Why Pair Donors?
In a paired differential-abundance study, each donor contributes a sample under each of two conditions, such as before and after treatment. The comparison is made within donor, rather than treating all samples from one condition as unrelated to all samples from the other. This matters because donors can differ substantially in their baseline cell-type composition. A donor who tends to have a high proportion of a cell type in both conditions need not obscure a consistent within-donor change.
The independent biological units remain the donors. Cells measured from one donor-condition sample are not independent biological replicates. As in other donor-level abundance analyses, first choose the denominator and outcome scale that match the question. A proportion of all profiled cells, a proportion within a parent population, and a log-ratio to a reference cell type answer different questions. Pairing changes how the conditions are compared; it does not resolve the choice of denominator or remove compositional dependence.
For a proportion outcome, \(\bar{D}\) is an average change in proportion, often reported in percentage points. For a logit or log-ratio outcome, it is an average change on that transformed scale. These are not interchangeable effect measures. For example, an increase from 5% to 10% is a 5-percentage-point change, while the corresponding log-odds change differs from the change from 45% to 50%.
Paired Contrasts and Their Uncertainty
A common analysis treats the donor-level contrasts \(D_1,\ldots,D_n\) as independent observations and estimates their mean. A paired t procedure is appropriate when donors are independent of one another, the pairs are correctly matched, and the distribution of contrasts is approximately normal (or the number of pairs is large enough for the mean to be adequately approximated by a normal distribution). With few pairs, severe skewness or influential contrasts can make the normal approximation unreliable. The method concerns the distribution of the differences, not the separate distributions in the two conditions.
To test \(H_0:\mu_D=0\), where \(\mu_D\) is the population mean within-donor contrast, use \(t=\bar{D}/SE(\bar{D})\) with \(df=n-1\). A two-sided test compares the statistic with a t distribution under the null. A confidence interval is \(\bar{D}\pm t^*SE(\bar{D})\), where \(t^*\) is the appropriate t quantile. This inference is about the mean donor-level contrast on the selected outcome scale.
Worked Example: Mean Change in a Cell-Type Proportion
Suppose six hypothetical donors have the following proportions of a specified cell type, using the same denominator in both conditions: 8% to 10%, 12% to 16%, 5% to 6%, 20% to 25%, 14% to 17%, and 9% to 12%. The paired changes, in percentage points, are \(2,4,1,5,3,3\).
The average change is \(\bar{D}=(2+4+1+5+3+3)/6=18/6=3.00\) percentage points. The deviations from 3 are \(-1,1,-2,2,0,0\); their squared sum is 10. Thus \(s_D=\sqrt{10/5}=\sqrt{2}=1.414\) percentage points, and \(SE(\bar{D})=1.414/\sqrt{6}=0.5774\) percentage points.
The paired statistic is \(t=3.00/0.5774=5.196\), with \(df=5\). The two-sided p-value is approximately 0.0035. At the 5% level, this is evidence against a zero mean paired change, subject to the model assumptions. Using \(t^*=2.571\), the 95% confidence interval is \(3.00\pm2.571(0.5774)\), or approximately 1.52 to 4.48 percentage points. Under the paired t model, this interval procedure has 95% long-run coverage for the population mean change. It does not mean there is a 95% probability that this fixed parameter lies in this particular interval.
The inference treats the six donor contrasts as independent and assumes their distribution is sufficiently close to normal for t inference. With only six pairs, the differences should be inspected for strong skewness or an unusually influential donor; the arithmetic alone does not verify that assumption.
Paired Regression on a Chosen Scale
A regression representation makes the within-donor comparison explicit. For a donor-level outcome \(Y_{ij}\), with \(j=0,1\) denoting condition, a fixed-donor-intercept model is \(Y_{ij}=\alpha_i+\beta C_{ij}+\epsilon_{ij}\), where \(C_{ij}=0\) in condition 0 and \(C_{ij}=1\) in condition 1. The intercept \(\alpha_i\) allows each donor to have a distinct baseline. The coefficient \(\beta\) is the condition contrast on the scale of \(Y\), under the model’s common-effect specification.
Subtracting the two observations for the same donor removes \(\alpha_i\). In a balanced design with one observation per donor-condition and ordinary least squares, the fitted \(\beta\) is the mean of the donor differences. If \(Y\) is a logit, \(\beta\) is a change in log odds; exponentiating it gives a ratio of odds. If \(Y\) is a log-ratio between cell types, \(\beta\) is a change in that log-ratio, and its exponent is a multiplicative change in the ratio. In either case, interpretation depends on the outcome definition.
Worked Example: A Paired Log-Odds Contrast
Consider three hypothetical donor pairs with proportions of a cell type, relative to all cells in a prespecified parent population, changing from 0.10 to 0.15, from 0.20 to 0.28, and from 0.30 to 0.40. For illustration, use each donor’s observed logit proportion, \(Y=\log[p/(1-p)]\), as the outcome in an equally weighted paired linear analysis. These proportions are nonzero and below one, so their logits are defined.
