Two-Part Models for Outcomes With Many Zeros

Why Separate the Two Parts?

Many economic outcomes have a substantial number of exact zeros and a continuous or otherwise varying range of positive values. A household may make no charitable donation during a period, while households that donate give different amounts. A firm may report no spending in a category, while firms with positive spending have widely varying expenditures. In such settings, the question of whether the outcome is positive is distinct from the question of how large it is when positive.

A two-part model represents these features with a participation equation and a separate positive-outcome model. The first describes the probability of a positive outcome given covariates. The second describes the mean outcome among observations with positive outcomes, again given covariates. Combining these components yields the expected outcome for the full population at those covariate values.

This setup applies when the outcome is nonnegative and zero whenever participation does not occur. A zero is treated as an observed outcome, not automatically as a missing value. The model does not require that the zeros be generated by the same latent process as the positive values.

Definition and Model Specification

Definition: Let \(Y\) be a nonnegative outcome and \(X\) a vector of observed covariates. Define the participation indicator \(D=1\{Y>0\}\), which equals one for a positive outcome and zero otherwise. A two-part model specifies (1) the participation probability \(\pi(x)=P(D=1\mid X=x)\) and (2) the positive-outcome conditional mean \(m(x)=E(Y\mid D=1,X=x)\). The model-implied unconditional conditional mean is \(E(Y\mid X=x)=\pi(x)m(x)\), provided \(Y=0\) whenever \(D=0\).

The first part is a binary-response model. For example, a logit specification is

P(D=1\mid X=x)=\pi(x) =\frac{\exp(x^\top\beta)}{1+\exp(x^\top\beta)}.

This equation ensures that the fitted participation probability lies between zero and one. A probit specification is another standard option. The positive-outcome part specifies \(E(Y\mid D=1,X=x)\), not the mean over all observations. It may use a linear conditional-mean form \(m(x)=z^\top\gamma\), where \(z\) can contain covariates or transformations of them, provided the implied mean is positive over the covariate range of interest. A log-link mean specification \(m(x)=\exp(z^\top\gamma)\) is one way to ensure positivity.

The two equations need not use exactly the same covariates or functional forms. Their roles must, however, remain distinct: \(\pi(x)\) concerns the chance of observing a positive outcome, and \(m(x)\) concerns the average size conditional on a positive outcome. If a log model is instead fitted to the logarithm of the positive outcome, simply exponentiating its fitted conditional mean generally does not give the conditional mean in outcome units; the positive-part model must deliver or be converted to the desired mean.

Deriving the Unconditional Mean

The product formula follows by conditioning on whether the outcome is positive. For a given \(x\), the event \(D=0\) has probability \(1-\pi(x)\), and on that event \(Y=0\). On the event \(D=1\), the conditional mean of \(Y\) is \(m(x)\). The law of total expectation therefore gives:

\begin{aligned} E(Y\mid X=x) &=E(Y\mid D=1,X=x)P(D=1\mid X=x)\\ &\quad+E(Y\mid D=0,X=x)P(D=0\mid X=x)\\ &=m(x)\pi(x)+0\cdot[1-\pi(x)]\\ &=\pi(x)m(x). \end{aligned}

This is a decomposition of the conditional mean, not an independence assumption. It holds because the outcome is zero for the nonparticipating group and because the two conditional quantities are defined at the same covariate value \(x\). The product is expressed in the units of \(Y\): a probability is dimensionless, while \(m(x)\) has outcome units.

Worked Example: Combining a Participation Rate and a Positive Mean

Suppose, hypothetically, that at a specified set of household characteristics, a model gives a 0.40 probability of making a positive donation. Among households with positive donations and these characteristics, the modeled mean donation is 150 dollars. Find the expected donation across all households with these characteristics.

Here \(\pi(x)=0.40\) and \(m(x)=150\) dollars. The unconditional conditional mean is

E(Y\mid X=x)=\pi(x)m(x)=0.40(150\text{ dollars})=60\text{ dollars}.

Thus the expected donation is 60 dollars per household at these covariate values. This does not mean every household donates 60 dollars, nor does it mean that the average among donors is 60 dollars. The positive-outcome mean is 150 dollars; the lower all-household mean incorporates the 60% probability of a zero.

Working Through a Logit Participation Equation

In an applied specification, the participation probability is usually calculated from covariates through a link function rather than supplied directly. The following example shows how the logit probability enters the product. All quantities are hypothetical.

Worked Example: Expected Spending at a Given Covariate Value

Suppose \(X\) is a scalar covariate and the participation equation is

\operatorname{logit}\{\pi(x)\}=-1+0.4x,

where \(\operatorname{logit}(q)=\log\{q/(1-q)\}\). Suppose the positive-outcome conditional mean is \(m(x)=5+2x\) dollars over the range considered. Calculate the unconditional expected outcome at \(x=2\).

