for missing continuous outcomes often begins by shifting their reference predictions. But two scenarios can have the same average missing outcome while differing in how dispersed individual outcomes are around that average. For example, a treatment could be assumed to lower the mean outcome among participants with missing measurements, while the uncertainty or heterogeneity of those outcomes could be assumed either unchanged or larger. A location-scale delta model represents these assumptions with separate parameters.

The location parameter describes a conditional mean shift. The scale parameter describes a change in residual variability relative to a reference distribution. Neither parameter is learned merely by comparing observed and missing outcomes: the missing outcomes are unavailable. The model makes these sensitivity assumptions explicit so their consequences can be examined.

Definition: Let $X=x$ denote the covariates used in a reference model, and let $R=0$ indicate that the outcome $Y$ is missing. Write $m(x)$ for the reference conditional mean, $s(x)>0$ for the reference conditional residual standard deviation, and $U$ for a reference standardized residual with mean zero and variance one. A location-scale delta model specifies $$ Y_{\mathrm{mis}}=m(x)+\delta+\lambda s(x)U,\qquad \lambda>0. $$ Here $\delta$ is the location shift, in the units of $Y$, and $\lambda$ is the residual-scale multiplier, which has no units. A positive $\delta$ shifts the missing-outcome distribution upward relative to the reference; $\lambda>1$ increases its residual standard deviation, and $0<\lambda<1$ decreases it.

The reference distribution is the chosen starting point for the sensitivity analysis. It supplies $m(x)$, $s(x)$, and the distribution of $U$; for example, it might be a model-based distribution used under a missing-at-random reference assumption. The location-scale adjustment does not itself justify that reference. It asks what follows if that reference distribution is shifted and rescaled in the specified way.

The standardized residual $U$ must be centered and scaled relative to the reference model: $E(U\mid X=x)=0$ and $\operatorname{Var}(U\mid X=x)=1$. The model can be applied within a defined treatment arm or missingness pattern, using the corresponding reference quantities and sensitivity parameters. In the examples below, parameters are constant within the stated group to keep the calculations transparent.

What the Two Parameters Assume

The parameters describe different departures from the reference distribution, and their interpretations depend on keeping those departures separate.

Location assumption: $\delta$ is the difference between the missing-outcome conditional mean and the reference conditional mean at the same $x$. It does not, by itself, change the residual standard deviation.
Scale assumption: $\lambda$ is the ratio of the missing-outcome residual standard deviation to the reference residual standard deviation at the same $x$. With centered residuals, it does not, by itself, change the conditional mean.
Shape assumption: rescaling $U$ preserves its standardized distributional shape. Thus this model does not independently change features such as skewness or tail shape. A location-scale model is more flexible than a shift alone, but it is not an unrestricted change to the full distribution.

The conditional moments follow directly from the definition, provided the reference residual has finite variance and the stated centering and scaling hold. The derivation also shows why the two parameters have distinct roles.

\begin{aligned} E(Y_{\mathrm{mis}}\mid X=x,R=0) &=m(x)+\delta+\lambda s(x)E(U\mid X=x)\\ &=m(x)+\delta,\\[4pt] \operatorname{Var}(Y_{\mathrm{mis}}\mid X=x,R=0) &=\lambda^2s(x)^2\operatorname{Var}(U\mid X=x)\\ &=\lambda^2s(x)^2. \end{aligned}

The conditional residual standard deviation is therefore $\lambda s(x)$, not $\lambda^2s(x)$; the square appears in the variance. If the reference residual distribution has cumulative distribution function $G$, then for $\lambda>0$,

F_{\mathrm{mis}}(y\mid x,R=0) =G\!\left(\frac{y-m(x)-\delta}{\lambda s(x)}\right).

This expression makes the shape assumption explicit: the adjusted distribution uses the same standardized reference law $G$, with its center shifted and its residual scale changed. It does not assert that every missing outcome is shifted by exactly $\delta$; rather, the conditional distribution is transformed.

