Why Mix Poisson Rates?
For a particular cell and gene, a Poisson model treats the expected count as fixed: conditional on rate \(\lambda\), the count varies through sampling alone. But cells may differ in their underlying rates, even when they are drawn from the same population or share measured covariates. If those rates vary across cells, a single fixed-rate Poisson distribution can be too restrictive. One way to represent that variation is to let the rate itself be random, then average the Poisson probabilities over its distribution.
This construction is called gamma-Poisson mixing. It produces a negative binomial count distribution. The derivation matters because it shows what the model means: the negative binomial is not just a different formula for counts, but a marginal distribution arising when Poisson rates vary according to a gamma distribution.
The parameter \(b\) is the gamma rate, not its scale: the gamma density has exponential factor \(e^{-b\lambda}\). This distinction is important when substituting values. The mixing assumption describes a population distribution of latent rates; it is not, by itself, a claim that a particular cell’s rate changes randomly from one molecule to the next.
Deriving the Marginal Probability
For a specified count \(y\), the law of total probability gives the marginal probability as an integral over all possible rates. Multiplying the conditional Poisson probability by the gamma density gives an integrand whose power of \(\lambda\) and exponential factor can be combined.
To evaluate the remaining integral, use the gamma-integral identity \(\int_0^\infty \lambda^{c-1}e^{-d\lambda}\,d\lambda=\Gamma(c)/d^c\), for \(c>0\) and \(d>0\). Here \(c=y+a\) and \(d=b+1\), both positive under the model assumptions. Substitution yields
This is a negative binomial probability mass function in a shape-and-probability form: its shape is \(a\), and the probability factor attached to the shape is \(b/(b+1)\). These labels are stated here only to identify the result of the derivation; the same distribution can be written using other parameter conventions. The integral itself establishes the probability law without requiring a choice among those conventions.
The resulting probabilities sum to one. In fact, each conditional Poisson distribution sums to one for every \(\lambda>0\), and the gamma density also integrates to one. Therefore, summing the mixture probabilities over the nonnegative integer counts gives \(\sum_y P(Y=y)=\int_0^\infty \sum_y P(Y=y\mid\Lambda=\lambda)f_\Lambda(\lambda)\,d\lambda=1\). The interchange is valid because all terms are nonnegative.
Worked Example: Evaluating a Mixed Count Probability
Suppose, hypothetically, that cell-specific rates follow a gamma distribution with shape \(a=2\) and rate \(b=1\), and counts are conditionally Poisson. Find the marginal probability of observing \(Y=1\).
The assumptions match the gamma-Poisson construction: \(a\) and \(b\) are positive, and the conditional count is Poisson given the rate. Substitute \(y=1\), \(a=2\), and \(b=1\) into the derived mass function:
Under this hypothetical mixture, the probability of a count of one is \(0.2500\). This is a marginal probability: it averages the conditional probability of one count over the full distribution of possible rates.
Interpreting the Mixture
The conditional model describes sampling variation at a given rate. The gamma distribution adds variation among rates. Consequently, the marginal count distribution reflects both sources of variation. Using the law of total variance, which separates conditional variation from variation in conditional means, gives \(\operatorname{Var}(Y)=E(\Lambda)+\operatorname{Var}(\Lambda)\). For the stated gamma shape-rate convention, \(E(\Lambda)=a/b\) and \(\operatorname{Var}(\Lambda)=a/b^2\). Thus, in this mixture, \(\operatorname{Var}(Y)=a/b+a/b^2\), whereas \(E(Y)=a/b\). The additional term comes from heterogeneity in the latent rates. This illustrates how mixing can produce greater count variability than a Poisson model with one fixed rate.
That interpretation depends on treating rates as varying across the units represented by the mixture. In single-cell work, the units might be cells for a given gene, after accounting for the chosen exposure or covariates. The model does not establish why their rates vary, nor does a good fit prove that the true biological mechanism is gamma-distributed.
Worked Example: Including an Exposure Factor
Suppose a hypothetical observation has exposure \(s=2\), and its rate per exposure unit is \(\Lambda\sim\operatorname{Gamma}(a=2,b=3)\). Conditional on \(\Lambda=\lambda\), let \(Y\sim\operatorname{Poisson}(s\lambda)\). Find the marginal probability that \(Y=2\).
