Likelihood Contributions Of Zero Counts

Why a Single Zero Has a Likelihood Contribution

A zero count is an observation, not by itself a diagnosis of why a transcript was not detected. Under a count model, it contributes a probability to the likelihood just as a positive count does. That contribution can favor some parameter values over others: for example, a zero may be more likely under a smaller expected count, or under a more dispersed negative binomial distribution.

Here the focus is the contribution made by an observed count to a negative binomial likelihood, and how to interpret changes in that contribution across parameter values. A likelihood compares how well specified parameter values account for the observed data. It does not, by itself, give the probability that a parameter value is true.

Definition: Let \(Y\) be a count with a negative binomial distribution parameterized by its mean \(\mu>0\) and dispersion \(\phi>0\), with \(\operatorname{Var}(Y)=\mu+\phi\mu^2\). For an observed count \(y\), the likelihood contribution is the probability \(P(Y=y\mid\mu,\phi)\), viewed as a function of \(\mu\) and \(\phi\) while \(y\) is held fixed. Counts are dimensionless; \(\mu\) is an expected count and \(\phi\) is a dispersion parameter.

In this parameterization, the negative binomial distribution has size \(1/\phi\) and success probability \(1/(1+\phi\mu)\). Its probability mass function is

P(Y=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, \qquad y=0,1,2,\ldots.

The same formula can be motivated as a Poisson count whose rate varies across observations according to a gamma distribution. Integrating over that rate produces the negative binomial probability above. This representation is one reason the model accommodates greater variability than a Poisson model with a fixed rate.

Zero and Positive Count Contributions

For a zero observation, the factor raised to \(y\) equals one, and the gamma-function ratio cancels. The likelihood contribution simplifies to

L_0(\mu,\phi) = P(Y=0\mid\mu,\phi) = (1+\phi\mu)^{-1/\phi}.

For a positive count, all factors in the probability mass function remain. For a set of conditionally independent counts \(y_1,\ldots,y_n\), with parameters \(\mu_i\) and \(\phi\), the joint likelihood is the product of their individual contributions. Equivalently, its log is the sum of the log contributions. Independence here is a modeling assumption conditional on the parameters and any included covariates; it is not guaranteed merely because observations come from different cells.

L(\mu_1,\ldots,\mu_n,\phi) = \prod_{i=1}^{n} P(Y_i=y_i\mid\mu_i,\phi), \qquad \ell = \sum_{i=1}^{n}\log P(Y_i=y_i\mid\mu_i,\phi).

In particular, each zero contributes \(-\log(1+\phi\mu_i)/\phi\) to the log likelihood. This contribution is not the log likelihood for an entire dataset unless the dataset contains only that one observation.

Worked Example: Calculating Zero and Positive Contributions

Suppose a hypothetical cell has a count modeled with \(\mu=2\) expected counts and \(\phi=0.5\). Calculate the contributions for an observed zero and an observed count of three.

Here \(1/\phi=2\), and \(1+\phi\mu=1+(0.5)(2)=2\). For zero:

P(Y=0\mid 2,0.5)=2^{-2}=\frac{1}{4}=0.25.

For \(y=3\), the gamma-function ratio is \(\Gamma(5)/(\Gamma(2)3!)=24/(1\cdot6)=4\), and the two probability factors are \((1/2)^2\) and \((1/2)^3\). Therefore:

P(Y=3\mid 2,0.5) = 4\left(\frac12\right)^2\left(\frac12\right)^3 = \frac18=0.125.

Under these parameters, the probability of the zero observation is 0.25, while the probability of exactly three counts is 0.125. These are probabilities for specified outcomes under the model; neither number alone indicates whether the model fits a larger dataset well.

How a Zero Changes Relative Support

A likelihood ratio compares the contributions under two candidate parameter values for the same observed data. If a zero is observed, the ratio for candidate means \(\mu_A\) and \(\mu_B\), at a fixed dispersion, is

\frac{L_0(\mu_A,\phi)}{L_0(\mu_B,\phi)} = \left( \frac{1+\phi\mu_A}{1+\phi\mu_B} \right)^{-1/\phi}.

