Mean-Dispersion Relationships Across Genes

Why Compare Dispersion Across Genes?

Single-cell count data can vary substantially from cell to cell, even for genes with similar average expression. A useful summary must therefore describe both the scale of expression and the variability around it. The negative binomial model represents this variability with a gene-specific dispersion parameter. Looking across genes, however, raises a practical problem: each gene’s dispersion is inferred from a finite number of cells, and that estimate can be especially noisy when the gene has few observations or a low mean.

A mean-dispersion relationship summarizes how dispersion tends to vary with mean expression across genes. It is a cross-gene pattern, not a claim that every gene follows exactly the same curve. Understanding that distinction helps explain why a trend can be informative even when individual gene-wise estimates are unreliable.

Definition: For gene $g$, let $Y_{ig}$ be the count in cell $i$, with mean $\mu_g$ under comparable cell exposures. In the negative binomial mean-dispersion parameterization, the count variance is $$ \operatorname{Var}(Y_{ig})=\mu_g+\alpha_g\mu_g^2, $$ where $\alpha_g\geq 0$ is the dispersion. The term $\mu_g$ is the Poisson sampling contribution; the additional term $\alpha_g\mu_g^2$ represents variation beyond that contribution. A mean-dispersion relationship describes how the $\alpha_g$ values tend to change with $\mu_g$ across genes.

The dispersion is dimensionless. When $\alpha_g=0$, the variance equals the mean, as in the Poisson model. Larger positive values indicate more variance than the Poisson model predicts at that mean. In single-cell data, this excess variability can reflect multiple sources; the dispersion parameter alone does not identify which source is responsible.

Why Gene-Wise Dispersion Estimates Are Unstable

In practice, the population mean and variance are unknown. For a gene observed in $n$ cells, the sample mean $\bar{x}_g$ and sample variance $s_g^2$ fluctuate across samples of cells. Substituting these into the variance formula gives the moment-based expression

\widetilde{\alpha}_g=\frac{s_g^2-\bar{x}_g}{\bar{x}_g^2}.

This expression is useful for understanding instability; it is not a guarantee that a noisy gene-wise estimate is suitable for downstream use. Its numerator is a difference between two estimated quantities, and its denominator is the squared estimated mean. Small changes in the sample variance or mean can therefore cause a large change in $\widetilde{\alpha}_g$, particularly when the mean is small. Finite samples can even produce a negative value, although the model’s dispersion parameter cannot be negative.

A short sensitivity calculation makes the low-mean problem clear. Holding the sample mean fixed, the change in the moment expression for a change in the sample variance is

\frac{\partial \widetilde{\alpha}_g}{\partial s_g^2}=\frac{1}{\bar{x}_g^2}.

At a mean of 1, a change of 1 in $s_g^2$ changes the expression by 1. At a mean of 5, the same variance change shifts it by only $1/25=0.04$. The sample variance itself is also uncertain: for independent observations with a finite fourth central moment $\mu_4$, its variance is

\operatorname{Var}(s^2)=\frac{1}{n}\left(\mu_4-\frac{n-3}{n-1}\sigma^4\right).

Thus a small number of cells can leave substantial sampling variation in the sample variance, and the conversion from variance to dispersion can amplify that variation. The formula also makes clear that precision depends on more than the mean: the distribution’s higher moments matter.

Worked Example: Same Mean, Very Different Dispersion Estimates

Suppose two hypothetical genes are observed in eight cells. Their counts are $0,0,1,1,2,3,4,5$ for Gene A and $1,1,1,2,2,2,3,4$ for Gene B. Both have total count 16, so each sample mean is $16/8=2$.

For Gene A, the sum of squared deviations from 2 is $4+4+1+1+0+1+4+9=24$. Therefore,

s_A^2=\frac{24}{8-1}=3.429,\qquad \widetilde{\alpha}_A=\frac{3.429-2}{2^2}=0.357.

For Gene B, the sum of squared deviations is $1+1+1+0+0+0+1+4=8$. Therefore,

s_B^2=\frac{8}{7}=1.143,\qquad \widetilde{\alpha}_B=\frac{1.143-2}{2^2}=-0.214.

The two estimates differ sharply despite the same mean and the same number of cells. The negative value for Gene B is not a valid negative dispersion; it shows that its observed sample variance fell below its sample mean. With only eight observations, sampling fluctuation can produce that result. These data do not establish that Gene A is intrinsically more variable than Gene B.

What a Mean-Dispersion Trend Summarizes

Across many genes, gene-wise estimates can be plotted against their corresponding mean expressions. A trend represents the typical dispersion at a given mean, often through a smooth curve or a small number of mean-expression groups. It helps summarize broad patterns that are hard to see in noisy individual estimates. For example, a trend may show that higher-mean genes tend to have lower dispersion, or that dispersion is roughly constant over part of the mean range.

There is no universal rule that dispersion must decrease as mean expression rises. The shape depends on the data, the experimental setting, and the way expression and exposure are handled. A trend is an empirical summary of the genes under consideration, not a biological law. Individual genes may sit above or below it, and such departures can be meaningful rather than errors to be automatically removed.

Worked Example: Summarizing a Pattern Across Genes

Consider hypothetical gene-wise summaries grouped by their sample means. Each listed variance is a sample variance; the resulting moment expression is calculated as $(s^2-\bar{x})/\bar{x}^2$.