Worked Calculations

Worked Example: A Lower Mean and Greater Residual Spread

Suppose a hypothetical study measures a continuous symptom score in points. For one covariate group, the reference model gives $m(x)=50$ points and $s(x)=10$ points. Consider $\delta=-4$ points and $\lambda=1.5$.

The assumed missing-outcome conditional mean is $$ 50+(-4)=46\text{ points}. $$ The assumed residual standard deviation is $$ 1.5(10)=15\text{ points}, $$ and the residual variance is $15^2=225$ points squared. Equivalently, the variance calculation is $\lambda^2s(x)^2=1.5^2(10^2)=2.25(100)=225$ points squared.

If $U=1$, the corresponding transformed value is $50-4+1.5(10)(1)=61$ points. If $U=-1$, it is $50-4+1.5(10)(-1)=31$ points. These values illustrate the shift and rescaling; they are not predictions that all missing participants have either value. Relative to the reference, this scenario has a mean four points lower and residual spread 50% larger.

Worked Example: Isolating a Scale Change

In a second hypothetical setting, assume the reference conditional mean is 12 hours and the reference residual standard deviation is 3 hours. Consider $\delta=2$ hours and $\lambda=0.5$.

The adjusted conditional mean is $12+2=14$ hours. The adjusted residual standard deviation is $0.5(3)=1.5$ hours, and its variance is $1.5^2=2.25$ hours squared. A shift-only model with the same $\delta=2$ but $\lambda=1$ would also have mean 14 hours, but its residual standard deviation would remain 3 hours and its variance would remain $3^2=9$ hours squared.

Thus the scale assumption, not the location shift, accounts for the smaller spread in the first scenario. The model does not say that the mean is reduced by half: $\lambda=0.5$ halves the residual standard deviation around the adjusted mean.

Worked Example: A Tail Probability Under a Normal Reference

Suppose a hypothetical reference distribution for a missing outcome is normal with mean 100 units and standard deviation 12 units. For this example only, assume the standardized reference residual $U$ is standard normal. Set $\delta=-5$ units and compare $\lambda=1.2$ with $\lambda=0.8$. The normal reference assumption makes the probability calculations possible; it is not required for the location-scale moment formulas.

For $\lambda=1.2$, the adjusted mean is $100-5=95$ units and the adjusted standard deviation is $1.2(12)=14.4$ units. The probability of an outcome below 80 units is $$ P(Y_{\mathrm{mis}}<80\mid x,R=0) =\Phi\!\left(\frac{80-95}{14.4}\right) =\Phi(-1.0417)\approx0.149. $$

For $\lambda=0.8$, the adjusted mean remains 95 units, but the standard deviation is $0.8(12)=9.6$ units. Therefore, $$ P(Y_{\mathrm{mis}}<80\mid x,R=0) =\Phi\!\left(\frac{80-95}{9.6}\right) =\Phi(-1.5625)\approx0.0592. $$

Under these hypothetical normal models, the larger residual scale assigns more probability to outcomes far below the mean, including values below 80 units. These are probabilities under the specified sensitivity distributions, not quantities identified from the observed outcomes alone.

Assumptions, Checks, and Alternatives

A location-scale adjustment is meaningful only relative to its reference model. Its main assumptions can be stated separately:

The reference conditional mean, residual standard deviation, and standardized residual distribution are appropriate for the covariates and group to which the adjustment is applied.
The missing-outcome conditional mean differs from the reference mean by the chosen $\delta$, and the residual standard deviation differs by the chosen positive multiplier $\lambda$.
The standardized residual distribution retains the reference shape after adjustment. In particular, this specification does not add a separate parameter for a change in skewness or tail behavior.

The first assumption concerns the reference model; the next two are sensitivity assumptions about missing outcomes. The observed outcomes can inform assessment of the reference model among observed cases, but they cannot reveal whether the missing cases actually have the chosen $\delta$, $\lambda$, or standardized shape. These parameters should be understood as scenario choices, not as estimates that the observed data have uniquely determined.

