Two-Part Models For Semi-Continuous Variables

Why Use Two Parts?

A semi-continuous variable has a point mass at zero and a continuous distribution over positive values. Examples include a cost that is zero for people with no expenditure and positive otherwise, or a duration that is zero when an event does not occur during a defined period. A single continuous model can blur these two features: it may assign probability to impossible negative values and fail to represent the concentration of observations exactly at zero.

For MICE, the practical solution is to model two questions separately: is the value positive? and, if it is, how large is the positive value? The resulting imputation preserves the possibility of a genuine zero while representing the positive distribution with a model suited to its shape.

Definition: Let $Y$ be a variable whose values are either zero or positive. Define $Z=I(Y>0)$, where $I(\cdot)$ equals 1 when its condition holds and 0 otherwise. A two-part model specifies (1) $q(x)=P(Z=1\mid X=x)$, the probability of a positive value given predictors $X$, and (2) the conditional distribution of $Y$ given $Y>0$ and $X=x$. Thus $Y=0$ when $Z=0$; when $Z=1$, $Y$ is drawn from the positive-value distribution.

Model and Assumptions

In a MICE update, $X$ denotes the predictors currently available for the target variable. The first component is a model for the binary status $Z$. For example, it may use a logistic regression, with $q(x)=\operatorname{logit}^{-1}(x^\mathsf{T}\beta)$. The second component is fitted to observed positive $Y$ values only, conditional on the same relevant predictors. Its distribution must produce positive values, either by its support or by a suitable transformation.

For example, a model for $\log(Y)$ among positive observations can use a normal regression. A draw on the log scale is exponentiated to give a positive value on the original scale. Alternatively, a model with a positive support can be used directly. The appropriate choice depends on the positive values’ scale, skewness, and scientific meaning; using a normal model for the positive component without checking its support can still produce invalid values.

The two parts should be fitted and used under several conditions. Zero must be recorded distinctly and meaningfully; a missing or censored positive value must not be silently recoded as zero. Both parts need adequate information for the predictor patterns at which imputations are required. The status model should represent variation in the probability of positivity, and the positive-value model should represent the distribution among positive cases. As with other imputation models, the included predictors and missingness assumptions must be appropriate to the analysis; this construction alone does not resolve missing-data assumptions.

For proper stochastic imputation, account for uncertainty in the fitted models as well as variation in individual outcomes. In each completed data set, draw model parameters (or otherwise incorporate their uncertainty), then make a status draw. If that draw is positive, make a predictive draw from the positive component. The model parameters and predictive variation should be redrawn as appropriate across imputations, rather than replacing every missing value with a fitted mean or probability.

Fit the status component.
For observed cases, form $Z=I(Y>0)$ and model its conditional probability using the predictors in the MICE update.

Fit the positive component.
Use observed cases with $Y>0$ to model the distribution of $Y$ conditional on positivity and predictors. Choose a form that respects positivity.

Draw the missing value.
For a missing case, draw $Z$ from its fitted binary distribution. Set $Y=0$ if $Z=0$; otherwise draw $Y$ from the fitted positive distribution.

Worked Predictive Draw

Worked Example: Imputing a Missing Monthly Cost

Suppose $Y$ is monthly cost in dollars, which is zero for some people and positive for others. For illustration, assume the status model is $\operatorname{logit}\{q(x)\}=-0.2+0.5x$, and the positive-component model is $\log(Y)=2.0+0.1x+\epsilon$. In an actual MICE run, parameter uncertainty would also be represented; here the displayed coefficients and residual draw are hypothetical values for one predictive draw.

For a case with $x=1$, the status linear predictor is $-0.2+0.5(1)=0.3$. Therefore,

q(1)=\frac{1}{1+\exp(-0.3)}\approx 0.5744.

Draw $U=0.62$ from a uniform distribution on $[0,1]$. Because $0.62>0.5744$, this draw selects $Z=0$, so the imputed cost is $\$0$. If instead the status draw were $U=0.40$, then $0.40<0.5744$ and the draw would select $Z=1$. With a hypothetical residual draw $\epsilon=0.2$, the positive-component draw would be

Y=\exp\{2.0+0.1(1)+0.2\}=\exp(2.3)\approx \$9.974.

These are two possible stochastic imputations for the same predictor pattern, not two values to average. The first draw is zero; the second is positive because its separately drawn status is positive.

What Distribution Does the Model Imply?

The two components combine into a mixture distribution: conditional on $X=x$, probability $1-q(x)$ is concentrated exactly at zero, and probability $q(x)$ is assigned to the positive distribution. This gives simple expressions for the mean and variance, useful for understanding what the model represents.

Result: If the positive-component conditional mean and variance at $x$ are $\mu_+(x)$ and $\sigma_+^2(x)$, then

E(Y\mid X=x)=q(x)\mu_+(x),

\operatorname{Var}(Y\mid X=x) =q(x)\sigma_+^2(x)+q(x)\{1-q(x)\}\mu_+^2(x).

To justify the formulas, $Y=ZY_+$, where $Y_+$ denotes a draw from the positive component when $Z=1$. Since $Y=0$ when $Z=0$, its conditional expectation is $q(x)\mu_+(x)$. Also, $E(Y^2\mid X=x)=q(x)\{\sigma_+^2(x)+\mu_+^2(x)\}$. Subtracting the square of the expectation gives

\begin{aligned} \operatorname{Var}(Y\mid X=x) &=q(x)\{\sigma_+^2(x)+\mu_+^2(x)\}-q^2(x)\mu_+^2(x)\\ &=q(x)\sigma_+^2(x)+q(x)\{1-q(x)\}\mu_+^2(x). \end{aligned}

The first variance term reflects spread among positive values. The second reflects variation between zero and positive status. Consequently, a variable can have substantial overall variability even when its positive values are tightly clustered.

