Hurdle Models Versus Zero-Inflated Models

Two Different Ways to Generate Zeros

A count distribution may assign substantial probability to zero, but a zero can arise through different mechanisms. In a zero-inflated model, a latent component can generate a certain zero; otherwise, a baseline count distribution generates the observation, which may itself also be zero. In a hurdle model, one process determines whether the count is zero or positive, and a positive count is then drawn from a baseline distribution restricted to positive values. The distinction is about how probability is allocated, not simply whether the data contain many zeros.

To compare the models precisely, use the same baseline count distribution for both. The formulas below apply at fixed covariate values and, if present, fixed random effects. After integrating over random effects, the marginal hurdle positive-count distribution is generally a mixture of the conditional truncated baselines, not the truncation of the marginal baseline. Mixing over random effects can change the marginal mass function, so comparisons must use the same conditioning or marginalization in both models.

Definition: Let \(f(y)=P(X=y)\) be a baseline probability mass function on the nonnegative integers, and let \(f_0=f(0)\). Assume \(f_0<1\), so the baseline assigns positive probability to positive counts. A zero-inflated model with structural-zero probability \(\pi\) assigns probability \(\pi\) to the extra-zero component and probability \(1-\pi\) to the baseline component. A hurdle model with crossing probability \(q\) assigns probability \(1-q\) to zero; conditional on crossing the hurdle, its positive count follows the baseline distribution conditioned on being positive. Here \(0\leq\pi\leq1\) and \(0\leq q\leq1\).

Writing the two probability mass functions makes the mechanisms visible. For a zero-inflated model, a positive count can only come from the baseline component. For a hurdle model, all positive counts come from the truncated baseline distribution.

\begin{aligned} P_{\mathrm{ZI}}(Y=0)&=\pi+(1-\pi)f_0,\\ P_{\mathrm{ZI}}(Y=y)&=(1-\pi)f(y),\qquad y=1,2,\ldots;\\[4pt] P_{\mathrm{H}}(Y=0)&=1-q,\\ P_{\mathrm{H}}(Y=y)&=q\,\frac{f(y)}{1-f_0},\qquad y=1,2,\ldots. \end{aligned}

The zero-inflated zero probability includes both the certain-zero component and zeros from the baseline component. The hurdle zero probability is set directly by the probability of not crossing the hurdle. In either model, the positive-count probabilities sum to the model's probability of a positive outcome: \(1-\pi\) times \(1-f_0\) for the zero-inflated model, and \(q\) for the hurdle model.

Positive Counts and the Condition for Agreement

Whenever a positive observation is possible, the two models give the same conditional distribution of the count among positive observations, provided they use the same baseline distribution. In the zero-inflated model, this requires \(\pi<1\); in the hurdle model, it requires \(q>0\). In both cases also require \(f_0<1\), as specified above.

\begin{aligned} P_{\mathrm{ZI}}(Y=y\mid Y>0) &=\frac{(1-\pi)f(y)}{(1-\pi)(1-f_0)} =\frac{f(y)}{1-f_0},\qquad \pi<1,\\ P_{\mathrm{H}}(Y=y\mid Y>0) &=\frac{q f(y)/(1-f_0)}{q} =\frac{f(y)}{1-f_0},\qquad q>0. \end{aligned}

The cancellation in these expressions is legitimate only when the probability of \(Y>0\) is positive. At \(\pi=1\), the zero-inflated model gives probability one to zero; at \(q=0\), the hurdle model does the same. In either case \(P(Y>0)=0\), so the conditional distribution given \(Y>0\) is undefined, not the truncated baseline distribution.

The models can give the same unconditional distribution too, but only with a matching relationship between their parameters. Equating their zero probabilities, or equivalently their probabilities of being positive, gives the relationship below.

q=(1-\pi)(1-f_0).

