Zero-Truncated Count Distributions

Why Truncation Changes a Count Model

A count can be positive not because zero events are impossible, but because units with zero events never enter the recorded data. For example, a hypothetical registry might include only people who experienced at least one event during a defined study period. The observed counts then have support \(1,2,3,\ldots\), even if the underlying event process could have produced zero. Treating these records as an ordinary Poisson or negative binomial sample ignores the selection rule: it assigns probability to values that could not have been recorded and uses the wrong probabilities for the positive counts.

The key distinction is between zero truncation and a response that simply happens to contain no observed zeros. Truncation is a feature of the sampling or recording process: all zero-count units are excluded. If zeros were eligible for inclusion but none happened to occur in a small sample, the likelihood is not automatically zero-truncated.

This tutorial focuses on selecting between zero-truncated Poisson and zero-truncated negative binomial responses. The underlying Poisson and negative binomial families have already been introduced. The essential change is conditioning their likelihoods on the fact that only positive counts can appear in the data.

Zero-Truncated Poisson Model

Suppose the underlying count follows a Poisson distribution with mean \(\lambda>0\), so \(P(Y=y)=e^{-\lambda}\lambda^y/y!\) for \(y=0,1,\ldots\). Since \(P(Y=0)=e^{-\lambda}\), the probability that a unit has a positive count is \(1-e^{-\lambda}\). Dividing each positive-count probability by this inclusion probability gives the zero-truncated Poisson distribution:

The denominator is part of the probability model, not an optional adjustment. For any \(\lambda>0\), the probabilities over positive counts sum to one because \(\sum_{y=1}^{\infty}P(Y=y)=1-P(Y=0)\).

The conditional mean is larger than the underlying Poisson mean. Since \(E(Y)=\lambda\) and the only removed value is zero, \(E[Y\,I(Y>0)]=E(Y)=\lambda\). Therefore:

This relationship explains why the sample average of positive counts should not be interpreted as an estimate of the underlying \(\lambda\) without accounting for truncation. It estimates the conditional mean, which exceeds \(\lambda\).

Likelihood and Estimation With Positive Counts

For independent observed positive counts \(y_1,\ldots,y_n\), the zero-truncated Poisson likelihood is the product of their conditional probabilities. Its log likelihood is:

The last term appears once for each observed unit. Omitting it gives the ordinary Poisson likelihood, which assumes that zeros could have been sampled. Differentiating the truncated log likelihood and setting the score to zero gives the estimating equation \(\bar{y}=\lambda/(1-e^{-\lambda})\), where \(\bar{y}\) is the mean of the observed positive counts. This equation generally requires numerical solution; \(\bar{y}\) itself is not the estimate of \(\lambda\).

Zero-Truncated Negative Binomial Responses

The negative binomial alternative is useful when the underlying counts have more variability than a Poisson model allows. Using the NB2 parameterization, let the untruncated mean be \(\mu>0\) and the dispersion parameter be \(\alpha>0\), so that \(\operatorname{Var}(Y)=\mu+\alpha\mu^2\). Its probability of zero is \(P(Y=0)=(1+\alpha\mu)^{-1/\alpha}\). Conditioning on a positive count gives:

As in the Poisson case, the denominator is the probability of being observed under the specified underlying distribution. The conditional mean is \(\mu/[1-P(Y=0)]\), because the removed count is zero. The truncated data therefore do not have mean \(\mu\); \(\mu\) remains the mean of the underlying, untruncated negative binomial distribution. In a regression model, it is often clearest to describe the linear predictor as modelling this underlying mean and to separately calculate the implied conditional mean for included units.

Using Truncation in a Mixed Model

In a GLMM, a common specification uses a log link for the underlying mean, for example \(\log(\mu_{ij})=x_{ij}^{T}\beta+z_{ij}^{T}b_j\), where \(i\) indexes units, \(j\) indexes groups, \(\beta\) contains fixed effects, and \(b_j\) is a group-level random effect. For a truncated Poisson response, conditional on \(b_j\), the positive response has a Poisson distribution with mean \(\mu_{ij}\), conditioned on being greater than zero. A truncated negative binomial GLMM uses the corresponding negative binomial conditional distribution and its zero probability. The random-effects distribution and the usual conditional independence assumptions must also be specified.

Selection can affect how random effects must be handled. If groups are represented in the sample regardless of their outcomes and individual units are retained exactly when their counts are positive, one can form the conditional positive-count likelihood given each group’s \(b_j\), then integrate over the random-effects distribution. But if the sampling process first generates a population of units and only units with positive counts are retained, the random-effects composition among retained units is generally not the original population composition. A unit with a random effect that makes positive counts more likely is more likely to be selected.

For a single selected unit from a population with random effect \(b\) having density \(g(b)\), the marginal distribution of its observed positive count is:

where \(f(y\mid b)\) is the untruncated conditional count probability. The numerator averages the joint probability of a positive count equal to \(y\) over the source-population random effects; the denominator is the population probability of inclusion. This is generally not the same as averaging the within-\(b\) truncated probabilities using the unadjusted density \(g(b)\). The correct construction depends on whether selection occurs at the unit level, group level, or another stage of the design.

Choosing a Response and Checking the Model

A zero-truncated response is appropriate when the study mechanism rules out zero counts by design: for instance, only units with at least one event are enumerated. Conditional on the covariates and random effects, a zero-truncated Poisson model assumes the underlying Poisson count process is adequate; a zero-truncated negative binomial model allows the specified extra-Poisson variability. The selection rule must be known and correctly represented. Dependence among counts must be handled through the model structure, rather than assumed away merely because the outcomes are truncated.

Compare the response choices by asking what could have entered the data. Use an ordinary Poisson or negative binomial likelihood if zero-count units were eligible and recorded, even if the realized sample happens to contain few or no zeros. A zero-inflated model represents an additional process that produces structural zeros alongside a count process; it is not a substitute for truncation when zeros were excluded by design. A hurdle model represents a separate zero-versus-positive process together with a positive-count distribution, and is relevant when zeros are observed and the zero process is modelled. These alternatives answer different sampling or data-generating questions.

Useful checks include confirming from the protocol that zero-count units could not be recorded, checking that all analysed responses are positive, and comparing the model-implied positive-count distribution with the observed frequencies. Large discrepancies in the relative frequencies of ones, twos, and larger counts can signal a poor count family or an incorrect mean structure. In a mixed model, also examine whether the fitted random-effects structure and the assumed selection stage agree with how units and groups entered the sample. A truncation correction cannot repair unmodelled dependence, excess variation, or selection based on some rule other than \(Y>0\).

A common mistake is to fit an ordinary Poisson model to positive-only data and interpret its fitted mean as the underlying event rate. The ordinary likelihood assigns positive probability to zeros and fails to condition on inclusion. Another mistake is to treat the observed positive-count mean as the underlying mean; the derivation above shows that truncation raises the conditional mean. Use the truncated likelihood when inclusion requires \(Y>0\), and state whether reported means refer to the source population or to the observed positive-count units.