Checks are still useful. Confirm that residuals in the observed data do not show a serious mismatch with the reference model's conditional mean, spread, or shape. Then inspect the adjusted distributions over the covariate ranges used in the analysis: a constant multiplier may be implausible if reference residual variability changes across those ranges, and transformed values may be incompatible with the outcome's possible range. Such findings challenge the reference or the specified scenario; they do not identify the correct missing-outcome distribution.

The main alternatives have different assumptions. A location-only delta adjustment sets $\lambda=1$, allowing the mean to change while retaining reference residual variability. A scale-only adjustment sets $\delta=0$, changing residual spread while retaining the reference mean. The location-scale model allows both changes but still preserves standardized shape. A model that changes shape as well requires additional distributional assumptions; it is not obtained merely by choosing different values of $\delta$ and $\lambda$.

The method is most straightforward for outcomes where a continuous location-scale transformation is plausible. For bounded or highly discrete outcomes, the adjusted distribution can assign weight to impossible values or fail to respect the outcome scale. Also, if the missing-outcome spread changes in a way that depends on covariates, one constant $\lambda$ may not represent the intended scenario. These are reasons to revise the reference or sensitivity specification, not reasons to interpret $\lambda$ as a general measure of uncertainty about the analysis.

Common Mistake

Pitfall: “The location shift is $-4$, so the variance must also decrease by 4,” or “$\lambda=1.5$ makes the mean 50% larger.” Both statements confuse the parameters. The location shift adds $\delta$ to the conditional mean and leaves the residual variance unchanged when considered alone. The scale multiplier changes the residual standard deviation to $\lambda s(x)$ and the variance to $\lambda^2s(x)^2$; by itself, it leaves the mean unchanged because the reference residual is centered.

Another important limitation is interpretive: a model with a larger $\lambda$ describes greater residual variability under that scenario. It does not automatically represent greater sampling uncertainty in an estimate, nor does it establish that real missing outcomes are more variable. Those conclusions require separate evidence or assumptions.

Specify the reference.
State the covariates, reference conditional mean $m(x)$, residual standard deviation $s(x)$, and standardized residual distribution being used.

Choose the two departures.
Give $\delta$ in outcome units and $\lambda>0$ as a residual-scale ratio. Explain the intended missing-outcome scenario for each separately.

Check the implications.
Calculate the adjusted mean and residual spread, then examine whether the resulting distribution is plausible for the outcome and covariate group.

Check Your Understanding

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

A reference conditional mean is 30 mg/L and the reference residual standard deviation is 4 mg/L. With $\delta=3$ mg/L and $\lambda=2$, what are the adjusted conditional mean, residual standard deviation, and residual variance?
What does $\lambda=0.7$ mean, and what does it not mean?
Suppose the reference residual has mean zero and variance one. Explain why changing $\delta$ alone does not change the residual variance, while changing $\lambda$ alone does not change the conditional mean.
Can observed outcomes determine the correct $\delta$ and $\lambda$ for cases whose outcomes are missing? What can observed outcomes help assess?

Solutions. (1) The mean is $30+3=33$ mg/L. The residual standard deviation is $2(4)=8$ mg/L, and the variance is $8^2=64$ (mg/L)$^2$. (2) It multiplies the reference residual standard deviation by 0.7, reducing it by 30%; it does not multiply the mean by 0.7. (3) The model is $m(x)+\delta+\lambda s(x)U$. Since $E(U)=0$, the mean is $m(x)+\delta$; since $\operatorname{Var}(U)=1$, the variance is $\lambda^2s(x)^2$. Thus each parameter has the stated separate role. (4) No. The missing outcomes do not reveal their own distribution. Observed outcomes can help assess the reference model's fit among observed cases, but cannot identify the missing-outcome sensitivity parameters.