Deriving A Zero-Inflated Count Model

Why a Mixture Model?

A count of zero can arise in more than one way. The ordinary count process may produce a zero, or a separate process may prevent any count from being observed. A zero-inflated model represents these possibilities with a mixture: one component always produces zero, while the other generates counts from a familiar distribution such as the Poisson or negative binomial.

The model is a probability construction, not proof that a particular observed zero has a known biological cause. In single-cell data, a zero may be compatible with multiple mechanisms. The mixture assigns probabilities under its assumptions; it does not directly reveal which mechanism occurred for a given cell and gene.

Definition: Let \(Y\) be a nonnegative count. Let \(S\) be a latent indicator with \(S=1\) denoting the structural-zero component and \(S=0\) denoting the ordinary count component. Set \(P(S=1)=\pi\), where \(0\leq\pi\leq1\). If \(S=1\), set \(Y=0\); if \(S=0\), draw \(Y\) from an ordinary count distribution with probability mass function \(f(y)\). Thus, for every nonnegative integer \(y\), \(P(Y=y)=\pi I(y=0)+(1-\pi)f(y)\), where \(I(y=0)\) equals 1 when \(y=0\) and 0 otherwise.

The structural component contributes probability only at zero. Consequently, the mixture probability at zero is \(\pi+(1-\pi)f(0)\), while the probability at any positive count is \((1-\pi)f(y)\). The ordinary component can itself generate zeros, so the zero probability combines both sources.

P(Y=0)=\pi+(1-\pi)f(0), \qquad P(Y=y)=(1-\pi)f(y),\quad y=1,2,\ldots

These probabilities form a valid distribution: their sum is \(\pi+(1-\pi)\sum_{y=0}^{\infty}f(y)=\pi+(1-\pi)=1\), because \(f\) is a probability mass function. This check also shows why the ordinary component must be a properly normalized count distribution.

Deriving the Zero-Inflated Poisson Model

For the zero-inflated Poisson model, suppose that the ordinary component follows a Poisson distribution with mean \(\mu>0\). Its probability mass function is \(f(y)=e^{-\mu}\mu^y/y!\) for nonnegative integer \(y\). Substitution into the mixture formula gives the zero probability and all positive-count probabilities.

P(Y=y)= \begin{cases} \pi+(1-\pi)e^{-\mu}, & y=0,\\[4pt] (1-\pi)e^{-\mu}\dfrac{\mu^y}{y!}, & y=1,2,\ldots. \end{cases}

The parameters have distinct roles. The parameter \(\pi\) is the probability of entering the structural-zero component; \(\mu\) is the mean of the Poisson distribution conditional on entering the ordinary component. In general, \(\mu\) is not the marginal mean of \(Y\). Since \(Y=0\) whenever \(S=1\), and the conditional mean when \(S=0\) is \(\mu\), the law of total expectation gives \(E(Y)=(1-\pi)\mu\).

Worked Example: Zero-Inflated Poisson Probabilities and Likelihood

Suppose a hypothetical gene-count model has \(\pi=0.25\) and ordinary-component mean \(\mu=1.2\). Calculate the probabilities of counts 0, 1, and 2, then calculate the likelihood of the hypothetical observations \(0,0,1,2\), treating the observations as independent with these same parameters.

For a zero, combine the structural-zero and ordinary Poisson contributions: \(P(Y=0)=0.25+0.75e^{-1.2}\). Using \(e^{-1.2}\approx0.301194\), this is \(0.25+0.75(0.301194)=0.475896\).

For a count of 1, only the ordinary component contributes: \(P(Y=1)=0.75e^{-1.2}(1.2)\approx0.271075\). For a count of 2: \(P(Y=2)=0.75e^{-1.2}(1.2^2/2)\approx0.162645\).

The likelihood is the product of the probabilities for the four independent observations: \(L=(0.475896)^2(0.271075)(0.162645)\approx0.009985\). This is the likelihood of the observed data at the specified parameter values, not the probability that these parameter values are correct. The multiplication uses the unrounded probabilities to avoid unnecessary rounding error.

Likelihood for Independent Counts

For observations \(y_1,\ldots,y_n\) that are conditionally independent under common parameters, the likelihood is the product of their mixture probabilities. Each zero contributes the combined mixture probability; each positive count contributes the ordinary count probability multiplied by \(1-\pi\). Let \(n_0\) be the number of observed zeros and \(n_+=n-n_0\) the number of positive counts.

L(\pi,\mu\mid y_1,\ldots,y_n) = \prod_{i:y_i=0}\left[\pi+(1-\pi)e^{-\mu}\right] \prod_{i:y_i>0}\left[(1-\pi)e^{-\mu}\frac{\mu^{y_i}}{y_i!}\right].