For example, given a hurdle crossing probability \(q\), the matching zero-inflation probability is \(\pi=1-q/(1-f_0)\), provided \(0\leq q\leq1-f_0\). Under this relationship, the zero masses agree, and the positive masses agree because \(qf(y)/(1-f_0)=(1-\pi)f(y)\) for every \(y>0\). Thus the entire distributions agree. If \(q>1-f_0\), the hurdle model has fewer zeros than the baseline distribution, something a zero-inflated model using that same baseline cannot achieve.

Key result: With a common baseline mass function \(f\), the models have the same conditional distribution among positive counts whenever conditioning is defined. They have the same full distribution exactly when \(q=(1-\pi)(1-f_0)\). At the all-zero boundaries \(\pi=1\) and \(q=0\), both can put probability one at zero, but conditioning on a positive count is undefined.

Worked Example: Comparing Positive Counts With Different Zero Masses

Suppose the shared baseline is Poisson with mean \(\lambda=1.5\). Then \(f_0=e^{-1.5}=0.2231\), \(f(1)=e^{-1.5}(1.5)=0.3347\), and \(f(2)=e^{-1.5}(1.5^2/2)=0.2510\). Compare a zero-inflated model with \(\pi=0.25\) and a hurdle model with \(q=0.50\).

In the zero-inflated model, \(P(Y=0)=0.25+0.75(0.2231)=0.4173\), and \(P(Y=1)=0.75(0.3347)=0.2510\). Its positive-outcome probability is \(0.75(1-0.2231)=0.5827\). Thus, among positive counts, \(P(Y=1\mid Y>0)=0.2510/0.5827=0.4308\).

In the hurdle model, \(P(Y=0)=1-0.50=0.5000\), and \(P(Y=1)=0.50(0.3347)/(1-0.2231)=0.2154\). Conditional on a positive count, the probability of one is \(0.2154/0.50=0.4308\), agreeing with the zero-inflated result. The models therefore share the positive-count distribution but, with these parameter values, do not have the same overall distribution: their zero probabilities are 0.4173 and 0.5000.

Worked Example: Choosing Parameters for the Same Full Distribution

Let the baseline be Poisson with \(\lambda=2\). Then \(f_0=e^{-2}=0.1353\), \(f(1)=2e^{-2}=0.2707\), and \(1-f_0=0.8647\). Set the hurdle crossing probability to \(q=0.60\). The matching zero-inflation probability is \(\pi=1-q/(1-f_0)=1-0.60/0.8647=0.3061\), which is a valid probability.

The hurdle model gives \(P(Y=0)=1-0.60=0.4000\). The zero-inflated model gives \(P(Y=0)=0.3061+(1-0.3061)(0.1353)=0.4000\), to rounding. For a count of one, the hurdle probability is \(0.60(0.2707)/(0.8647)=0.1878\); the zero-inflated probability is \((1-0.3061)(0.2707)=0.1878\). The common conditional positive probability at one is \(0.2707/0.8647=0.3130\). The matching relation makes the zero masses and every positive mass agree, not just the displayed probability at one.

Worked Example: A Hurdle Model With Fewer Zeros Than the Baseline

Use a Poisson baseline with \(\lambda=1\), so \(f_0=e^{-1}=0.3679\), \(f(1)=e^{-1}=0.3679\), and \(1-f_0=0.6321\). Consider a hurdle model with \(q=0.80\). It has zero probability \(1-q=0.2000\), less than the baseline zero probability.

Its probability of a count of one is \(0.80(0.3679)/(0.6321)=0.4657\). No zero-inflated model with this same baseline can have zero probability below 0.3679: for any \(0\leq\pi\leq1\), \(\pi+(1-\pi)f_0=f_0+\pi(1-f_0)\geq f_0\). The observed probability of zero in this example therefore cannot be matched by adjusting \(\pi\) while keeping the baseline fixed.

At the boundary \(q=0\), the hurdle model instead puts probability one at zero, just as a zero-inflated model with \(\pi=1\) does. In both boundary cases the probability of a positive count is zero. The conditional probability of a count of one given \(Y>0\), for example, has no defined value because its conditioning event has probability zero.

How to Compare the Models and Check Their Limits

The following sequence keeps the mechanisms and calculations separate. It first fixes the baseline distribution, then compares the zero mass, and only then compares positive counts.