A ratio greater than one means that the observed zero is more likely under candidate A than candidate B. It is a comparison of likelihoods, not the probability of either candidate being correct. To turn likelihoods into posterior probabilities requires a prior distribution and a Bayesian calculation.

Worked Example: Comparing Two Candidate Means After a Zero

A hypothetical zero count is evaluated under \(\phi=0.5\) and two candidate means, \(\mu_A=1\) and \(\mu_B=4\). The zero probabilities are:

L_0(1,0.5)=(1+0.5\cdot1)^{-2}=1.5^{-2}=\frac49\approx0.4444, $$ $$ L_0(4,0.5)=(1+0.5\cdot4)^{-2}=3^{-2}=\frac19\approx0.1111.

The likelihood ratio for the lower mean relative to the higher mean is \((4/9)/(1/9)=4\). Thus, this zero is four times as likely under \(\mu=1\) as under \(\mu=4\), at the specified dispersion. It favors the lower mean in this pairwise comparison, but does not establish that the true mean is 1.

A zero also changes relative support for dispersion values. Holding \(\mu\) fixed, its log contribution is \(\ell_0= -\log(1+\phi\mu)/\phi\). To check how this changes with \(\phi\), differentiate:

\frac{\partial \ell_0}{\partial\phi} = \frac{\log(1+\phi\mu)-\phi\mu/(1+\phi\mu)}{\phi^2}.

For \(x>0\), \(\log(1+x)>x/(1+x)\). One justification is that the difference is zero at \(x=0\) and has derivative \(x/(1+x)^2>0\) for \(x>0\). Taking \(x=\phi\mu\) shows that the derivative above is positive. Therefore, at fixed positive \(\mu\), the zero probability increases with \(\phi\). This does not mean a zero always provides strong evidence for high dispersion: its influence depends on the other observations and on the range of candidate parameter values.

Worked Example: Comparing Two Dispersions After a Zero

For a hypothetical zero count with fixed mean \(\mu=2\), compare \(\phi_A=0.5\) with \(\phi_B=1\):

L_0(2,0.5)=(1+0.5\cdot2)^{-2}=2^{-2}=0.25, $$ $$ L_0(2,1)=(1+1\cdot2)^{-1}=3^{-1}\approx0.3333.

The likelihood ratio for \(\phi_B\) relative to \(\phi_A\) is \(0.3333/0.25\approx1.333\). This zero is about 1.333 times as likely under \(\phi=1\) as under \(\phi=0.5\), when the mean is fixed at 2. It does not by itself identify the dispersion, because positive counts and other zeros also contribute to the full likelihood.

Positive Counts and the Shape of the Evidence

Positive observations can pull relative support in a different direction from zeros. In this parameterization, the derivative of the log likelihood contribution for a count \(y\) with respect to the mean is

\frac{\partial}{\partial\mu} \log P(Y=y\mid\mu,\phi) = \frac{y-\mu}{\mu(1+\phi\mu)}.

The denominator is positive. At a fixed parameter value, a count above \(\mu\) gives a positive slope in the log likelihood as \(\mu\) increases; a count below \(\mu\) gives a negative slope. For \(y=0\), the derivative is \(-1/(1+\phi\mu)\), so a zero locally favors decreasing the mean. A positive count does not necessarily favor increasing the mean: its direction depends on whether it lies above or below the candidate mean.

Worked Example: Combining a Zero With Positive Counts

Consider hypothetical independent counts \(0,1,3\), each modeled with the same \(\phi=0.5\). Compare candidate means \(\mu=2\) and \(\mu=1\), without fitting any additional parameters.

At \(\mu=2\), the individual contributions are \(1/4\), \(1/4\), and \(1/8\), so the joint likelihood is

L(2,0.5)=\frac14\cdot\frac14\cdot\frac18=\frac{1}{128}=0.0078125.

