Why Bounds Must Enter the Imputation Model
A normal linear model is convenient for imputing a continuous variable, but its predictive distribution assigns positive probability to every real number. That can produce impossible imputations: a proportion below 0, a concentration below a physical lower bound, or a value above a known maximum. Replacing such draws by the nearest boundary may appear to fix the problem, but it changes the distribution in an arbitrary way and can create a pile-up exactly at the bound.
A truncated model instead specifies a continuous distribution only on the allowable interval and renormalizes its density there. Its draws respect the bounds without being forced onto them. The bounds must have a scientific or measurement basis; truncation is not a general-purpose repair for an unsuitable model.
This model is appropriate when the conditional distribution is reasonably represented by a normal distribution within the allowable range, the bounds are known and correctly applied, and the predictors capture the relevant structure in the conditional mean. For inference from incomplete data, the imputation model must also be compatible with the analysis model and the assumed missingness mechanism; conditional-model assumptions alone do not establish that the missingness mechanism is ignorable. For a variable with a known support, observed values should be inside that support, apart from explainable recording or rounding issues.
Density and Drawing Rule
Write \(\phi\) for the standard normal density and \(\Phi\) for its cumulative distribution function. Define standardized bounds \(a=(L-\mu(x))/\sigma\) and \(b=(U-\mu(x))/\sigma\). The probability that an untruncated normal draw falls inside the allowable interval is \(\Phi(b)-\Phi(a)\). Dividing the ordinary normal density by this probability gives the conditional density:
The denominator is essential: without it, the density over \((L,U)\) would integrate to less than one. To generate a draw, first sample \(V\) uniformly between \(\Phi(a)\) and \(\Phi(b)\), then transform back to the outcome scale:
This rule follows because the CDF of the restricted distribution at \(y\) is \(\{\Phi((y-\mu(x))/\sigma)-\Phi(a)\}/\{\Phi(b)-\Phi(a)\}\). Setting that CDF equal to \(W\), rearranging, and applying \(\Phi^{-1}\) gives the stated draw. The result is strictly between the bounds with probability one when \(L<U\).
In MICE, uncertainty in the regression parameters and residual scale must be represented as well as the random residual draw. Use the parameter-drawing approach for continuous regression imputation, then generate the missing value from the appropriately truncated predictive distribution using those draws. If the observed response itself comes from a truncated sampling distribution, fitting an ordinary normal regression to the observed values and truncating only the imputations may give inappropriate parameter estimates. A truncated likelihood accounts for the normalizing probability \(\Phi(b_i)-\Phi(a_i)\) for each predictor pattern \(i\).
Worked Examples
Worked Example: Imputing a Proportion Between Zero and One
Suppose a hypothetical continuous proportion \(Y\) is missing for a record with predictors \(x\). For illustration, a parameter draw gives \(\mu(x)=0.52\) and \(\sigma=0.18\), and the known bounds are 0 and 1. Assume the conditional normal shape is adequate within this interval and the proportion is not concentrated at either boundary.
Standardize the bounds: \(a=(0-0.52)/0.18=-2.8889\) and \(b=(1-0.52)/0.18=2.6667\). Using \(\Phi(a)\approx0.001933\) and \(\Phi(b)\approx0.9962\), the probability retained by truncation is \(0.9962-0.001933=0.994267\), approximately \(0.9943\).
For a hypothetical uniform draw \(W=0.25\), the target standard-normal probability is \(V=0.001933+0.25(0.994267)=0.25050\), approximately. Since \(\Phi^{-1}(0.25050)\approx-0.6729\), the imputed value is \(Y=0.52+0.18(-0.6729)\approx0.3989\).
The result is inside \((0,1)\), and it is not rounded to the nearest boundary. The calculation is conditional on the parameter values stated for this example; in an actual multiple-imputation run, the model parameters would also be drawn to reflect their uncertainty.
Worked Example: A Laboratory Result Below the Detection Limit
Imagine a hypothetical laboratory assay with a known physical lower bound of 0 and a detection limit of 0.20 concentration units. A result is recorded as “below detection,” so the latent concentration is known to lie in \([0,0.20)\), rather than being an ordinary unclassified missing value. Suppose the conditional normal model for this record has \(\mu(x)=0.52\) units and \(\sigma=0.18\) units.
The standardized interval is \(a=(0-0.52)/0.18=-2.8889\) and \(b=(0.20-0.52)/0.18=-1.7778\). Its probability under the untruncated normal model is approximately \(\Phi(-1.7778)-\Phi(-2.8889)=0.03772-0.001933=0.03579\).
For \(W=0.60\), set \(V=0.001933+0.60(0.03579)=0.023407\). The corresponding standard-normal quantile is approximately \(\Phi^{-1}(0.023407)=-1.9881\), giving \(Y=0.52+0.18(-1.9881)\approx0.1621\) concentration units.
This draw is positive and below 0.20, as the assay status requires. The interval restriction uses information in the “below detection” status. It would be wrong to use that interval for a result that was simply missing for an unrelated reason.
Worked Example: Truncation on a Finite Measurement Range
Consider a hypothetical continuous laboratory measure known to lie between 0 and 1.2 units. For one record, suppose the conditional model has \(\mu(x)=0.80\) units and \(\sigma=0.50\) units. These bounds and a normal shape within them are assumed appropriate for this illustration.
