Why the Average Is Not the Whole Result
A study may estimate an average response across donors, but that average does not describe how similar the donors are to one another. Two sets of donors can have the same average outcome while one set is tightly clustered and the other contains very different responses. In single-cell studies, this distinction matters because the biological interpretation of an average can depend on whether it is typical of most donors or conceals substantial donor-to-donor variation.
The target here is between-donor heterogeneity: the spread of donor-level outcomes in the population of donors relevant to the study. This is not the same as variation among cells within a donor, nor is it the uncertainty in estimating a mean from a finite sample of donors. Earlier material on separating within- and between-donor variance and on generalizing beyond observed donors provides useful context; here the focus is on defining, estimating, and interpreting the spread itself.
For example, \(Y_i\) might be a donor-specific mean expression value for a gene in a defined cell type, or a donor-specific treatment contrast on a specified expression scale. The outcome and scale must be fixed before interpreting its spread. A standard deviation of 1 on a log expression scale does not mean the same thing as a standard deviation of 1 on a raw-count scale.
Estimating Spread Across Observed Donors
Suppose \(m\) donors are sampled from the target population and a donor-level outcome is observed for each. Write those outcomes as \(y_1,\ldots,y_m\), their sample mean as \(\bar{y}\), and their sample variance as \(s_D^2\):
When the donors are independent and representative draws from the target population, \(s_D^2\) estimates \(\tau^2\), and \(s_D\) estimates \(\tau\). The sample variance uses \(m-1\) in the denominator because the sample mean is estimated from the same donor outcomes. Its units are the squared units of \(Y_i\); the standard deviation returns to the original units.
A key distinction is that \(s_D\) describes the spread of donor outcomes, whereas the standard error of the sample mean describes uncertainty in estimating \(\mu_D\). Under independent, identically distributed donors with finite variance, the standard error is estimated by \(s_D/\sqrt{m}\). More donors can reduce uncertainty in the estimated mean without making the underlying donors more alike.
Worked Example: Same Outcome Scale, Two Different Summaries
Suppose five hypothetical donors have estimated outcomes of 2, 4, 5, 7, and 12 units. Treat these as directly observed donor outcomes for this calculation.
Mean. The average is \(\bar{y}=(2+4+5+7+12)/5=30/5=6\) units.
Between-donor spread. The deviations from 6 are \(-4,-2,-1,1,6\). Their squared sum is \(16+4+1+1+36=58\). Therefore, \(s_D^2=58/(5-1)=14.5\) units squared, and \(s_D=\sqrt{14.5}=3.808\) units.
Uncertainty in the mean. The estimated standard error is \(SE(\bar{y})=3.808/\sqrt{5}=1.703\) units. The value 3.808 describes the observed donor-to-donor spread; 1.703 describes the estimated sampling uncertainty in the average under the stated sampling assumptions. They answer different questions.
A confidence interval for the mean is not a summary of donor heterogeneity. For example, under independent normally distributed donor outcomes, a \(t\) interval for \(\mu_D\) uses \(s_D/\sqrt{m}\) and has \(m-1\) degrees of freedom. Its width concerns uncertainty in the population mean, not the range in which individual donor outcomes are expected to fall.
Uncertainty in an Estimate of the Spread
A sample standard deviation is itself uncertain, especially when the number of donors is small. If the donor outcomes are independent draws from a normal population with variance \(\tau^2\), then \((m-1)s_D^2/\tau^2\) follows a chi-square distribution with \(m-1\) degrees of freedom. This result gives a confidence interval for the variance. Let \(\chi^2_{q,df}\) be the \(q\) quantile of the chi-square distribution with \(df\) degrees of freedom. A \(100(1-\alpha)\%\) interval is:
The denominators appear in this order because the larger chi-square quantile produces the lower endpoint. Taking square roots of both endpoints gives an interval for the standard deviation. This interval relies on normality of the donor-level outcomes; with few donors, departures from normality or one unusual donor can have a substantial effect.
