Worked Continuous-Variable Model Choices

Choosing a Conditional Model From the Variable’s Shape

In MICE, a continuous variable is imputed from a conditional model: the distribution of its missing values given the other variables in the imputation model. A familiar normal linear model is a useful starting point, but it can be inappropriate if it assigns probability to impossible values, poorly represents skewness, or treats a genuine point mass as though it were part of a smooth distribution. Model choice therefore depends on both the observed distribution and the variable’s possible values.

The aim is not to make every marginal histogram look normal. The relevant question is whether a model gives plausible conditional distributions across predictor patterns. The four choices considered here are a normal model on the original scale, a normal model after transformation, a normal model restricted to known bounds, and a two-part model for a point mass at zero plus positive values.

Definition: Let \(Y_j\) be a continuous variable with missing values and let \(X\) denote the predictors used to impute it. A conditional imputation model specifies \(P(Y_j\mid X)\) for the missing entries, estimated from the observed entries of \(Y_j\). A model is suitable only if its conditional distributions are plausible for the variable’s support and its observed relationship with \(X\).

These models are alternatives, not a ranking from simplest to best. The normal model is parsimonious when its assumptions are adequate. A transformation changes the scale on which the model is specified. Truncation conditions a distribution on a known range. A two-part model represents a discrete probability at zero separately from the continuous distribution among positive values.

Four Worked Model Choices

Worked Example: Normal Model on the Original Scale

Suppose a hypothetical clinic records a continuous biomarker concentration in milligrams per liter. Values may be negative in principle after a background correction, and the observed values span both sides of zero. In a model with one predictor \(x\), assume the fitted conditional model is \(Y_j\mid x\sim N(12+1.5x,3^2)\). For a participant with \(x=2\), the conditional mean is \(12+1.5(2)=15\) mg/L. If the predictive standard-normal draw is \(z=0.50\), a residual draw gives

Y_j^{\mathrm{imp}}=12+1.5(2)+3(0.50)=16.5\ \text{mg/L}.

This specification is plausible if, within relevant predictor patterns, the conditional mean is adequately linear, the residual spread is approximately constant, and residuals are reasonably represented by a normal distribution. The model also has support over the whole real line, which is not a problem here because negative values are possible under the measurement definition. If the biomarker were necessarily nonnegative, this same model could generate impossible imputations. A good overall fit would not fix that support mismatch.

Worked Example: Transforming a Positive, Skewed Variable

Consider a hypothetical measure of annual expenditure, which is positive and strongly right-skewed. Suppose the conditional model is normal for the natural logarithm: \(\log(Y_j)\mid x\sim N(1.50,0.40^2)\) for a particular predictor pattern. A standard-normal draw \(z=0.50\) gives

\log(Y_j^{\mathrm{imp}})=1.50+0.40(0.50)=1.70, \qquad Y_j^{\mathrm{imp}}=\exp(1.70)\approx 5.474.

The imputed value is about 5.474 expenditure units, is positive by construction, and reflects variation on the log scale. At this predictor pattern, \(\exp(1.50)\approx4.482\) is the conditional median on the original scale, not the conditional mean. Under the stated lognormal model the conditional mean is

E(Y_j\mid x)=\exp\left(1.50+\frac{0.40^2}{2}\right) =\exp(1.58)\approx4.855.

This distinction matters when interpreting summaries, but the mean correction is not a replacement for drawing an imputed value: each missing entry should reflect predictive variation, including residual and parameter uncertainty as required by the imputation procedure. The log model requires strictly positive observed values. Residual patterns on the log scale and the fit across predictor patterns should be examined; a log transformation does not guarantee a good conditional model.

Worked Example: Respecting Known Bounds With Truncation

Suppose a hypothetical test score is continuous and physically restricted to the interval from 0 to 100. For one predictor pattern, consider a normal distribution with mean 50 and standard deviation 20, restricted to that interval. Write \(\Phi\) for the standard normal cumulative distribution function. A draw from a standard uniform variable \(u\) can be converted to a truncated-normal draw by solving

u= \frac{ \Phi\left((y-50)/20\right)-\Phi(-2.5) }{ \Phi(2.5)-\Phi(-2.5) }.

For \(u=0.80\), \(\Phi(-2.5)\approx0.006210\) and \(\Phi(2.5)\approx0.993790\). Thus the target untruncated cumulative probability is

0.006210+0.80(0.993790-0.006210)\approx0.79627.

The standard normal quantile at approximately 0.79627 is \(z\approx0.828\). Therefore the imputed score is \(50+20(0.828)\approx66.56\) points, which lies within the permitted range. This model is appropriate only if the bounds are genuine limits on possible values and the truncated conditional normal distribution is a reasonable shape within them. Truncating a model does not repair an unsuitable mean structure or a badly misspecified residual distribution.

Worked Example: A Point Mass at Zero and Positive Values

Suppose a hypothetical health-cost variable has many exact zeros and positive amounts among people with costs. Assume the zero is a real category of the measurement process, not a rounded small positive value. For a particular predictor pattern, let the fitted probability of a positive value be \(q=0.40\). Draw \(u\) uniformly on \([0,1]\), using the convention that \(u\leq q\) selects the positive part. With \(u=0.22\), select a positive value. Suppose the positive-part model is \(\log(Y_j)\mid Y_j>0,x\sim N(1.50,0.40^2)\); using \(z=0.50\), the imputation is \(\exp(1.50+0.40(0.50))\approx5.474\) cost units. With \(u=0.72\), select the zero part and impute 0 cost units.

The two pieces answer separate questions: whether the value is positive, and how large it is when positive. The example assumes a suitable model for the probability of positivity and a suitable conditional model among positive observations. The positive-part distribution must be checked using only positive observations, while the frequency of zeros and its relationship to predictors should also be checked. If zero instead means “below a detection limit,” treating it as an ordinary point mass may misrepresent the measurement mechanism.

