Choosing a Model From What the Counts Show
For a gene measured across cells, a count model is a working description of how the observed counts vary, not a label for the gene’s biology. A Poisson model can be a useful baseline, but its mean-variance relationship may be too restrictive. A negative-binomial model allows overdispersion, while a zero-inflated model adds a process that can produce extra zeros. The practical task is to choose a model whose assumptions are defensible and whose fitted distribution reproduces the count features that matter.
A large zero fraction alone does not decide the question. A low-mean Poisson variable can produce many zeros, and a negative-binomial variable can produce more zeros than a Poisson with the same mean. Similarly, a zero-inflated model may match the zero fraction while failing to reproduce positive counts or the upper tail. The introductory material on comparing count-model fits and on predictive checks provides the background for the workflow here: compare fitted predictions with the observed data, not model names in isolation.
Models and Their Assumptions
Let \(X_i\) be the count for a gene in cell \(i\), and suppose for now that the cells being analyzed are sufficiently comparable to share the same conditional count distribution. The models below treat counts as conditionally independent given the model and any included covariates. Dependence or unmodeled differences among cell groups can make a single pooled distribution misleading.
These definitions do not establish that a zero corresponds to a biological gene-off state, nor that the extra-zero component represents a particular technical event. Such interpretations need separate justification. A zero-inflated model is a distributional option; its extra component is not automatically an observed or identified biological state.
A useful initial summary is the sample mean \(\bar{x}\), sample variance \(s^2\), fraction of zeros, and one or more upper-tail counts, such as the number of cells with \(X_i\geq5\). For a negative-binomial method-of-moments starting value, when \(s^2>\bar{x}\), one can use \(\hat{\alpha}=(s^2-\bar{x})/\bar{x}^2\). This is a rough dispersion estimate, not a substitute for fitting and checking the model. If \(s^2\leq\bar{x}\), that formula does not yield a positive overdispersion estimate; it is not evidence for fitting a negative-binomial model with negative dispersion.
Why the Mean and Variance Are Only a Starting Point
The Poisson assumption implies equality of mean and variance. That equality gives a quick diagnostic, but agreement does not prove the Poisson model is adequate, and disagreement does not reveal its cause. A negative-binomial distribution can represent overdispersion through variation in its rate; cell heterogeneity, capture variation, or other unmodeled structure can also produce apparent overdispersion. Count summaries are therefore screening tools. The fitted distribution should also be checked against features such as the zero fraction, moderate counts, and upper tail.
The reason zero counts are informative but not decisive follows directly from the model probabilities. Under a Poisson model, \(P(X=0)=e^{-\mu}\). Under a negative-binomial model in the mean-dispersion parameterization, \(P(X=0)=(1+\alpha\mu)^{-1/\alpha}\). A fitted model’s expected number of zeros among \(n\) comparable cells is \(nP(X=0)\); its expected number at or above a threshold \(c\) is \(nP(X\geq c)\). These are predictions to compare with data, not guarantees that the realized counts will equal their expectations.
Worked Example: Comparing Poisson and Negative-Binomial Predictions
Suppose a hypothetical gene has counts in \(n=1{,}000\) comparable cells. The observed mean is 1.34, the sample variance is 3.14, 410 cells have zero counts, and 64 have counts of at least 5. These are stipulated summaries of a hypothetical data set.
Check dispersion. The Poisson model would have variance 1.34, much smaller than the observed 3.14. A method-of-moments negative-binomial dispersion is
This is consistent with substantial overdispersion relative to Poisson and provides a reasonable starting point for fitting a negative-binomial model. It does not, by itself, establish that the negative-binomial distribution fits the whole count profile.
Poisson predictions. Set \(\mu=1.34\). The predicted number of zeros is
For counts of at least 5, sum the Poisson probabilities from 0 through 4 and take the complement. For this mean, \(P(X\leq4)\approx0.987985\), so
The observed 410 zeros and 64 counts at least 5 both exceed these Poisson predictions substantially. The Poisson model does not capture the observed dispersion or these two parts of the distribution.
Negative-binomial predictions. For a fitted negative-binomial candidate with mean near 1.34 and dispersion near 1, suppose numerical evaluation gives \(P(X\leq4)\approx0.93816\). Then
That upper-tail prediction is close to the observed 64, unlike the Poisson prediction of 12.01. This supports the negative-binomial model over the Poisson for this feature. The zero count must still be checked under the fitted negative-binomial model, as must other parts of the distribution. If those checks also look reasonable, the simpler negative-binomial model is a defensible choice; this comparison alone does not justify adding a zero-inflation component.
A Predictive Comparison in Practice
The main comparison is between the observed distribution and distributions generated by fitted candidate models. A transparent check is to calculate fitted expected counts in bins chosen before inspecting the results: zero, one, two, three to four, and five or more, for example. A model can match the mean while missing the zeros or tail, so a single summary is rarely enough. Posterior predictive checks and simulation-based checks, covered in the earlier material on those subjects, make this comparison more complete by showing how much variation in the summaries is expected under the fitted model.
Model complexity matters, but it is secondary to adequacy. A zero-inflated model has an additional parameter and may improve fit by accommodating zero counts. It is not a general remedy for every failure of a Poisson model: if the negative-binomial model already reproduces zeros and the positive-count distribution, an extra zero component may be unnecessary. Conversely, if a model matches zeros but systematically misses the positive tail, the problem is not solved simply because its zero prediction is good. Likelihood-based criteria can help compare fitted candidates, but should be interpreted alongside predictive checks and the purpose of the analysis.
