Negative Binomial Hurdle Models

Why Use a Negative Binomial Hurdle Model?

A count outcome can have two distinct features: whether any events occur, and how many occur when the count is positive. A negative binomial hurdle model represents these features with separate probability components. A Bernoulli process determines whether the outcome is zero or positive; conditional on being positive, the count follows a negative binomial distribution truncated to exclude zero.

This construction is useful when the positive counts are more variable than a Poisson model allows, while the chance of observing a zero needs its own parameter. For example, a hypothetical clinic might record the number of visits made by each patient in a fixed period. The model can represent both the chance of making at least one visit and the distribution of visit counts among patients who made at least one.

Definition: Let \(Y\) be a nonnegative integer-valued outcome. A negative binomial hurdle model has a hurdle probability \(h=P(Y>0)\). With probability \(1-h\), \(Y=0\). With probability \(h\), \(Y\) is drawn from a negative binomial distribution conditional on being positive. Thus the positive-count distribution is a zero-truncated negative binomial distribution.

Model Definition and Probability Formula

Use a negative binomial distribution parameterized by its mean \(\mu>0\) and shape parameter \(\kappa>0\). In this parameterization, the untruncated parent distribution has mean \(\mu\) and variance \(\mu+\mu^2/\kappa\). For integer \(y\geq 0\), its probability mass function is

f_{\mathrm{NB}}(y;\mu,\kappa) = \frac{\Gamma(y+\kappa)}{\Gamma(\kappa)y!} \left(\frac{\kappa}{\kappa+\mu}\right)^\kappa \left(\frac{\mu}{\kappa+\mu}\right)^y.

In particular, the probability of zero under the untruncated parent distribution is \(f_0=(\kappa/(\kappa+\mu))^\kappa\). Removing zero leaves total probability \(1-f_0\). Therefore the conditional probability of a positive count \(y\) is the parent probability divided by \(1-f_0\).

Model formula: For \(0<h<1\), \(\mu>0\), \(\kappa>0\),

P(Y=0)=1-h, \qquad P(Y=y)=h\frac{f_{\mathrm{NB}}(y;\mu,\kappa)}{1-f_0}, \quad y=1,2,\ldots, \qquad f_0=\left(\frac{\kappa}{\kappa+\mu}\right)^\kappa.

The hurdle parameter \(h\) is the probability of a positive count. The parameters \(\mu\) and \(\kappa\) describe the untruncated negative binomial parent distribution used to construct the positive-count distribution. In particular, \(\mu\) is not the mean of the observed positive counts: conditioning on positivity changes the distribution. This distinction matters when assigning numerical values or interpreting model output.

Why the Truncation Factor Is Required

A probability distribution for positive counts must sum to one over \(y=1,2,\ldots\). The untruncated negative binomial probabilities over those values sum to \(1-f_0\), not to one. Dividing each by \(1-f_0\) renormalizes them. Multiplying the resulting conditional probabilities by \(h\) then assigns total probability \(h\) to all positive counts.

\sum_{y=1}^{\infty}P(Y=y) = h\sum_{y=1}^{\infty} \frac{f_{\mathrm{NB}}(y;\mu,\kappa)}{1-f_0} = h\frac{1-f_0}{1-f_0} = h.

Adding \(P(Y=0)=1-h\) gives total probability \(h+(1-h)=1\). This normalization check is a quick way to detect a missing truncation factor or a mistaken zero probability. The expected count also provides a useful check: since the parent negative binomial has mean \(\mu\), its positive-count conditional mean is \(\mu/(1-f_0)\). Consequently,

E(Y)=h\frac{\mu}{1-f_0}.

This expression weights the average among positive counts by the chance \(h\) of having a positive count at all.

Worked Example: A Geometric Parent Distribution

Suppose a hypothetical count outcome has \(h=0.60\), \(\mu=2\), and \(\kappa=1\). Calculate the probabilities of zero, one, and two counts.

With \(\kappa=1\), the parent negative binomial has \(f_0=(1/(1+2))^1=1/3\). Its probabilities for one and two are \(f_{\mathrm{NB}}(1)= (1/3)(2/3)=2/9\) and \(f_{\mathrm{NB}}(2)=(1/3)(2/3)^2=4/27\). The positive-count normalizing denominator is \(1-f_0=2/3\).

