Deriving Dispersion as a Mean-Adjusted Variability Measure

Why Adjust Variability for the Mean?

A gene with a high average count often has a high variance even when its variation is ordinary for that level of expression. Ranking genes by raw variance can therefore put highly expressed genes near the top simply because their counts are larger. A variance-to-mean ratio offers a simple adjustment: it compares the observed variability with the scale of the mean.

This ratio is useful as a descriptive measure, not as a complete model of variability. It removes the linear mean dependence expected under a Poisson model, but it does not guarantee that genes with different means are fully comparable under more general count models. The earlier material on mean-variance relationships in single-cell counts motivates this distinction.

Definition: For a count-valued random variable $X$ with positive mean $\mu=E(X)$, the variance-to-mean ratio, also called the index of dispersion, is $$ D=\frac{\operatorname{Var}(X)}{E(X)}=\frac{\operatorname{Var}(X)}{\mu}. $$ For gene $g$, measured across $n$ cells on a comparable expression scale, its sample estimate is $$ d_g=\frac{s_g^2}{\bar{x}_g}, $$ where $\bar{x}_g$ is the sample mean and $s_g^2$ is the sample variance across those cells. This estimate is defined only when $\bar{x}_g>0$ and $n\ge 2$.

The ratio is dimensionless when the numerator and denominator are measured on the same count scale. It is important to distinguish this descriptive ratio from a model-specific “dispersion parameter.” For example, a negative-binomial model may use a parameter that controls excess variance; its parameter is not generally equal to $d_g$.

Deriving the Mean Adjustment

The motivation follows from the variance functions of common count models. Under a Poisson model, the mean and variance are equal. Thus, a larger mean directly produces a larger variance, even though the variance-to-mean ratio remains constant. For an overdispersed negative-binomial model, variance also increases with the mean, but not just linearly.

X\sim\operatorname{Poisson}(\mu) \quad\Longrightarrow\quad \operatorname{Var}(X)=\mu \quad\Longrightarrow\quad D=\frac{\mu}{\mu}=1.

The negative-binomial parameterization used here is $\operatorname{Var}(X)=\mu+\mu^2/\theta$, where $\theta>0$. Dividing this variance by the mean gives

D =\frac{\mu+\mu^2/\theta}{\mu} =1+\frac{\mu}{\theta}.

This derivation shows both what the ratio accomplishes and what it does not. It removes the Poisson model’s linear dependence on the mean: if the Poisson assumptions hold, the population ratio is one regardless of $\mu$. But under the negative-binomial model, the ratio still increases with $\mu$, unless $\theta$ changes in a compensating way. So “mean-adjusted” means divided by the mean, not fully independent of the mean under every plausible count model.

Justification for the sample calculation: If the observations for a gene are independent draws from a distribution with mean $\mu$ and variance $\sigma^2$, then the sample mean estimates $\mu$, and the sample variance $s^2$, with denominator $n-1$, estimates $\sigma^2$. Their ratio $s^2/\bar{x}$ is a plug-in estimate of $\sigma^2/\mu$. It is not generally an unbiased estimator of that ratio, because a ratio of random estimates need not have expectation equal to the ratio of their expectations.

Worked Example: How the Ratio Behaves Under Count Models

Suppose two hypothetical genes have population means of 4 and 16 counts per cell. First assume their counts follow Poisson distributions. For the gene with mean 4, the variance is 4 and its ratio is $4/4=1$. For the gene with mean 16, the variance is 16 and its ratio is $16/16=1$. The raw variance is four times as large for the higher-mean gene, while the ratio is the same.

Now suppose instead that both genes follow negative-binomial distributions with $\theta=2$. At mean 4, the variance is $$ 4+\frac{4^2}{2}=4+8=12, $$ so the ratio is $12/4=3$. At mean 16, the variance is $$ 16+\frac{16^2}{2}=16+128=144, $$ so the ratio is $144/16=9$. Here the higher-mean gene also has a larger ratio. Dividing by the mean removed one factor of mean from the quadratic term, but did not eliminate the remaining mean dependence.

The example supports a restrained interpretation: the ratio can make a raw-variance comparison more informative, especially against a Poisson-like baseline, but it does not automatically make genes with different means directly comparable under all count distributions.

Calculating Gene-Wise Ratios

In practice, calculate each gene’s mean and sample variance across the same set of cells, then divide the variance by the mean. The calculation is descriptive: it summarizes the spread of observed values relative to their average. The expression values should be on a comparable scale across cells. If cell library sizes differ substantially and have not been addressed, differences in sequencing depth can contribute to the apparent variation. A ratio does not itself normalize depth or remove technical effects.

Worked Example: Comparing Raw Variance and Dispersion Rankings

Consider hypothetical counts for four genes in four equal-depth cells. The table gives the observed counts in each cell. Assume these values are directly comparable across cells; for each gene, use the sample variance with denominator $n-1=3$.