Specify the gene and the cell population or conditions. Consider whether pooling unlike cell groups could create apparent dispersion or excess zeros.
Use Poisson as a baseline; compare it with negative binomial when variance exceeds the mean. Consider zero inflation only when a fitted baseline leaves a relevant zero discrepancy and the added component is scientifically interpretable.
Check expected counts for zeros, several count ranges, and the upper tail. Use simulation or predictive checks to judge whether the discrepancies are larger than ordinary sampling variation.
Prefer a simpler model when its assumptions are plausible and its predictive behavior is adequate for the scientific question. Record the unresolved limitations rather than treating selection as certainty.
Worked Decisions and Their Limits
Worked Example: A Sparse Gene That Does Not Need Zero Inflation
In a second hypothetical data set, \(n=500\) cells have mean count \(\bar{x}=0.40\), variance \(s^2=0.42\), and 335 zeros. The observed zero fraction is \(335/500=0.670\).
Under a Poisson model with \(\mu=0.40\), the zero probability is \(e^{-0.40}\approx0.6703\), so the predicted number of zeros is \(500(0.6703)=335.2\). The observed 335 is essentially that prediction.
The variance is also close to the Poisson mean. A negative-binomial moment estimate, if used as a diagnostic, is
For this dispersion, the negative-binomial zero probability is \((1+0.125(0.40))^{-1/0.125}=1.05^{-8}\approx0.6768\), or \(500(0.6768)=338.4\) predicted zeros. Both candidate predictions are close to the observed count. The zero fraction gives no reason to add a zero-inflation parameter. A full check of the positive counts is still needed before choosing Poisson, but this example shows why sparsity alone is not evidence of zero inflation.
Worked Example: A Zero-Inflated Fit That Still Misses the Tail
Consider a third hypothetical data set with \(n=500\), observed mean 0.60, 350 zeros, and 12 counts of at least 5. A fitted zero-inflated Poisson candidate has baseline rate \(\lambda=1.6\) and extra-zero probability \(\pi=0.625\). Its mean is \((1-\pi)\lambda=0.375(1.6)=0.60\), matching the observed mean.
Its zero probability is
It predicts \(500(0.700711)=350.4\) zeros, close to the observed 350. But for a positive threshold, only the Poisson component contributes. Since \(P(\operatorname{Poisson}(1.6)\geq5)\approx0.023681\), the fitted model predicts
counts of at least 5, compared with 12 observed. Thus this candidate reproduces the zeros and mean but underpredicts the upper tail. It is not adequate merely because its zero prediction is close. The negative-binomial alternative should be fitted and checked against the same summaries before choosing a model; the information given here does not establish which model best reproduces the full distribution.
Diagnostics, Alternatives, and Failure Modes
A useful diagnostic compares observed and fitted expected counts across several bins, not only at zero. Sparse data make individual bins noisy, so combine ranges when expected counts are small and use simulated datasets from the fitted model to assess the size of discrepancies. Repeated underprediction of high counts suggests that the fitted model does not capture the tail. A residual pattern across cell groups or conditions may instead indicate that one common distribution is pooling unlike populations.
| Candidate | What it allows | What to check | Limitation |
|---|---|---|---|
| Poisson | Mean equals variance under the model | Dispersion, zeros, and tail predictions | Can be too restrictive when counts are overdispersed |
| Negative binomial | Variance above the mean through \(\alpha\) | Zeros and the full positive-count distribution | Extra dispersion does not guarantee the correct distributional shape |
| Zero-inflated Poisson or negative binomial | An additional source of zeros plus a baseline count distribution | Zero prediction and predictions for positive counts and the tail | More parameters; the extra-zero component may be weakly identified or lack a justified interpretation |
The key failure modes are choosing by the observed zero percentage alone, treating mean-variance agreement as proof of Poisson fit, or adding zero inflation to fix a mismatch that occurs elsewhere in the distribution. Another limitation is that a fitted single-gene model may summarize a mixture of cell states rather than a single homogeneous process. A good predictive fit is evidence that a model reproduces selected features of these data; it does not prove the model’s biological mechanism or guarantee adequate predictions for every future analysis.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A fitted Poisson model predicts 270 zeros and 15 counts of at least 5. The data contain 275 zeros and 60 counts of at least 5. Which feature is the clearest evidence of poor fit, and what should be checked next?
- For \(\bar{x}=0.8\) and \(s^2=1.6\), calculate the negative-binomial method-of-moments dispersion estimate.
- A zero-inflated model matches the observed zero fraction but predicts far fewer large positive counts than observed. Is the model adequate on that evidence?
- Why is a high observed zero fraction not sufficient by itself to justify a zero-inflated model?
Solutions. (1) The upper tail is the clearest discrepancy: 60 observed versus 15 predicted. Check predictive distributions across other count ranges and compare a negative-binomial fit; do not infer from this discrepancy alone that zero inflation is needed. (2) \(\hat{\alpha}=(1.6-0.8)/0.8^2=0.8/0.64=1.25\). (3) No. The model matches zeros but misses a relevant part of the positive-count distribution, so compare alternatives and assess overall predictive fit. (4) Low-mean Poisson and negative-binomial distributions can both produce many zeros; judge zeros relative to predictions from fitted baseline models and check the rest of the distribution.