The zero probability is \(P(Y=0)=1-0.60=0.40\). For one count, \(P(Y=1)=0.60(2/9)/(2/3)=0.60(1/3)=0.20\). For two counts, \(P(Y=2)=0.60(4/27)/(2/3)=0.60(2/9)=0.1333\).

These are probabilities in the hurdle distribution, not the parent negative binomial. For example, the parent probability of one is \(2/9\), whereas the hurdle probability of one is \(0.20\). The latter includes both truncation and the probability of crossing the hurdle.

Calculating Specific Counts and Ranges

For any positive integer \(y\), first find its untruncated negative binomial probability, then divide by \(1-f_0\), and finally multiply by \(h\). For a range of positive counts, it is often simpler to sum the corresponding parent probabilities before dividing. The probability of a positive-count event \(A\) is

P(Y\in A)=h\frac{\sum_{y\in A}f_{\mathrm{NB}}(y;\mu,\kappa)}{1-f_0}, \qquad A\subseteq\{1,2,\ldots\}.

For instance, \(P(Y\geq 2)\) equals \(h\) times the parent probability of \(Y\geq 2\), divided by \(1-f_0\). Equivalently, the conditional positive-count probability of one can be subtracted from one: \(P(Y\geq 2)=h\left[1-f_{\mathrm{NB}}(1;\mu,\kappa)/(1-f_0)\right]\).

Worked Example: Probabilities with Shape Two

For a hypothetical process, let \(h=0.75\), \(\mu=3\), and \(\kappa=2\). Find \(P(Y=0)\), \(P(Y=1)\), \(P(Y=2)\), and \(P(Y\geq 2)\).

The parent zero probability is \(f_0=(2/(2+3))^2=0.16\), so the truncation denominator is \(1-f_0=0.84\). For one count, the negative binomial coefficient is \(\Gamma(3)/(\Gamma(2)1!)=2\), giving \(f_{\mathrm{NB}}(1)=2(0.4)^2(0.6)=0.192\). For two counts, the coefficient is \(\Gamma(4)/(\Gamma(2)2!)=3\), giving \(f_{\mathrm{NB}}(2)=3(0.4)^2(0.6)^2=0.1728\).

Thus \(P(Y=0)=1-0.75=0.25\), \(P(Y=1)=0.75(0.192/0.84)=0.1714\), and \(P(Y=2)=0.75(0.1728/0.84)=0.1543\). For the tail, subtract the conditional probability of one from the probability of crossing the hurdle: \(P(Y\geq 2)=0.75[1-(0.192/0.84)]=0.5786\).

The probabilities check: \(P(Y=0)+P(Y=1)+P(Y\geq2)=0.25+0.1714+0.5786=1.0000\). The tail includes \(Y=2\), so it must be at least as large as \(P(Y=2)\), as it is here.

Worked Example: A Small Shape Parameter

Suppose a hypothetical count process has \(h=0.40\), \(\mu=1.5\), and \(\kappa=0.5\). Calculate the probabilities of zero, one, and at least two counts.

Here \(\kappa/(\kappa+\mu)=0.5/2=0.25\), so \(f_0=(0.25)^{0.5}=0.50\) and \(1-f_0=0.50\). For one count, the coefficient is \(\Gamma(1.5)/(\Gamma(0.5)1!)=0.5\). Therefore \(f_{\mathrm{NB}}(1)=0.5(0.25)^{0.5}(0.75)=0.1875\).

The hurdle probabilities are \(P(Y=0)=1-0.40=0.60\) and \(P(Y=1)=0.40(0.1875/0.50)=0.15\). The probability of at least two is the remaining positive-count probability: \(P(Y\geq2)=0.40[1-(0.1875/0.50)]=0.40(0.625)=0.25\). They sum to \(0.60+0.15+0.25=1.00\). The small shape parameter permits substantial variation in the parent counts; the truncation and hurdle calculations are still the same.

Assumptions and Model Checks

The probability formulas apply when the outcome is a count with possible values \(0,1,2,\ldots\), the hurdle probability is between zero and one, and the positive outcomes follow the specified zero-truncated negative binomial distribution. The parent negative binomial uses a common \(\mu\) and \(\kappa\) for the observations being described. If a model includes predictors or latent effects, these formulas describe probabilities conditional on their specified values; the numerical examples above have no such terms.