Gene	Cell 1	Cell 2	Cell 3	Cell 4
A	2	4	6	8
B	8	10	12	14
C	10	14	18	22
D	1	2	3	4

For gene A, the mean is $(2+4+6+8)/4=5$. The squared deviations from 5 sum to $9+1+1+9=20$, so $s_A^2=20/3=6.667$, and $$ d_A=\frac{6.667}{5}=1.333. $$

For gene B, the mean is 11, and its squared deviations also sum to $20$. Thus $s_B^2=20/3=6.667$, but $d_B=6.667/11=0.6061$. The two genes tie on raw variance, but gene A has the larger variance-to-mean ratio.

For gene C, the mean is 16. Its squared deviations sum to $36+4+4+36=80$, so $s_C^2=80/3=26.67$ and $d_C=26.67/16=1.667$. For gene D, the mean is $2.5$, its squared deviations sum to $2.25+0.25+0.25+2.25=5$, and $s_D^2=5/3=1.667$. Its ratio is $1.667/2.5=0.6667$.

The raw-variance ranking is C, then A and B tied, then D. The ratio ranking is C, A, D, B. In particular, B’s raw variance matches A’s, but its higher mean gives it a lower ratio. These rankings describe these four observations; with only four cells, they should not be treated as stable population rankings.

Assumptions, Diagnostics, and Failure Modes

The arithmetic requires only a positive sample mean and at least two observations. Interpreting the ratio as a useful gene comparison requires more care. The cells should be measured on a comparable scale, and the set of cells should be relevant to the feature-selection question. Strong differences in depth, batch, or cell composition can inflate variability unrelated to the gene’s within-population expression variation. A pooled ratio can also combine distinct biological subgroups: that may indicate meaningful cell-type specificity, or it may obscure the question being asked.

Check the inputs and the resulting rankings rather than treating the ratios as self-validating. Useful checks include:

Confirm that all genes were summarized across the same cells and on a comparable expression scale.
Inspect the gene means alongside the ratios. If ratios systematically rise with mean, division by the mean has not removed the observed mean dependence.
Look for a small number of cells with unusually large counts. Such observations can dominate a sample variance, especially when $n$ is small.
Check how many zeros each gene has. A gene expressed in very few cells can have a high ratio driven by a few nonzero observations.
Do not calculate a ratio for a gene with zero sample mean; it would require division by zero. A gene with positive mean but zero sample variance has ratio zero.

Worked Example: Diagnosing Sensitivity to One Cell

Suppose a hypothetical gene has counts $2,2,2,10$ in four equal-depth cells. Its mean is $16/4=4$. The deviations are $-2,-2,-2,6$, whose squares sum to $4+4+4+36=48$. Therefore $s^2=48/3=16$, and $d=16/4=4$.

If the last count were 6 instead, the values would be $2,2,2,6$, with mean $12/4=3$. The squared deviations are $1,1,1,9$, so $s^2=12/3=4$ and $d=4/3=1.333$. A change in one cell’s count has reduced the estimated ratio from 4 to 1.333.

This is a diagnostic warning, not a reason to delete the cell automatically. The high value could reflect a measurement problem or a genuine cell subgroup. The appropriate response is to inspect the observation and the study context, then interpret the ranking with that uncertainty in mind.

What the Ratio Does—and Does Not—Offer

Raw variance is simple to calculate, but it remains on the squared expression scale and commonly rises with the mean. The variance-to-mean ratio rescales that variance by the average expression and has a convenient Poisson reference value of one. The squared coefficient of variation, $s^2/\bar{x}^2$, divides by the mean twice; it measures relative variability on a different scale and can become especially large for genes with very small means. Neither ratio is universally preferable: the choice depends on the variability pattern and the feature-ranking objective.

A single ratio also cannot account for a changing mean-variance relationship, differences in measurement precision, or uncertainty in each gene’s estimate. More elaborate approaches model expected variability across expression levels or compare observations with a fitted count model. Those approaches answer related but distinct questions. The variance-to-mean ratio is valuable as a transparent calculation and an initial diagnostic, not as proof that a gene is biologically variable or as a significance test.

Key takeaway: For each gene with positive mean, $d_g=s_g^2/\bar{x}_g$ adjusts observed variance by the expression mean. It can prevent a raw-variance ranking from automatically favoring high-mean genes, but it does not remove every mean effect, technical effect, or source of sampling instability.

Check Your Understanding

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

A count distribution has mean 7 and follows a Poisson model. What are its variance and variance-to-mean ratio?
Two genes have sample means 4 and 12 and the same sample variance of 9. Calculate both ratios and explain how their raw-variance and ratio rankings compare.
Under the negative-binomial variance model $\operatorname{Var}(X)=\mu+\mu^2/\theta$, derive the variance-to-mean ratio.
Why is a gene’s ratio undefined when its sample mean is zero?
Name one reason a high gene-wise ratio might not represent stable, within-population biological variability.

Solutions. (1) The variance is 7 and the ratio is $7/7=1$. (2) The ratios are $9/4=2.25$ and $9/12=0.75$; the genes tie in raw variance, while the lower-mean gene has the larger ratio. (3) Dividing by $\mu>0$ gives $(\mu+\mu^2/\theta)/\mu=1+\mu/\theta$. (4) The calculation divides by the sample mean, so a zero denominator makes the ratio undefined. (5) A high ratio may be driven by one unusually large observation, sparse expression in only a few cells, or differences in cell composition or technical measurement.