Taking logarithms changes the products into sums. For the zero-inflated Poisson model with common parameters, the log likelihood is

\ell(\pi,\mu) = n_0\log\!\left[\pi+(1-\pi)e^{-\mu}\right] +n_+\log(1-\pi)-n_+\mu +\left(\sum_{i:y_i>0}y_i\right)\log\mu -\sum_{i:y_i>0}\log(y_i!).

The factors involving \(n_+\) arise because only positive observations contribute the ordinary Poisson mass directly; a zero's probability is a sum of two possible component probabilities. The expression assumes independent observations and common \(\pi\) and \(\mu\). If cell-specific parameters are specified, the same construction applies observation by observation, with the corresponding \(\pi_i\) and ordinary-distribution parameters in each factor.

Worked Example: Constructing a Zero-Inflated Poisson Likelihood

Use the hypothetical observations \(0,0,1,2\) and parameters \(\pi=0.25,\mu=1.2\). There are \(n_0=2\) zeros and two positive counts. The zero contribution is \(0.475896^2\), and the positive-count contributions are \(0.271075\) and \(0.162645\), so \(L=(0.475896)^2(0.271075)(0.162645)\approx0.009985\).

Equivalently, substituting into the product form gives \(L=[0.25+0.75e^{-1.2}]^2[0.75e^{-1.2}(1.2)][0.75e^{-1.2}(1.2^2/2)]\). The two expressions agree because the first bracket is the mixture probability for each zero and each remaining bracket is the mixture probability for its positive count. This product is a likelihood value; it is not a model-selection criterion by itself.

Extending the Mixture to a Negative Binomial Count

Overdispersed counts can use a negative binomial ordinary component instead of a Poisson component. Use the mean-dispersion parameterization in which the ordinary count has mean \(\mu>0\), dispersion \(\phi>0\), and variance \(\mu+\phi\mu^2\). Its probability mass function can be written using the gamma function as follows. When \(\phi\) tends to zero, this distribution approaches the Poisson distribution with mean \(\mu\).

f_{\mathrm{NB}}(y\mid\mu,\phi) = \frac{\Gamma(y+1/\phi)}{\Gamma(1/\phi)y!} \left(\frac{1}{1+\phi\mu}\right)^{1/\phi} \left(\frac{\phi\mu}{1+\phi\mu}\right)^y, \quad y=0,1,2,\ldots.

At \(y=0\), the last factor equals 1, so the ordinary negative binomial zero probability is \(f_{\mathrm{NB}}(0)=(1+\phi\mu)^{-1/\phi}\). Applying the same latent mixture yields the zero-inflated negative binomial (ZINB) probabilities:

P(Y=y)= \begin{cases} \pi+(1-\pi)(1+\phi\mu)^{-1/\phi}, & y=0,\\[4pt] (1-\pi)f_{\mathrm{NB}}(y\mid\mu,\phi), & y=1,2,\ldots. \end{cases}

For independent observations with common parameters, the ZINB likelihood is

L(\pi,\mu,\phi\mid y_1,\ldots,y_n) = \prod_{i:y_i=0}\left[\pi+(1-\pi)(1+\phi\mu)^{-1/\phi}\right] \prod_{i:y_i>0}(1-\pi)f_{\mathrm{NB}}(y_i\mid\mu,\phi).

Its log likelihood is the sum of the logarithm of the zero-mixture probability for every zero and \(\log(1-\pi)+\log f_{\mathrm{NB}}(y_i\mid\mu,\phi)\) for every positive count. This is the same mixture logic as for the zero-inflated Poisson; only the ordinary component changes. The marginal mean is again \((1-\pi)\mu\), while \(\mu+\phi\mu^2\) describes the variance within the ordinary component, not the full mixture variance.

Worked Example: A Zero-Inflated Negative Binomial Likelihood

Consider hypothetical independent counts \(0,1,2,0\), with \(\pi=0.20,\mu=2,\phi=0.50\). Here \(1/\phi=2\), and the ordinary negative binomial probabilities have \(f_{\mathrm{NB}}(0)=0.25\), \(f_{\mathrm{NB}}(1)=0.25\), and \(f_{\mathrm{NB}}(2)=0.1875\). For example, the formula gives \(f_{\mathrm{NB}}(1)=2(0.5)^2(0.5)=0.25\), and \(f_{\mathrm{NB}}(2)=3(0.5)^2(0.5)^2=0.1875\).