First evaluate the logit index:

$$ -1+0.4(2)=-0.2. $$

Convert this index to a probability using the inverse-logit formula:

\pi(2)=\frac{\exp(-0.2)}{1+\exp(-0.2)} \approx 0.450166.

The positive conditional mean at \(x=2\) is

m(2)=5+2(2)=9\text{ dollars}.

Therefore,

E(Y\mid X=2)=\pi(2)m(2) =\frac{\exp(-0.2)}{1+\exp(-0.2)}(9) \approx 4.051494\text{ dollars}.

To three decimal places, the expected outcome is approximately 4.051 dollars, or approximately 4.05 dollars to the nearest cent. The probability was kept unrounded in the final multiplication so that intermediate rounding would not change the reported result. This is the mean over all observations at \(x=2\), including those with zero outcomes.

Both Parts Can Contribute to Differences in Expected Outcomes

Two groups can differ in their unconditional expected outcomes because their participation probabilities differ, because their positive conditional means differ, or because both differ. A comparison of the product alone does not reveal which component accounts for the difference. Consider the following hypothetical model predictions for two covariate profiles.

Worked Example: Comparing Two Covariate Profiles

Suppose profile A has participation probability \(\pi_A=0.30\) and positive conditional mean \(m_A=20\) dollars. Profile B has \(\pi_B=0.50\) and \(m_B=24\) dollars. Calculate the unconditional means and decompose their difference.

For profile A,

E(Y\mid A)=0.30(20)=6\text{ dollars}.

For profile B,

E(Y\mid B)=0.50(24)=12\text{ dollars}.

The difference is \(12-6=6\) dollars. One exact decomposition, changing participation first and then the positive mean, is

\begin{aligned} \pi_Bm_B-\pi_Am_A &=(\pi_B-\pi_A)m_A+\pi_B(m_B-m_A)\\ &=(0.50-0.30)(20)+(0.50)(24-20)\\ &=4+2=6\text{ dollars}. \end{aligned}

Under this ordering, 4 dollars of the difference is associated with the change in participation probability evaluated at profile A’s positive mean, and 2 dollars with the change in positive mean evaluated at profile B’s participation probability. A different ordering can allocate the interaction differently, although the total difference remains 6 dollars. These are descriptive model-based comparisons; without an appropriate causal design, they do not establish that changing a covariate would cause either component to change.

Assumptions and Interpretation

The product identity is exact for the stated outcome structure, but useful model predictions still depend on the specifications. Before interpreting fitted values, check the following:

Outcome structure: \(Y\) is nonnegative, \(D=1\{Y>0\}\), and \(Y=0\) whenever \(D=0\). If the outcome can be negative, or if \(D\) is defined by a different rule, this particular decomposition needs modification.
Participation equation: the chosen binary-response specification represents \(P(D=1\mid X=x)\) adequately over the covariate values being interpreted.
Positive-outcome equation: the second model represents \(E(Y\mid D=1,X=x)\) on the original outcome scale, with positive predictions in the relevant range.
Target population: both components refer to the same covariate profile and target population. Combining a probability for one population with a positive mean for another does not produce the intended conditional mean.

A two-part specification does not by itself make covariate effects causal. It describes conditional probabilities and conditional means given the variables included in the model. A causal interpretation requires additional design or identification assumptions beyond the product formula.

Common Mistake: Confusing the Positive Mean With the Overall Mean

A frequent error is to report \(m(x)\) as the expected outcome for the full population. That omits observations with zero outcomes. If the probability of a positive outcome is 0.25 and the conditional positive mean is 80 units, the overall conditional mean is \(0.25(80)=20\) units, not 80 units. The value 80 is the mean among the positive-outcome group only.

A second error is to multiply the participation probability by a prediction that is already the unconditional mean. That counts the participation probability twice. Keep the meaning of each prediction explicit: \(\pi(x)\) is a probability, \(m(x)\) is a mean conditional on \(D=1\), and \(\pi(x)m(x)\) is the unconditional conditional mean.

Key takeaway: For a nonnegative outcome that is zero whenever \(D=0\), the two-part model combines participation and positive-outcome size as \(E(Y\mid X=x)=\pi(x)m(x)\). Interpret each component according to its conditioning event and report the product in the original outcome units.

Check Your Understanding

Answer each question by identifying the relevant component and showing the calculation where appropriate.

If the probability of a positive outcome is 0.35 and the conditional mean among positive outcomes is 40 units, what is the unconditional conditional mean?
In a two-part model, what population does \(m(x)=E(Y\mid D=1,X=x)\) describe?
A logit participation index at a covariate value equals zero. What is the corresponding participation probability?
Why does the product formula not require independence between participation and the outcome conditional on \(X\)?
Two profiles have the same participation probability but different positive conditional means. Which component accounts for their difference in unconditional means?