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Imputation Models by Variable Type · Tutorial 180 of 497

Interpreting Mean And Variance In Count Models

Learn how Poisson and negative-binomial dispersion assumptions shape the variability and tail behavior of imputed counts.

Intermediate 9 min read

What You'll Learn

  • State the conditional mean-variance assumptions of Poisson and NB2 count models
  • Derive the negative-binomial variance using a Poisson-Gamma representation
  • Calculate and compare count probabilities under different dispersion assumptions
  • Explain why a larger variance does not mean every imputed count will be larger
  • Diagnose overdispersion and distinguish it from a misspecified conditional mean
  • Recognize the limitations of choosing a count model from dispersion alone

Why the Mean–Variance Relationship Matters

A count model does more than predict an average. In multiple imputation, it also supplies a probability distribution from which a missing count can be drawn. Two models can give the same predicted count for a person and still produce very different imputations because they assign different amounts of probability to counts far from that prediction. The key distinction here is the model’s conditional mean–variance relationship: how the variance of the count is assumed to change when the predictors are held fixed.

The Poisson regression material establishes the log-linear conditional mean used for counts, and the earlier material on negative-binomial models introduces a model that allows overdispersion. This tutorial focuses on interpreting and comparing their variance assumptions and on the consequences for count draws. It does not treat an observed count’s marginal variance as if it were automatically the model’s conditional variance.

Definition: Let \(Y_i\) be a nonnegative integer count for individual \(i\), and let \(\boldsymbol{x}_i\) denote that individual’s observed predictors. The conditional mean is \(\mu_i=E(Y_i\mid\boldsymbol{x}_i)\), and the conditional variance is \(\operatorname{Var}(Y_i\mid\boldsymbol{x}_i)\). A Poisson regression assumes \(\operatorname{Var}(Y_i\mid\boldsymbol{x}_i)=\mu_i\). The NB2 negative-binomial regression assumes \(\operatorname{Var}(Y_i\mid\boldsymbol{x}_i)=\mu_i+\alpha\mu_i^2\), where \(\alpha>0\) is a dispersion parameter.

In both models, a common mean specification is \(\log(\mu_i)=\boldsymbol{x}_i^\top\boldsymbol{\beta}\), so \(\mu_i=\exp(\boldsymbol{x}_i^\top\boldsymbol{\beta})\). The Poisson assumption is often called equidispersion: at fixed predictors, the variance equals the mean. NB2 allows the variance to exceed the mean, with the excess increasing quadratically as the mean grows. If the count is observed over an exposure period \(t_i\), an offset \(\log(t_i)\) can be included in the log mean; that changes the mean specification but not the distinction between these variance functions.

These are conditional statements. Counts can have different means because individuals have different predictors. Consequently, the variance across all individuals may exceed the overall mean even when each person’s count follows a Poisson distribution conditional on their predictors. Evidence for overdispersion should therefore be assessed relative to the fitted conditional model, not inferred simply by comparing the raw sample variance and mean.

Where the Negative-Binomial Variance Comes From

One useful justification for the NB2 variance is a Poisson model with a random individual-specific rate. Conditional on a latent rate \(\Lambda_i\), suppose \(Y_i\mid\Lambda_i,\boldsymbol{x}_i\sim\operatorname{Poisson}(\Lambda_i)\). Let \(\Lambda_i\mid\boldsymbol{x}_i\) have a Gamma distribution with mean \(\mu_i\) and variance \(\alpha\mu_i^2\). This Gamma variation represents extra rate heterogeneity beyond the observed predictors.

The law of total expectation gives \(E(Y_i\mid\boldsymbol{x}_i)=E(\Lambda_i\mid\boldsymbol{x}_i)=\mu_i\). For the variance, apply the law of total variance: the expected conditional Poisson variance contributes \(\mu_i\), and variation in the latent rate contributes \(\alpha\mu_i^2\).

$$ \begin{aligned} \operatorname{Var}(Y_i\mid\boldsymbol{x}_i) &=E\{\operatorname{Var}(Y_i\mid\Lambda_i,\boldsymbol{x}_i)\mid\boldsymbol{x}_i\} +\operatorname{Var}\{E(Y_i\mid\Lambda_i,\boldsymbol{x}_i)\mid\boldsymbol{x}_i\}\\ &=E(\Lambda_i\mid\boldsymbol{x}_i)+\operatorname{Var}(\Lambda_i\mid\boldsymbol{x}_i)\\ &=\mu_i+\alpha\mu_i^2. \end{aligned} $$

This gives a complete reason for the NB2 variance function under the stated mixture representation. The associated negative-binomial distribution can be parameterized by size \(r=1/\alpha\) and probability \(q=r/(r+\mu_i)\). Under this convention, its probability mass at \(y=0,1,2,\ldots\) is

$$ P(Y_i=y\mid\boldsymbol{x}_i) =\frac{\Gamma(y+r)}{\Gamma(r)y!} \left(\frac{r}{r+\mu_i}\right)^r \left(\frac{\mu_i}{r+\mu_i}\right)^y, \qquad \operatorname{Var}(Y_i\mid\boldsymbol{x}_i) =\mu_i+\frac{\mu_i^2}{r}. $$

Since \(r=1/\alpha\), small \(\alpha\) means less extra dispersion, while large \(\alpha\) means more. As \(\alpha\) approaches zero, the NB2 variance approaches \(\mu_i\), and the negative-binomial distribution approaches the Poisson distribution with that mean. Software may report a size, shape, or dispersion parameter using a different convention, so its definition must be checked before interpreting a numerical estimate.