Therefore \(P(Y=0)=0.20+0.80(0.25)=0.40\), \(P(Y=1)=0.80(0.25)=0.20\), and \(P(Y=2)=0.80(0.1875)=0.15\). The likelihood for \(0,1,2,0\) is \(L=(0.40)(0.20)(0.15)(0.40)=0.004800\). The two zeros each use the mixture probability \(0.40\); using only the ordinary negative binomial zero probability \(0.25\) would omit the structural-zero contribution.

What the Mixture Says About an Observed Zero

Although the component indicator \(S\) is unobserved, its conditional probability given a zero follows from Bayes' rule. The joint probability of a structural zero and an observed zero is \(\pi\), while the overall probability of a zero is \(\pi+(1-\pi)f(0)\). Thus

P(S=1\mid Y=0)=\frac{\pi}{\pi+(1-\pi)f(0)}.

For any positive count, \(P(S=1\mid Y=y)=0\) under this model, because the structural component cannot generate positive counts. This conditional probability for a zero is model-based: it depends on the assumed mixture and its parameters, and is not an observed label.

Worked Example: Conditional Structural-Zero Probability

For the zero-inflated Poisson parameters \(\pi=0.25,\mu=1.2\), the probability of a zero is \(0.25+0.75e^{-1.2}\approx0.475896\). Therefore, \(P(S=1\mid Y=0)=0.25/0.475896\approx0.5253\).

Under this specified model, about 52.53% is the conditional probability that a zero came from the structural component. The remaining probability, about 47.47%, is the conditional probability that it came from the ordinary Poisson component. These are probabilities conditional on observing zero and on the model parameters; they do not identify the mechanism for an individual observed cell.

Assumptions, Comparisons, and Failure Modes

The likelihood derivations require that the ordinary component be a specified, normalized count distribution; the structural component produce only zero; and observations be independent conditional on the parameters used in their probability factors. The common-parameter formulas further assume the same \(\pi\) and ordinary-distribution parameters for all observations. More elaborate models can let parameters vary with measured covariates, but their likelihood still uses the same zero-versus-positive mixture construction.

The zero-inflated Poisson and zero-inflated negative binomial differ in the distribution assigned to the ordinary component. The former uses a Poisson mean-variance relationship; the latter allows overdispersion through \(\phi\). Setting \(\pi=0\) recovers the corresponding ordinary count model. A zero-inflated model should not be used automatically just because the data contain zeros: the mixture adds a parameter and a latent interpretation that may not be supported by the data or study design.

A practical warning is that mixture parameters can be difficult to distinguish. A high \(\pi\) can raise the zero probability, while changes in \(\mu\) or, in the negative binomial case, \(\phi\), also change the ordinary component's zero probability and positive-count distribution. Weak information in the observed counts can therefore make different parameter combinations fit similarly. Convergence problems, estimates near parameter boundaries, or large uncertainty are reasons to examine the fit carefully; they are not evidence that one latent mechanism has been established.

A related conceptual error is to call every zero assigned to the ordinary component a “sampling zero” with certainty, or every zero assigned to the structural component a known dropout event. The model distinguishes latent components mathematically, but the observed count alone does not reveal its component. A zero-inflated mixture is also not simply a negative binomial with a larger mean or dispersion: it changes the probability mass at zero by adding a separate point-mass component.

Key takeaway: To derive a zero-inflated count model, assign probability \(\pi\) to a component that forces \(Y=0\), and probability \(1-\pi\) to an ordinary count distribution. The zero probability is a sum of component probabilities; each positive-count probability is the ordinary probability multiplied by \(1-\pi\). The likelihood is the product of these probabilities across observations.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

For a zero-inflated Poisson model with \(\pi=0.10\) and \(\mu=2\), calculate \(P(Y=0)\) to three significant figures.
In a ZINB model, suppose the ordinary negative binomial probability of zero is 0.30 and \(\pi=0.20\). Calculate the mixture probability of zero.
Write the likelihood for the hypothetical independent observations \(0,2\) under a zero-inflated Poisson model with parameters \(\pi,\mu\).
Under a zero-inflated model, can a positive count have come from the structural-zero component? Explain.

Solutions. 1. \(P(Y=0)=0.10+0.90e^{-2}\approx0.10+0.1218=0.2218\), or \(0.222\) to three significant figures. 2. \(P(Y=0)=0.20+0.80(0.30)=0.440\). 3. The likelihood is \([\pi+(1-\pi)e^{-\mu}][(1-\pi)e^{-\mu}\mu^2/2!]\): the first factor is for the zero and the second for the positive count 2. 4. No. The structural component forces \(Y=0\), so a positive count can only come from the ordinary component under this model.