The exposure multiplies the conditional rate. In the mixture integral, the Poisson factor is \(e^{-s\lambda}(s\lambda)^y/y!\), so the exponential factors combine as \(e^{-(b+s)\lambda}\). Applying the same gamma-integral identity gives
For the specified values, \(b+s=3+2=5\). Therefore,
The factor \(s\) changes the count’s effective exposure, not the gamma distribution assigned to the rate per exposure unit. When \(s=1\), the expression reduces to the original mass function.
Worked Example: Comparing a Mixed Rate With a Fixed Rate
Consider a hypothetical gamma-Poisson model with \(a=4\) and \(b=2\). The gamma-distributed rate has mean \(a/b=2\). Compare the probability of observing \(Y=3\) under this mixture with the probability under a fixed-rate Poisson model whose rate is also 2.
For the mixture, substitute \(y=3\), \(a=4\), and \(b=2\):
For the fixed-rate Poisson model, the rate is 2 for every unit, so
The two models have the same mean rate, but their probabilities for a particular count differ: here the mixed model assigns probability \(0.1463\) to a count of three, compared with \(0.1804\) under the fixed-rate model. This single comparison does not imply that the mixed model assigns a larger probability to every count. It shows why matching the mean alone does not make the models equivalent.
Assumptions, Diagnostics, and Limits
The gamma-Poisson derivation is exact under its assumptions. In particular, the conditional distribution must be Poisson with rate proportional to the modeled exposure, and the rates across the units being mixed must follow the specified gamma distribution. The rate is positive, and the shape and rate parameters satisfy \(a>0\) and \(b>0\). If counts are dependent across units in ways not represented by their rates, or if the rate distribution is poorly described by a gamma density, the marginal negative binomial model need not describe the data well.
A useful first diagnostic is to compare observed count variability with what a fixed-rate Poisson model would predict, as in the earlier material on diagnosing the Poisson mean-variance assumption. Evidence of excess variability motivates considering rate heterogeneity; it does not identify the gamma distribution as its cause. Model checks should also examine whether fitted probabilities describe the observed count distribution across its range, and whether fit varies systematically with cell-level features or exposure. A mismatch can indicate an inadequate rate distribution, omitted structure, dependence, or other features the simple mixture does not capture.
The main alternative in this comparison is a fixed-rate Poisson model. It is simpler and appropriate when a common conditional rate is a defensible approximation. Gamma-Poisson mixing is useful when variation in rates across units is part of the model, but it adds assumptions and parameters. Neither model automatically accounts for every source of variability. In particular, a negative binomial fit alone cannot determine whether the variation is biological, technical, or due to unmodeled covariates.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- For \(a=1\), \(b=2\), and no exposure multiplier beyond \(s=1\), calculate \(P(Y=0)\) under the gamma-Poisson mixture.
- Write the marginal probability formula when \(Y\mid\Lambda=\lambda\sim\operatorname{Poisson}(s\lambda)\) and \(\Lambda\sim\operatorname{Gamma}(a,b)\), using the shape-rate convention.
- In words, distinguish a conditional Poisson probability from a marginal mixture probability.
- Does evidence of excess count variability prove that the latent rates follow a gamma distribution? Explain.
Solutions. (1) Substitute into the derived formula: \(P(Y=0)=\frac{\Gamma(1)}{0!\Gamma(1)}(\frac{2}{3})^1(\frac{1}{3})^0=\frac{2}{3}=0.6667\). (2) The formula is \(P(Y=y)=\frac{\Gamma(y+a)}{y!\Gamma(a)}(\frac{b}{b+s})^a(\frac{s}{b+s})^y\), for \(y=0,1,2,\ldots\). (3) A conditional probability is calculated at one specified latent rate \(\lambda\); a marginal probability averages those conditional probabilities over the gamma distribution of rates. (4) No. Excess variability can motivate a model with rate heterogeneity, but does not establish that the rates have a gamma distribution; that assumption requires separate assessment against the data and scientific context.