At \(\mu=1\), the negative binomial has success probability \(2/3\). The contributions are \(4/9\) for zero, \(8/27\) for one, and \(16/243\) for three. Thus:

L(1,0.5)=\frac49\cdot\frac{8}{27}\cdot\frac{16}{243} =\frac{512}{59049}\approx0.008670.

The likelihood ratio \(L(2,0.5)/L(1,0.5)=59049/65536\approx0.901\) is less than one, so these three observations jointly favor \(\mu=1\) over \(\mu=2\), slightly. The zero alone favors \(\mu=1\): its contribution ratio is \((1/4)/(4/9)=9/16=0.5625\). The count of three favors \(\mu=2\), with contribution ratio \((1/8)/(16/243)=243/128\approx1.898\). The joint comparison makes clear why a zero should be interpreted alongside the full observed count pattern.

Assumptions, Diagnostics, and Common Pitfalls

The calculations above are valid for the stated negative binomial model and parameterization. For multiple observations, multiplying their contributions requires conditional independence given the modeled means and dispersion. If cells share donor-level or other unmodeled effects, treating all cell counts as independent can overstate how much information the data contain. The mean may also depend on modeled covariates or exposure; the \(\mu\) in a contribution must be the mean appropriate to that observation, not automatically a common mean for every cell.

Inspect whether the fitted means and dispersion give plausible probabilities across the observed count range, not only for zeros. A model can assign a reasonable probability to zero counts yet poorly represent the positive tail, or vice versa. Systematic differences between observed and model-implied counts across cells or count levels are a warning that the assumed mean structure, dispersion, or dependence structure may be inadequate. A likelihood contribution is conditional on the chosen model; it cannot diagnose every source of non-detection.

The Poisson model is a limiting alternative as \(\phi\) approaches zero, with contribution \(e^{-\mu}\mu^y/y!\). A zero-inflated model instead assigns additional probability through a mixture component for extra zeros. Those alternatives change the likelihood being evaluated; a zero's negative binomial contribution should not be mistaken for an estimate of the probability that a cell is biologically off. Which model best describes a dataset requires comparing the models against the complete data, not choosing one from a single zero.

Mistake: “The zero probability is 0.25, so there is a 25% probability that \(\mu=2\) is correct.” The value 0.25 is the probability of observing zero if the specified model parameters hold. It is not a probability assigned to those parameters.
Mistake: “A zero proves high dispersion or a structural zero.” The negative binomial zero probability can rise with dispersion, but one observation cannot distinguish that explanation from a lower mean or other mechanisms. Use the full count pattern and an appropriate model.
Mistake: “Multiply the probabilities across cells because each cell is a separate row.” A product likelihood assumes conditional independence. Shared sample effects or other dependence may require a model that represents that structure.

Key takeaway: A zero contributes \((1+\phi\mu)^{-1/\phi}\) to the negative binomial likelihood. Likelihood ratios show how that observation changes relative support for specified parameter values; they are not posterior probabilities, and they do not explain the biological cause of the zero.

Check Your Understanding

Answer each question, then compare with the solutions below.

For \(\mu=3\) and \(\phi=0.5\), calculate the negative binomial contribution of an observed zero.
At \(\phi=0.5\), a zero is observed. Compare its likelihood under \(\mu=1\) and \(\mu=2\), and state which candidate it favors.
Why is a zero-count likelihood of 0.2 not a 20% probability that the model parameters are true?
State the assumption needed to multiply individual count contributions into a joint likelihood.

Solutions. (1) \(P(Y=0\mid3,0.5)=(1+0.5\cdot3)^{-2}=2.5^{-2}=0.16\). (2) The probabilities are \(4/9\approx0.4444\) for \(\mu=1\) and \(1/4=0.25\) for \(\mu=2\); the ratio is \((4/9)/(1/4)=16/9\approx1.778\), so the zero favors \(\mu=1\). (3) The value is the probability of that observation conditional on the specified parameters, not a probability distribution over parameter values. (4) The observations must be conditionally independent given the model parameters and included covariates.