Specify the baseline.
Write \(f(y)\) and \(f_0\), and verify \(f_0<1\). The positive-count truncation is not defined if the baseline itself cannot produce a positive count.

Identify the zero mechanism.
For a zero-inflated model, calculate \(\pi+(1-\pi)f_0\). For a hurdle model, calculate \(1-q\). Do not treat \(\pi\) and \(q\) as the same kind of probability.

Check whether positive conditioning is valid.
Use the conditional positive distribution only when \(\pi<1\) for the zero-inflated model or \(q>0\) for the hurdle model.

Compare the full distributions if needed.
Check \(q=(1-\pi)(1-f_0)\). A match gives equal zero and positive probabilities; failure of this relation means the full distributions differ.

This comparison assumes the same baseline distribution and the same conditioning level in both models. In a GLMM, formulas may be conditional on group-specific random effects. If those effects are integrated out, the marginal count distribution is a mixture; the simple formulas still apply only when \(f\), \(f_0\), and the component probabilities are defined for that marginal comparison. A conditional equivalence at fixed random effects should not automatically be described as an equivalence after marginalization if the mixing differs between models.

There are also limits to what the observed count distribution can reveal. When the parameter relationship holds, the two constructions produce exactly the same probabilities for every observable count. The counts alone then do not distinguish whether zeros arose from a separate latent component or from a hurdle process. Conversely, the fact that a hurdle model can represent fewer zeros than its baseline does not establish that this mechanism generated any particular data set; it is a statement about the model's probability range.

Common Mistakes and Practical Checks

Calling every zero in a zero-inflated model structural. The baseline component also generates zeros, with probability \((1-\pi)f_0\). The total zero probability is \(\pi+(1-\pi)f_0\).
Confusing hurdle crossing probability with zero probability. Here \(q=P(Y>0)\), so the hurdle model's zero probability is \(1-q\).
Assuming equal positive-count distributions imply equal full distributions. They imply agreement after conditioning on \(Y>0\), not equal probabilities of being positive. Check \(q=(1-\pi)(1-f_0)\) for full-distribution equality.
Using the conditional formula at an all-zero boundary. At \(\pi=1\) or \(q=0\), \(P(Y>0)=0\); a conditional positive distribution is undefined.
Ignoring the baseline's zero mass. With a fixed baseline, a zero-inflated model cannot have fewer zeros than \(f_0\), while a hurdle model can if \(q>1-f_0\).

Key takeaway: Hurdle and zero-inflated models share the same truncated baseline distribution among positive counts when positive outcomes can occur. Their zero probabilities and probabilities of positive counts differ in general; the full distributions coincide only under the parameter relation \(q=(1-\pi)(1-f_0)\).

Check Your Understanding

Answer each question using the definitions and formulas above. Solutions follow the questions.

A baseline count distribution has \(f_0=0.20\). A zero-inflated model has \(\pi=0.25\). What are its zero probability and positive-outcome probability?
For that same baseline, what value of \(q\) makes a hurdle model have the same full distribution?
Why can a zero-inflated model with \(f_0=0.30\) not have zero probability 0.20?
At \(\pi=1\), is \(P(Y=1\mid Y>0)\) equal to \(f(1)/(1-f_0)\)? Explain.

Solutions. (1) The zero probability is \(0.25+0.75(0.20)=0.4000\); the positive-outcome probability is \(1-0.4000=0.6000\), equivalently \(0.75(1-0.20)=0.6000\). (2) The matching hurdle probability is \(q=(1-\pi)(1-f_0)=0.75(0.80)=0.6000\). (3) Its zero probability is \(f_0+\pi(1-f_0)=0.30+0.70\pi\), which is at least 0.30 for \(\pi\geq0\), so it cannot be 0.20. (4) No. When \(\pi=1\), \(P(Y>0)=0\), so conditioning on \(Y>0\) is undefined; cancellation in the conditional-distribution formula is not valid.

Hurdle Models Versus Zero-Inflated Models

What You'll Learn