Worked Example: Comparing Variances at the Same Mean

Suppose two hypothetical fitted models both predict a mean count of \(4\) events for an individual with a particular predictor pattern. One is Poisson; the other is NB2 with \(\alpha=0.25\). Compare their conditional standard deviations.

Under the Poisson model, the conditional variance equals the mean:

$$ \operatorname{Var}(Y\mid\boldsymbol{x})=4, \qquad SD(Y\mid\boldsymbol{x})=\sqrt{4}=2. $$

Under NB2, substitute \(\mu=4\) and \(\alpha=0.25\):

$$ \operatorname{Var}(Y\mid\boldsymbol{x}) =4+0.25(4^2)=4+4=8, \qquad SD(Y\mid\boldsymbol{x})=\sqrt{8}\approx2.828. $$

Both models have the same conditional expected count, \(4\) events. The NB2 model assigns a conditional standard deviation about \(2.828\) events rather than \(2\). This does not mean a negative-binomial draw must be larger than a Poisson draw. It means that the NB2 distribution is more variable across repeated draws under the same predictor pattern.

What Dispersion Changes in Imputed Counts

For a missing count, a fitted mean alone is not an imputation. A count imputation draws an integer from the model’s conditional distribution. Under a Poisson model, the draw has mean and variance \(\mu_i\). Under NB2, the mean remains \(\mu_i\), but the variance becomes \(\mu_i+\alpha\mu_i^2\). With greater dispersion, draws are less tightly concentrated around the mean; the distribution can assign more probability both to low counts and to unusually high counts.

The following calculation compares probabilities, not just variances. It holds the mean fixed at \(4\) and uses the NB2 value \(\alpha=0.25\), so \(r=4\). The NB probability of zero is \((r/(r+\mu))^r\), while the Poisson probability of zero is \(e^{-\mu}\).

$$ \begin{aligned} P_{\text{Poisson}}(Y=0)&=e^{-4}\approx0.01832,\\ P_{\text{NB2}}(Y=0)&=\left(\frac{4}{4+4}\right)^4 =\left(\frac12\right)^4=0.06250. \end{aligned} $$

For a count of \(8\), the Poisson probability is \(e^{-4}4^8/8!\approx0.02977\). Under NB2, the probability is \(\binom{11}{8}(1/2)^{12}=165/4096\approx0.04028\). In this example the negative-binomial model assigns more probability to both zero and eight, while preserving a mean of four. This illustrates how extra dispersion spreads probability away from the center; it does not establish that every possible tail probability must increase for every mean and dispersion value.

Worked Example: Calculating Negative-Binomial Draw Probabilities

A hypothetical count-imputation model predicts \(\mu=2.5\) events for a record with missing outcome data. Suppose the fitted NB2 dispersion is \(\alpha=0.5\). Find the probability of imputing zero events and the probability of imputing one event, treating the model parameters as fixed for this calculation.

The size is \(r=1/\alpha=2\), and the NB probability parameter is \(q=r/(r+\mu)=2/(2+2.5)=4/9\). Thus \(1-q=5/9\). The zero probability is:

$$ P(Y=0\mid\boldsymbol{x})=q^r =\left(\frac49\right)^2 =\frac{16}{81} \approx0.1975. $$

For one event, the negative-binomial mass formula gives:

$$ \begin{aligned} P(Y=1\mid\boldsymbol{x}) &=\frac{\Gamma(3)}{\Gamma(2)1!} \left(\frac49\right)^2\left(\frac59\right)\\ &=2\cdot\frac{16}{81}\cdot\frac59 =\frac{160}{729} \approx0.2195. \end{aligned} $$

These are probabilities of the two specific possible imputed values under the stated fitted distribution. They do not mean that the eventual imputation is a fractional count: the draw is one integer outcome. In an actual multiple-imputation procedure, uncertainty in estimated model parameters may also be represented by parameter draws before drawing the missing count. The calculation here isolates the effect of the count distribution conditional on a fixed mean and dispersion.

Assumptions, Diagnostics, and Comparisons

Interpreting either variance function depends on a sufficiently suitable conditional mean model. The usual regression formulation assumes that the counts are nonnegative integers, that the log mean is modeled appropriately using the predictors, and that records are conditionally independent under the fitted model unless a dependence structure is explicitly included. Poisson regression adds the conditional equality of mean and variance. NB2 instead assumes a particular quadratic increase in variance with the mean, often with one common \(\alpha\) across records. Neither assumption is guaranteed by the count scale alone.