Worked Example: An Interval for Between-Donor Variance
Use the five hypothetical outcomes from the preceding example. They give \(m=5\), \(s_D^2=14.5\) units squared, and \((m-1)s_D^2=58\) units squared. Assume independent normally distributed donor outcomes. For a 95% interval with \(df=4\), use \(\chi^2_{0.975,4}=11.143\) and \(\chi^2_{0.025,4}=0.484\).
Substitution gives a lower endpoint of \(58/11.143=5.205\) units squared and an upper endpoint of \(58/0.484=119.8\) units squared. Thus, the 95% interval for \(\tau^2\) is approximately \([5.205,119.8]\) units squared. Taking square roots gives an interval for \(\tau\) of approximately \([2.281,10.95]\) units.
This wide interval reflects the limited information about population spread in only five donors. Under the model assumptions, the procedure has 95% coverage over repeated samples; it does not mean there is a 95% probability that the fixed population variance lies in this particular computed interval.
When Donor-Level Outcomes Are Noisy Estimates
In single-cell data, a donor-level outcome may be estimated from a finite number of sampled cells rather than observed without error. Denote the estimate for donor \(i\) by \(Z_i\), and write \(Z_i=Y_i+e_i\), where \(Y_i\) is the donor's underlying outcome and \(e_i\) is estimation error. Suppose the errors have mean zero, are independent across donors and independent of the underlying outcomes, and have known variances \(v_i\). Then:
Consequently, the across-donor variance of the estimates can reflect both true donor heterogeneity and estimation noise. If donors have unequal cell counts or different precision, their \(v_i\) values may differ. Treating the observed spread of \(Z_i\) as if it were entirely biological can therefore overstate the spread of underlying donor outcomes.
For independent estimates with a common mean and the stated error model, the expected sample variance of the \(Z_i\) values is \(\tau^2\) plus the average measurement-error variance. This motivates a method-of-moments correction:
This is a simple correction, not a universal replacement for a model that handles donor-specific precision. In particular, the corrected estimate can be negative in a finite sample. Since a variance cannot be negative, one may report a boundary estimate of zero, but should also recognize that the data then provide little evidence for resolving true heterogeneity from estimation noise. Uncertainty in the \(v_i\) values also affects the correction.
The justification follows from the definition of sample variance. If \(Z_i\) have a common mean and independent variances \(q_i=\tau^2+v_i\), then \(\sum_i(Z_i-\bar{Z})^2=\sum_i Z_i^2-m\bar{Z}^2\). Taking expectations, the common mean terms cancel, leaving \(\sum_i q_i-(1/m)\sum_i q_i\). Dividing by \(m-1\) gives:
Subtracting the average \(v_i\) therefore removes the expected contribution of estimation noise under these assumptions. The derivation does not make noisy donor estimates equivalent to error-free outcomes; it identifies what the correction estimates on average.
Worked Example: Correcting Observed Spread for Estimation Noise
Suppose five hypothetical donor-level estimates are 1.2, 1.8, 2.0, 2.7, and 3.3 units. Assume their independent estimation errors have known variance \(v_i=0.25\) units squared for every donor, and assume errors are independent of the underlying donor outcomes.
The mean of the estimates is \(2.2\) units. Their deviations are \(-1.0,-0.4,-0.2,0.5,1.1\), whose squared sum is \(1.00+0.16+0.04+0.25+1.21=2.66\). Thus, \(s_Z^2=2.66/4=0.665\) units squared.
The average estimation-error variance is \(0.25\) units squared, so the corrected estimate is \(\widehat{\tau^2}_{\mathrm{corr}}=0.665-0.25=0.415\) units squared. The corresponding estimated between-donor standard deviation is \(\sqrt{0.415}=0.644\) units. This correction is interpretable only given the assumed error variances and independence conditions; it is not evidence that each donor's true outcome is known precisely.
Comparing Heterogeneity Without Losing the Scale
A standard deviation gives an absolute measure of spread in the outcome's units. This is usually the clearest first summary. A coefficient of variation, \(s_D/\bar{y}\), expresses spread relative to the mean, but is only sensible when the outcome has a meaningful zero and the mean is sufficiently far from zero. It can be unstable or misleading for outcomes that can be negative, are centered near zero, or use an arbitrary transformed scale.