Why Truncation Changes the Draw

For a normal distribution with mean \(\mu_x\) and standard deviation \(\sigma_x\), restricted to known bounds \(L\) and \(U\), the conditional density is the original normal density divided by the probability of falling between the bounds. Its cumulative distribution on that interval is

F_T(y\mid x)= \frac{ \Phi\left((y-\mu_x)/\sigma_x\right)- \Phi\left((L-\mu_x)/\sigma_x\right) }{ \Phi\left((U-\mu_x)/\sigma_x\right)- \Phi\left((L-\mu_x)/\sigma_x\right) }, \qquad L\leq y\leq U.

This expression follows by subtracting the normal probability below \(L\) from the probability below \(y\), then dividing by the total probability between \(L\) and \(U\). Setting \(F_T(y\mid x)=u\) produces a draw from the conditional distribution within the permitted range. Merely replacing values below \(L\) by \(L\), or above \(U\) by \(U\), is not equivalent: it creates extra probability at the boundaries. The inverse-CDF draw instead samples from the normal distribution conditional on being inside the interval.

A Practical Selection and Checking Strategy

Before fitting, identify the actual support: which values are possible under the variable’s definition and measurement process? Then inspect the distribution and its relationship to predictors. These checks guide the conditional model, but are not formal proof that assumptions hold. The model should be assessed on observed values and across predictor patterns relevant to missing cases.

Establish support.
Decide whether negative values, exact zeros, or values beyond fixed bounds are possible. Distinguish genuine zeros and bounds from censoring, rounding, or data errors.

Inspect conditional shape.
Examine residual spread, skewness, and patterns against predictors. A marginally skewed variable may have more nearly normal conditional residuals, or the reverse.

Choose the simplest plausible model.
Use the original-scale normal model when its support and conditional residual behavior are credible; consider a transformation for positive skew, truncation for genuine bounds, or a two-part model for a genuine point mass plus continuous positive values.

Check generated imputations.
Compare observed and imputed distributions, ranges, boundary frequencies, and relationships with predictors. Check conditional patterns rather than relying only on pooled marginal plots.

A diagnostic warning is not automatically evidence that the chosen distribution family is wrong. Curvature in residuals may indicate an inadequate mean function; changing normal to lognormal does not alone address that. Changing residual spread may call for a model that accounts for heteroscedasticity rather than simply a transformation. Conversely, a model can have an acceptable average fit and still produce impossible values or an inaccurate tail.

The main alternatives differ in what they assume and guarantee. An original-scale normal model is straightforward but permits any real value. A transformed model changes the scale and guarantees positivity for a log transform, but imposes a particular relationship and distribution on that scale. A truncated model respects known bounds while retaining a chosen underlying shape, but requires defensible bounds and a suitable truncated distribution. A two-part model represents exact zeros separately, but requires adequate information for both the positive-probability and positive-value components.

No one of these choices establishes that missingness is ignorable, makes an incomplete predictor pattern informative, or corrects a faulty measurement definition. If diagnostics show serious lack of fit, possible responses include revising the predictor or mean structure, using a better-supported distributional specification, and conducting sensitivity analyses. Compare imputations with observed data without treating the observed cases as a perfect reference: their covariate patterns may differ from those of cases with missing values.

Common Mistakes and Limitations

Choosing only from the marginal histogram. The conditional distribution given predictors is what the imputation model specifies. Inspect residual and predictor relationships as well.
Clipping invalid normal draws. Replacing negative values by zero or capping values at a bound changes the distribution and creates artificial boundary mass. Use a model designed for the support.
Assuming transformation solves every problem. A log transform handles positivity, but does not guarantee linearity, constant variance, or adequate tails on the transformed scale.
Using two parts for every variable with zeros. First determine whether zeros are genuine or reflect censoring, rounding, or a detection limit; those mechanisms may require different modeling.

All four approaches are conditional model specifications, and their success depends on the observed data supporting the fitted relationships. Sparse positive cases can make a two-part model unstable; extreme truncation can leave little probability mass to draw from; and a transformation may be undefined for some observed values. Such limitations should be reported and assessed rather than hidden by the fact that an algorithm returned completed data.

Check Your Understanding

For each question, select or interpret a model using the support and distributional information provided. Solutions follow the questions.

A continuous laboratory measure can take negative values after background correction. Conditional residuals appear roughly symmetric with stable spread. Which model is a reasonable initial choice?
A positive outcome has a long right tail. A log-scale conditional model has mean 1.20 and standard deviation 0.30. For a predictive draw \(z=1\), calculate the imputed value.
A variable is known to lie between 10 and 30. Why is replacing normal draws below 10 by 10 not the same as drawing from a truncated normal model?
A cost measure has exact zeros and positive values. Its fitted positive probability for one case is 0.25. Under the convention that \(u\leq q\) selects a positive value, what part is selected by \(u=0.10\), and by \(u=0.60\)?
Name one diagnostic that could reveal a problem even when every imputed value is within the variable’s support.

Solutions. (1) An original-scale normal conditional model is a reasonable initial choice, subject to checking the conditional mean and residual assumptions. Its support is compatible with possible negative values. (2) The log draw is \(1.20+0.30(1)=1.50\), so the imputed value is \(\exp(1.50)\approx4.482\) outcome units. (3) Clipping adds probability mass at 10; a truncated-normal draw samples from the normal distribution conditional on lying between 10 and 30 and does not create that artificial mass. (4) \(u=0.10\leq0.25\) selects the positive part; \(u=0.60>0.25\) selects zero. (5) Examples include systematic residual curvature, an incorrect imputed tail or boundary frequency, or an implausible relationship between imputed values and predictors.