Useful checks compare observed counts with distributions or residual patterns expected under the fitted model. For example, examine whether Pearson residual variability is systematically larger than expected, whether observed and fitted count frequencies differ substantially across ranges of fitted means, and whether high counts occur more often than the model predicts. Such checks can signal a mismatch, but no single diagnostic proves that NB2 is the correct explanation. An incorrect mean structure, omitted predictors, clustering, unequal exposure, or influential observations can also produce apparent overdispersion.

A common diagnostic summary is the Pearson dispersion ratio, the Pearson chi-squared statistic divided by its residual degrees of freedom. A ratio substantially above one can indicate more residual variation than the fitted Poisson model accounts for. It is a clue, not a formal verdict: its interpretation depends on the model, sample size, and data structure. Residual patterns and the substantive data-generating context matter as well.

ApproachConditional varianceImplication for count drawsMain caution
Poisson regression\(\mu_i\)Draws follow a Poisson distribution with fitted mean \(\mu_i\)Can understate variability when there is conditional overdispersion
NB2 regression\(\mu_i+\alpha\mu_i^2\)Draws allow extra variability that grows quadratically with the meanThe assumed variance function may not describe the data’s source of variation
Quasi-Poisson variance model\(\phi\mu_i\)Can represent variance larger than the mean through \(\phi\)A variance specification alone does not define a full count probability distribution for drawing imputations

The quasi-Poisson comparison highlights why estimating a larger standard error is not the same as specifying a predictive distribution for imputation. A count imputation requires a distribution that assigns probabilities to integer values. NB2 supplies one; a quasi-Poisson variance relationship by itself does not. Conversely, choosing NB2 only because a dispersion summary is high can be misleading if the problem is an inadequate mean model or dependence between records.

Common pitfall: “The negative-binomial variance is larger, so every missing count imputed from it will be higher.” This is wrong. Variance describes spread across repeated draws, not the direction of a particular draw. At the same fitted mean, an NB2 distribution can produce more low counts as well as more high counts. For an individual missing value, the realized imputation is random.

In practice, the dispersion assumption matters because it affects the distribution of completed data sets and therefore summaries calculated from them. If the Poisson assumption is too restrictive, imputations may be too concentrated and may underrepresent plausible variation in missing counts. If NB2 is used when its extra dispersion is unnecessary or poorly matched, it can assign an unsuitable distribution to the missing values. Model choice should be guided by conditional fit, predictor coverage, and the scientific structure of the count—not by the raw variance alone.

Dispersion is not a complete account of count behavior. A fitted mean and variance do not guarantee that the full distribution is well calibrated, and a single NB2 parameter may not capture every pattern in the data. When a distinct count-generating process is scientifically plausible, a more specialized model may be needed; the appropriate model depends on that process and is not determined solely by observing many zeros or a large sample variance.

1
Specify the conditional mean.
State how predictors determine \(\mu_i\), and include relevant exposure information where appropriate.
2
State the dispersion assumption.
For Poisson, use \(\operatorname{Var}(Y_i\mid\boldsymbol{x}_i)=\mu_i\); for NB2, use \(\mu_i+\alpha\mu_i^2\) and define the parameterization.
3
Check conditional fit.
Inspect residual variation and observed-versus-predicted count behavior, considering mean-model errors and dependence as possible explanations.
4
Draw from the chosen predictive distribution.
For each missing count, use the fitted conditional distribution; do not substitute the mean for a random count draw.

Check Your Understanding

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

  1. At a fixed predictor pattern, a Poisson model gives \(\mu=6\). What are its conditional variance and standard deviation, in count units?
  2. An NB2 model has \(\mu=3\) and \(\alpha=0.4\). Calculate its conditional variance.
  3. Why can an NB2 model with the same mean as a Poisson model produce more zero counts and more high counts?
  4. A data set’s raw variance is larger than its raw mean. Does that alone establish conditional overdispersion? Explain.
  5. Why is a quasi-Poisson variance relationship by itself insufficient to specify a count-imputation draw?

Solutions. (1) The variance is \(6\) counts squared, and the standard deviation is \(\sqrt{6}\approx2.449\) counts. (2) Substitute into the NB2 formula: \(3+0.4(3^2)=3+3.6=6.6\) counts squared. (3) NB2 spreads probability more broadly around the same mean; the increased variance can put more probability in lower and upper parts of the distribution, though it does not make every tail probability larger in every setting. (4) No. Different conditional means across individuals can make the marginal variance exceed the marginal mean even under conditional Poisson models; assess variability relative to the fitted conditional model. (5) A variance relationship specifies spread but does not, by itself, give probabilities for each possible integer count. Imputation requires a predictive distribution from which to draw an integer value.

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