For independent observations with common parameters, the likelihood contribution from an observed count \(y_i\) is \(1-h\) if \(y_i=0\), and \(h f_{\mathrm{NB}}(y_i;\mu,\kappa)/(1-f_0)\) if \(y_i>0\). This follows directly by applying the model probability formula to each observation. It also shows why the two parts of the model address different features: zero versus positive status contributes through \(h\), while the sizes of positive counts contribute through the truncated negative binomial probabilities.

Check fitted probabilities in both parts rather than relying only on an overall fit statistic. Among observations, compare the observed and predicted proportions that are zero. Among positive counts, compare frequencies at small values and in the upper tail with the fitted zero-truncated negative binomial distribution. Systematic underprediction of large positive counts may indicate that its dispersion or distributional shape is inadequate; a good match to the zero proportion alone does not establish a good fit for positive counts. If observations are dependent, treating them as independent can also invalidate uncertainty calculations, even if the marginal count probabilities appear plausible.

How It Differs from Alternatives

A negative binomial hurdle model is not the same construction as a zero-inflated negative binomial model. In the hurdle model, the positive-count component is explicitly conditioned to exclude zero, and the hurdle parameter sets the total probability of a positive count. In a zero-inflated negative binomial model, the count-generating negative binomial component can itself generate zero. These models therefore assign probabilities through different mechanisms, even when both can fit data with many zeros.

A Poisson hurdle model uses the same zero-versus-positive structure but uses a zero-truncated Poisson distribution for the positive counts. The negative binomial version is useful when the positive-count distribution needs more variability than a Poisson distribution provides. Model choice should follow the data-generating assumptions and the fit of the positive counts, not just the observed number of zeros. Counts alone do not prove that a particular unobserved process generated the zeros.

Common Mistakes and Limitations

Leaving out the truncation denominator. Using \(h f_{\mathrm{NB}}(y)\) for \(y>0\) makes the positive probabilities sum to \(h(1-f_0)\), not \(h\). Divide by \(1-f_0\).
Using the parent mean as the positive-count mean. The mean among positive counts is \(\mu/(1-f_0)\), not \(\mu\). The overall mean is \(h\mu/(1-f_0)\).
Confusing the hurdle probability with the zero probability. Here \(h=P(Y>0)\), so the zero probability is \(1-h\).
Assuming the distributional form is validated by matching zeros. The hurdle component controls zero frequency, but the positive-count negative binomial must also fit the observed positive counts.

The model is limited if the positive counts do not resemble a zero-truncated negative binomial or if the single hurdle probability is too restrictive for the setting. It also separates the two probability components mathematically; this separation does not by itself establish two distinct causal processes in the real world.

Key takeaway: A negative binomial hurdle distribution assigns \(1-h\) to zero and distributes the remaining probability \(h\) across positive counts using a renormalized negative binomial. For every positive-count probability, the essential calculation is \(h\) times the parent negative binomial probability divided by \(1-f_0\).

Check Your Understanding

Use the model definition above. Each problem has its solution immediately afterward so that the calculations can be checked.

If \(h=0.7\), what is \(P(Y=0)\)? Solution: \(P(Y=0)=1-h=1-0.7=0.30\).
For a parent negative binomial with \(f_0=0.20\), what denominator normalizes its positive-count probabilities? Solution: \(1-f_0=1-0.20=0.80\).
Suppose \(h=0.5\), \(f_0=0.25\), and \(f_{\mathrm{NB}}(1)=0.15\). Calculate \(P(Y=1)\). Solution: \(P(Y=1)=0.5(0.15/0.75)=0.5(0.20)=0.10\).
With the same \(h\) and \(f_0\) as in question 3, what is \(P(Y\geq2)\) if \(f_{\mathrm{NB}}(1)=0.15\)? Solution: The conditional positive probability of one is \(0.15/0.75=0.20\). Therefore \(P(Y\geq2)=0.5(1-0.20)=0.40\).
A calculation gives positive hurdle probabilities summing to \(0.84\) when \(h=0.70\). Is it normalized? Solution: No. Positive probabilities must sum to \(h=0.70\), while the zero probability is \(0.30\); together they should sum to one. The result suggests a missing or incorrect truncation adjustment.