Worked Example: Equal Averages, Different Donor Spread
Consider two hypothetical groups of five donors each, with outcomes measured in the same units. Group A has outcomes 4, 5, 5, 6, and 4. Group B has outcomes 1, 3, 5, 7, and 8.
Both means equal \(24/5=4.8\) units. For Group A, the squared deviations from 4.8 sum to \(0.64+0.04+0.04+1.44+0.64=2.80\), so \(s_A^2=2.80/4=0.700\) units squared and \(s_A=0.837\) units.
For Group B, the squared deviations sum to \(14.44+3.24+0.04+4.84+10.24=32.80\), so \(s_B^2=32.80/4=8.20\) units squared and \(s_B=2.864\) units. The observed sample variances differ by a factor of \(8.20/0.700=11.71\), despite equal sample means. This describes the observed samples; five donors per group are not enough, by themselves, to establish the population variances precisely.
A group comparison may focus on the difference in population means, the difference in variances, or both, depending on the scientific question. These are distinct estimands. Comparing sample standard deviations is descriptive, and a formal claim about population differences requires an inferential procedure with assumptions suited to the design and distribution. A mean-only analysis can miss heterogeneity, while a spread-only comparison does not establish a difference in average outcome.
Diagnostics, Alternatives, and Limitations
Before interpreting an estimate of heterogeneity, check what contributes to the donor-level values and whether the sampling assumptions are credible. Useful checks include plotting donor outcomes, checking for influential donors, examining donor-specific precision, and verifying that the outcomes share a common definition and scale. A highly skewed distribution or an extreme donor can make the sample variance and the normal-theory variance interval particularly sensitive. The normality assumption concerns the distribution of donor-level outcomes, not the distribution of individual cells.
When donor estimates have unequal precision, reporting their raw sample standard deviation alongside the estimated \(v_i\) values makes the limitation visible. A measurement-error correction is an option when those variances are defensible and the independence assumptions are reasonable. A hierarchical or other model-based approach may be preferable when precision varies substantially or when a full uncertainty analysis is required. A simple sample variance is easier to interpret, but it does not automatically accommodate noisy estimates, complex donor sampling, or dependence between donors.
- Do not confuse spread with precision. \(s_D\) describes variation among donors; \(SE(\bar{y})\) describes uncertainty in an estimated average.
- Do not treat cells as independent donor replicates. The number of cells does not change the number of independently sampled donors.
- Do not label all observed spread biological. Noisy donor-level estimates can add variability beyond true between-donor differences.
- Do not assume a stable variance estimate from few donors. Small donor samples can yield very wide intervals and sensitivity to individual outcomes.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- Three hypothetical donor outcomes are 2, 4, and 6 units. Calculate their mean, sample variance, sample standard deviation, and estimated standard error of the mean.
- The sample variance of five noisy donor estimates is 0.90 units squared. If the average known estimation-error variance is 0.40 units squared, what is the corrected method-of-moments estimate of between-donor variance and its corresponding standard deviation?
- In one sentence each, explain what the between-donor standard deviation and the standard error of the donor mean describe.
- What assumption is required for the chi-square confidence interval for the donor-level variance, and what does a 95% confidence level mean?
Solutions. (1) The mean is \((2+4+6)/3=4\) units. The squared deviations sum to \(4+0+4=8\), so the sample variance is \(8/(3-1)=4\) units squared and the sample standard deviation is \(2\) units. The estimated standard error is \(2/\sqrt{3}=1.155\) units. (2) The corrected variance is \(0.90-0.40=0.50\) units squared, giving a standard deviation of \(\sqrt{0.50}=0.707\) units. (3) The between-donor standard deviation describes the spread of donor outcomes in the target population; the standard error describes sampling uncertainty in an estimate of their mean. (4) The donor outcomes must be independent draws from a normal population for the stated chi-square result. A 95% confidence level means that the interval procedure covers the fixed population variance in 95% of repeated samples under those assumptions.