Worked Example: Mean and Variance With a Point Mass at Zero

Suppose, at a fixed predictor pattern, the probability of a positive value is $q=0.40$. Among positive values, assume the mean is $\mu_+=10$ units and the variance is $\sigma_+^2=9$ units$^2$. The unconditional mean is

E(Y)=q\mu_+=0.40(10)=4.00\text{ units}.

The variance is

\begin{aligned} \operatorname{Var}(Y) &=q\sigma_+^2+q(1-q)\mu_+^2\\ &=0.40(9)+0.40(0.60)(10^2)\\ &=3.6+24=27.6\text{ units}^2. \end{aligned}

Thus the standard deviation is $\sqrt{27.6}\approx 5.254$ units. The calculation includes both positive-value variation and the variability caused by having a 60% chance of an exact zero. Using only the positive-component variance would miss the latter contribution.

Worked Comparison Across Predictor Groups

Worked Example: Why the Two Components Both Matter

In hypothetical data, suppose group A has 10 people: 6 have zero expenditure, and the four positive values are 4, 6, 8, and 10 units. Group B also has 10 people: 2 have zero expenditure, and the eight positive values are 5, 7, 9, 11, 13, 15, 17, and 19 units. For this small illustration, calculate the observed positive fraction and positive mean separately in each group.

For group A, the positive fraction is $4/10=0.40$, and the mean among positive values is $(4+6+8+10)/4=28/4=7$ units. The implied group mean from the two-part calculations is $0.40(7)=2.8$ units.

For group B, the positive fraction is $8/10=0.80$, and the positive mean is $(5+7+9+11+13+15+17+19)/8=96/8=12$ units. Its implied group mean is $0.80(12)=9.6$ units.

The group difference reflects both a higher probability of positive expenditure and a larger positive mean in group B. A model for positive values alone would miss the difference in zero frequency; a model for zero status alone would miss the difference in positive amounts. These sample summaries illustrate the two components, but ten observations per group would generally be too little to justify treating them as stable estimates in a real analysis.

Diagnostics, Alternatives, and Limitations

Check each component on its own and then check the combined imputations. For the status component, compare observed and fitted positive proportions across important predictor patterns; inspect whether predictions are implausibly close to zero or one and whether some patterns have too few positive or zero cases. For the positive component, examine residual behavior or other suitable checks on the chosen scale, and compare its fitted distribution with observed positive values. In the completed data, inspect the fraction of zeros, the positive-value distribution, and relationships with key predictors. The imputed values should have a plausible point mass at zero and a plausible positive tail, not merely a plausible overall mean.

A common alternative is a single normal regression for $Y$. It may be convenient, but generally represents neither an exact point mass at zero nor a positive-only continuous distribution, and can generate negative values. A transformation-only approach can improve skewness or support among positive values, but does not by itself model the probability of zero. The two-part model directly represents both features. It is distinct from a model that posits a separate latent structural-zero mechanism: the basic two-part specification treats the observed zero status as one component and does not, by itself, distinguish different causes of zero.

The approach also has limits. Sparse data can make either component unstable, especially when few positive observations are available for the conditional amount model. A poor status model can misrepresent zero frequency, while a poor positive model can distort the range and tail of imputed amounts. Exponentiating a normal draw for $\log(Y)$ guarantees positivity, but the resulting distributional assumptions still need checking. For example, the exponentiated linear predictor without residual variation describes a conditional median under a lognormal model, not a predictive draw from the full positive distribution.

Common mistake: fit the positive model and use it for every missing case. This assumes every missing value is positive and discards the point mass. Draw status first, and use the positive model only when that draw is positive.
Common mistake: impute the fitted expected value. Replacing every missing value with $q(x)\mu_+(x)$ creates fractional values and suppresses both zero-status variation and positive-value variation. Make stochastic draws for completed data sets.
Common mistake: truncate a continuous model at zero and call it a two-part model. Truncation can enforce positive support, but it does not create a separate probability mass at exactly zero. Model status separately when zeros are a distinct feature.

Key takeaway: A semi-continuous imputation requires a model for whether the value is positive and a model for its magnitude conditional on being positive. Draw from both components, check each component’s fit, and assess the combined completed-data distribution.

Check Your Understanding

Answer each question using the two-part model definitions above. Brief solutions follow the questions.

If $q(x)=0.30$, $\mu_+(x)=8$ units, and $\sigma_+^2(x)=4$ units$^2$, what are the conditional mean and variance of $Y$?
A missing case has fitted $q(x)=0.65$. A uniform draw is $0.71$. What status is imputed, and is a positive-component draw needed?
Why is fitting the positive-value model only to observed positive cases appropriate for the second component?
Name one diagnostic for the zero-status component and one for the positive-value component.
Explain why imputing every missing value with $q(x)\mu_+(x)$ is not equivalent to drawing from the two-part model.

Solutions. (1) The mean is $0.30(8)=2.4$ units. The variance is $0.30(4)+0.30(0.70)(8^2)=1.2+13.44=14.64$ units$^2$. (2) Since $0.71>0.65$, the status draw is zero; no positive-value draw is needed. (3) The second component describes $Y$ conditional on $Y>0$, so its fitting data must have positive observed values. (4) Compare observed and fitted positive proportions across predictor patterns; for the positive component, inspect residual behavior on its modeled scale or compare its fitted distribution with observed positive values. (5) The product is a conditional expectation, not a predictive draw: it removes the exact-zero probability and the variation in positive amounts, and can produce a value that is neither zero nor representative of a particular draw.