Mean $\bar{x}$	Sample variances $s^2$	Moment dispersion values	Group average
2	3.43, 1.14, 2.80	0.358, -0.215, 0.200	0.114
5	8, 7, 10	0.120, 0.080, 0.200	0.133
10	15, 21, 17	0.050, 0.110, 0.070	0.0767

For the first group, for example, the three values are $(3.43-2)/4=0.358$, $(1.14-2)/4=-0.215$, and $(2.80-2)/4=0.200$. Their average is $(0.358-0.215+0.200)/3=0.114$, to the displayed precision. The corresponding averages for means 5 and 10 are $(0.120+0.080+0.200)/3=0.133$ and $(0.050+0.110+0.070)/3=0.0767$.

These summaries suggest lower typical dispersion at mean 10 than at mean 5, but they do not establish a smooth or universal decreasing relationship: the mean-5 group average is slightly higher than the mean-2 group average. The negative value at mean 2 also illustrates why individual estimates can be erratic. A trend should be interpreted as a broad description, with the spread of estimates and the number of genes supporting each part of the trend kept in view.

Check Cell Exposure Before Interpreting the Trend

The simple variance formula assumes comparable exposures or a model that accounts for differences in exposure. In sequencing data, cells can have different library sizes or capture opportunities. If expected counts vary because exposures differ, pooling raw counts can create extra variation even when the count distribution at each exposure is Poisson. A trend based on unadjusted means and variances may then reflect exposure differences as well as gene-level variability.

Worked Example: Unequal Exposures Can Mimic Overdispersion

Suppose a hypothetical gene has a baseline expected count of 4 for a cell with exposure 1. Three cells have exposures 0.5, 1, and 1.5, so their conditional Poisson means are 2, 4, and 6. Assume counts are independent conditional on those exposures and follow their stated Poisson distributions.

The average conditional mean is $(2+4+6)/3=4$. The average conditional variance is also 4, because a Poisson count’s variance equals its mean. But the conditional means vary across cells, with variance

\frac{(2-4)^2+(4-4)^2+(6-4)^2}{3}=\frac{8}{3}=2.667.

By the law of total variance, the variance across cells is the average conditional variance plus the variance of the conditional means:

\operatorname{Var}(Y)=4+2.667=6.667.

If these raw pooled moments were treated as though exposures were equal, the moment expression would be $(6.667-4)/4^2=0.1667$. Yet the example specified Poisson sampling within each exposure level, with no additional conditional dispersion. The apparent excess variance arises from mixing different expected counts. Exposure differences must therefore be accounted for before interpreting a cross-gene dispersion pattern.

Diagnostics, Alternatives, and Limitations

Useful checks focus on whether the trend represents the intended variability rather than artifacts of limited data or mismatched assumptions. Examine the number of cells and the range of means represented, and look at the spread of gene-wise estimates around the trend. A trend supported by many genes in one mean range but very few in another should not be treated as equally reliable everywhere. Check for negative moment values and extreme values, especially among low-mean genes; they are signs of unstable sample moments, not evidence for negative or arbitrarily large biological dispersion.

Also check that cell exposures and relevant design differences have been handled in a way consistent with the model. If residual variation changes systematically across cell groups or conditions, a single pooled trend can obscure that structure. A trend can be misleading when cells are not comparable, when a small number of unusual genes dominate a region, or when the mean itself is estimated imprecisely.

The main alternatives answer somewhat different questions. A constant-dispersion summary assumes a common dispersion across genes; it is simple, but can miss a systematic mean-dependent pattern. Gene-wise estimates preserve gene-specific differences, but are noisy when observations are limited. A smooth or grouped mean-dispersion trend describes the typical pattern while retaining the possibility of gene-specific departures. These are different levels of summary, not competing claims that every gene has identical variability.

A trend also has limits. It does not prove a particular biological mechanism, separate technical from biological sources of variation, or establish that a gene is unusually variable for a biological reason. It may be sensitive to the set of genes included and to choices about exposure adjustment and smoothing. The introductory material on moments of the negative binomial count model gives the variance interpretation used here; the key issue in this tutorial is that finite-cell estimates of those moments may be noisy across genes.

Common Pitfall

Wrong reasoning: “This gene’s estimated dispersion is below the trend, so the gene must have less biological variability.”
Correction: A gene below the trend has a lower observed estimate than the cross-gene summary at that mean. Sampling noise, exposure handling, and other sources of variability can affect that estimate. The comparison alone does not identify a biological cause.

Key takeaway: Gene-wise dispersion estimates can be unstable because they are built from noisy sample variances and means, with low means making the conversion especially sensitive. A mean-dispersion trend summarizes the typical cross-gene pattern, but its interpretation depends on adequate cell information, appropriate exposure handling, and attention to gene-specific departures.

Check Your Understanding

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

Under the negative binomial mean-dispersion parameterization, what variance corresponds to mean 3 and dispersion 0.2?
Why can a moment-based gene-wise dispersion expression be negative, and what does that imply?
If a gene’s mean is 2, how much does $\widetilde{\alpha}=(s^2-\bar{x})/\bar{x}^2$ change when $s^2$ increases by 0.4 and the mean is held fixed?
Can a decreasing mean-dispersion trend be treated as a universal law across genes? Explain.

Solutions. (1) $\operatorname{Var}(Y)=3+0.2(3^2)=3+1.8=4.8$ counts squared. (2) The sample variance can fall below the sample mean, making the numerator negative. Since model dispersion is nonnegative, the negative expression signals sampling fluctuation or a mismatch with the simple model, not negative variability. (3) The change is $0.4/2^2=0.1$. (4) No. A decreasing trend may describe the observed genes over a particular range, but the relationship depends on the data and analysis conditions, and individual genes can depart from it.