Capture Efficiency As Binomial Thinning

Why Capture Efficiency Changes Counts

A molecule can be present in a cell yet fail to be captured and sequenced. Even if every molecule has the same chance of capture, the observed count will therefore vary around the number present. A useful model treats capture as a separate sampling step: first a latent molecule count is generated, then each molecule is independently retained with probability \(q\). Here \(q\), between 0 and 1, is the capture probability.

This is binomial thinning. It connects a model for the number of molecules available to a model for the count actually observed. The connection is important because thinning changes the mean and variance in specific ways; it does not, by itself, imply an additional class of structural zeros.

The model assumes a fixed \(q\) for the counts under consideration, independent capture outcomes across molecules, no captured molecules that were not in \(X\), and no molecule counted more than once. The latent count \(X\) may itself vary across cells according to a Poisson or negative binomial distribution. The capture step is then applied to that count.

General Consequences of Thinning

Given \(X=x\), a binomial count has conditional mean \(qx\) and conditional variance \(q(1-q)x\). Averaging over the latent count gives the unconditional mean and variance of \(Y\):

The mean is reduced in proportion to capture probability. The variance has two contributions: variation in which molecules are captured, represented by \(q(1-q)E[X]\), and retained variation in the latent count, represented by \(q^2\operatorname{Var}(X)\). These expressions follow from the laws of total expectation and total variance:

The same result can be expressed using a probability-generating function. For a nonnegative integer-valued random variable \(X\), define \(G_X(s)=E[s^X]\). Conditional on \(X\), the generating function of \(Y\) is \((1-q+qs)^X\). Thus,

Substituting a model's generating function into this identity shows whether its distributional form is preserved by thinning.

Thinning a Negative Binomial Count

Use the mean-dispersion parameterization of the negative binomial: \(X\) has mean \(\mu\), dispersion \(\phi>0\), and variance \(\mu+\phi\mu^2\). One construction is a Poisson count with a gamma-distributed rate: conditional on a rate \(\Lambda\), \(X\mid\Lambda\sim\operatorname{Poisson}(\Lambda)\), with \(\Lambda\sim\operatorname{Gamma}(1/\phi,\text{scale}=\mu\phi)\). The gamma distribution has mean \(\mu\phi/\phi=\mu\) and variance \(\mu^2\phi\), giving the stated negative binomial mean and variance.

Conditional on \(\Lambda\), independent thinning of a Poisson count with probability \(q\) gives a Poisson count with rate \(q\Lambda\). Since multiplying a gamma random variable by \(q\) multiplies its scale by \(q\) while leaving its shape unchanged, \(q\Lambda\) has gamma shape \(1/\phi\) and scale \(q\mu\phi\). Consequently, the thinned count is negative binomial with mean \(q\mu\) and the same dispersion \(\phi\). The generating-function calculation gives the same result directly.

The second expression is the negative binomial generating function with mean \(q\mu\) and dispersion \(\phi\). Thus its variance and zero probability are

The dispersion is unchanged under this model, but the mean-dependent variance changes. It would be incorrect to multiply the original variance by \(q\): the thinning variance formula includes both the added molecule-sampling variation and the scaled latent variation.

Conditional Non-Detection and a Second Example

For a fixed latent count \(X=x\), the probability of non-detection is the probability that all \(x\) molecules fail capture: \(P(Y=0\mid X=x)=(1-q)^x\). This conditional probability depends on the latent count: it is 1 when \(x=0\), and decreases as \(x\) increases when \(0<q\leq 1\). Averaging over \(X\) gives the marginal zero probability \(P(Y=0)=E[(1-q)^X]=G_X(1-q)\). This identity makes clear why the observed zero rate depends on both capture efficiency and the distribution of latent counts.

Diagnostics, Alternatives, and Limits

Thinning is a sampling model, so its assumptions should be considered alongside its predictions. If \(q\) is known or controlled in a hypothetical dilution or capture experiment, the expected mean should scale as \(q\), while the distributional formulas above predict the variance and zero rate. Systematic disagreement across comparable samples can indicate that the fixed-\(q\), independent-capture model is inadequate. For example, capture probabilities might differ across cells, or capture outcomes might not be independent. A single excess-zero comparison does not identify which assumption failed.

Binomial thinning is distinct from two common alternatives. Deterministically multiplying a count by \(q\) would produce fractional values in general and omit the random molecule-to-molecule sampling step. A zero-inflated model instead adds a separate probability of a structural zero to a count distribution. Thinning can already produce many zeros, especially when counts are small or capture is inefficient; adding a zero-inflation component should not be justified merely by observing zeros.

The model also has limits. It assumes the same capture probability for each molecule being modeled, and it does not account for false counts, molecule duplication, or other sources of measurement error. When the probability varies across cells, a single fixed-\(q\) marginal distribution may no longer describe the observed counts; analyzing that variation is a separate modeling problem. The present result isolates what independent capture alone does to a count distribution.