Here \(a=(0-0.80)/0.50=-1.60\) and \(b=(1.20-0.80)/0.50=0.80\). Using \(\Phi(-1.60)\approx0.05480\) and \(\Phi(0.80)\approx0.7881\), the retained probability is approximately \(0.7881-0.05480=0.7333\).
For \(W=0.80\), \(V=0.05480+0.80(0.7333)=0.64144\), approximately. Since \(\Phi^{-1}(0.64144)\approx0.362\), the draw is \(Y=0.80+0.50(0.362)=0.981\) units, approximately. It lies inside the specified range and is not artificially placed at either endpoint.
Model Fitting, Diagnostics, and Alternatives
The truncated model is most useful when the bounds are genuine and the conditional distribution within them has a plausible shape. If the response is sampled from a truncated normal population, its conditional likelihood contribution for an observed value \(y_i\) is the normal density at \(y_i\) divided by the probability of falling within the bounds for that record. The normalizing probability depends on \(x_i\) whenever the conditional mean does, so it cannot generally be ignored during fitting.
Check the model before relying on its imputations. Verify that the units and bounds are correct, that no observed values fall outside them without explanation, and that conditional distributions and residual patterns do not show strong skewness, changing spread, or unexplained multiple modes. Inspect the implied probability \(\Phi(b_i)-\Phi(a_i)\): a very small value means the untruncated model places little mass in the allowable interval, so the restricted draws may be highly sensitive to model assumptions and numerical calculations. Compare observed and imputed distributions by predictor patterns, and examine whether estimates are sensitive to reasonable alternative models and bounds.
For very narrow intervals or extreme tails, direct numerical evaluation of CDF differences can lose precision. Use numerically stable truncated-normal routines rather than subtracting nearly equal probabilities. A small retained probability can also make rejection sampling inefficient: the expected number of untruncated proposals per accepted draw is \(1/\{\Phi(b)-\Phi(a)\}\). The inverse-CDF method avoids that inefficiency, but still requires accurate tail calculations.
| Approach | What it does | Key limitation |
|---|---|---|
| Truncated conditional model | Draws from the specified distribution conditional on the allowable interval. | Depends on a defensible conditional distribution and correctly known bounds. |
| Transformation | Maps a bounded variable to an unbounded scale and models it there. | Can be awkward at boundaries and may not give a suitable transformed-scale model. |
| Rejection sampling | Repeatedly draws from an untruncated model until a draw lies within bounds. | Can be very inefficient when the retained probability is small. |
| Clipping | Replaces out-of-range draws by a boundary value. | Creates an artificial point mass at the boundary and is not a draw from the truncated model. |
Truncation does not solve every bounded-outcome problem. A variable with genuine point masses at 0 or 1 is not adequately described by a continuous distribution on the open interval alone. Likewise, a poor conditional mean, severe skewness, or heteroscedasticity is not repaired merely by imposing bounds. A transformation or another conditional model may be more suitable; the choice should be guided by the outcome’s measurement process and observed conditional behavior. The earlier material on transforming skewed continuous variables is relevant when skewness, rather than support alone, is the main concern.
Common Mistakes and Practical Checks
- Clipping instead of truncating. Setting negative values to zero changes the distribution by creating excess boundary values. Draw from the normalized conditional distribution instead.
- Ignoring censoring information. A value reported below a detection limit is known to lie in an interval. Imputing it from the full distribution discards that information.
- Confusing bounds with a remedy for model misspecification. A truncated normal model still requires a plausible conditional shape and predictor structure.
- Ignoring parameter uncertainty. Repeatedly drawing only residual variation around a single fitted model generally understates imputation uncertainty. Include parameter and scale uncertainty using the continuous-model procedure.
- Using a continuous model for boundary masses. If exact zeros or ones occur as genuine outcomes, a model that only assigns continuous probability between the bounds cannot represent them.
Check Your Understanding
Answer each question, then compare with the solutions below.
- For \(Y\mid x\sim N(0.50,0.20^2)\) restricted to \((0,1)\), state the standardized bounds and the probability that an untruncated draw is retained, approximately.
- Why is replacing every negative normal draw by zero not equivalent to drawing from a normal distribution truncated below at zero?
- A hypothetical assay reports a result as below 0.10 units. What interval should be used for an imputation if the physical lower bound is 0, and what additional information does this interval encode?
- A truncated model retains only 0.02 of the untruncated normal probability. Give one computational concern and one modeling concern.
Solutions. (1) \(a=(0-0.50)/0.20=-2.50\) and \(b=(1-0.50)/0.20=2.50\). The retained probability is \(\Phi(2.50)-\Phi(-2.50)\approx0.9876\). (2) Clipping places all negative probability at exactly zero, creating a point mass there; truncation removes that probability and renormalizes the density over positive values. (3) Use \([0,0.10)\). It encodes that the latent concentration is below the detection limit, not merely missing without explanation. (4) Rejection sampling would require about \(1/0.02=50\) proposals per accepted draw on average. The small retained probability also signals that the underlying conditional model assigns little mass to the allowed interval, so its shape and parameter estimates warrant careful review.