The three logit differences are \(\log(0.15/0.85)-\log(0.10/0.90)=0.4626\), \(\log(0.28/0.72)-\log(0.20/0.80)=0.4418\), and \(\log(0.40/0.60)-\log(0.30/0.70)=0.4418\). Their mean is \(\hat{\beta}=(0.4626+0.4418+0.4418)/3=0.4487\) log-odds units.
The corresponding ratio of odds is \(\exp(0.4487)=1.566\). On this illustrative scale, the geometric mean odds in condition 1 are about 1.57 times the geometric mean odds in condition 0 across these pairs. This is not a 56.6-percentage-point increase, nor does it imply the same proportion change for every donor. The example treats the observed logits as outcomes and gives donors equal weight; it does not account for the differing precision of proportions estimated from different cell counts.
An alternative is to model the cell-type count directly. For a one-cell-type-versus-rest outcome, a paired binomial model can specify \(K_{ij}\sim\mathrm{Binomial}(N_{ij},p_{ij})\) and \(\mathrm{logit}(p_{ij})=\alpha_i+\beta C_{ij}\), where \(K_{ij}\) is the cell-type count and \(N_{ij}\) is the chosen denominator. Here \(\exp(\beta)\) is the condition odds ratio conditional on the donor intercepts. This likelihood uses the count denominators rather than treating estimated proportions as equally precise. Its binomial assumptions include an appropriate denominator and binomial sampling variation; extra variation across samples can invalidate the simple binomial variance. The count model and paired t analysis therefore differ in both assumptions and the uncertainty they represent.
Why Pairing Can Improve Precision
For paired outcomes, the variance of an individual difference is \(\mathrm{Var}(Y_1-Y_0)=\sigma_1^2+\sigma_0^2-2\mathrm{Cov}(Y_1,Y_0)\). When measurements from the same donor are positively correlated, the covariance term reduces the variance of the difference. Pairing can therefore remove variation due to stable donor-to-donor baseline differences. It does not guarantee a smaller standard error in every dataset: negative within-donor correlation, poor matching, or additional variability in the differences can erase the advantage.
This expression assumes finite variances and defines \(\rho\) as the correlation between the two condition outcomes within a donor. If both standard deviations are 10 percentage points and the within-donor correlation is 0.80, the difference variance is \(100+100-2(0.80)(10)(10)=40\) squared percentage points. If the conditions were independent with those same variances, the variance would be 200. This comparison explains the potential gain; it is not a promise that every observed paired study will have that variance reduction.
Worked Example: Comparing Paired and Unpaired Standard Errors
Suppose five hypothetical donors have proportions of a cell type of 10%, 20%, 30%, 40%, and 50% in condition 0, and 12%, 23%, 32%, 44%, and 55% in condition 1. The paired differences are 2, 3, 2, 4, and 5 percentage points. Their mean is \(16/5=3.20\) percentage points. The squared deviations from 3.20 sum to 6.80, so \(s_D=\sqrt{6.80/4}=1.304\) and the paired standard error is \(1.304/\sqrt{5}=0.583\) percentage points.
For comparison, the condition 0 mean is 30% and its sample variance is \(1000/4=250\) squared percentage points. The condition 1 mean is 33.2%, and its sample variance is \(1146.8/4=286.7\). An independent-samples standard error for the difference in means is \(\sqrt{250/5+286.7/5}=\sqrt{107.34}=10.36\) percentage points. The large difference between these standard errors arises because the paired values track one another closely: much of the between-donor level variation cancels in the differences.
The unpaired calculation discards the matching and is not the preferred analysis if these are genuinely matched donors. This numerical comparison is descriptive, not a guarantee of precision in another study. It also assumes that these outcomes are measured on the same denominator and that the donor is the independent unit.
Assumptions, Diagnostics, and Alternatives
Before interpreting a paired effect, verify that each condition has been matched to the correct donor and that the comparison is not confounded with an unmodeled feature that always changes with condition. For instance, if every condition 1 sample was processed in a different batch from every condition 0 sample, a condition effect cannot be distinguished from that batch difference using pairing alone. Check that the denominator is consistent with the scientific question and that the outcome is defined for every included pair. If a donor lacks one condition, the simple paired contrast is unavailable for that donor; dropping incomplete pairs can change the target population if missingness is systematic.
For paired t inference, inspect a plot of the donor differences and look for extreme values, strong skewness, or a pattern tied to donor characteristics. With few pairs, these checks have limited power to reveal departures from normality. For regression, inspect whether one donor dominates the estimate and whether the assumed common condition effect is a useful summary. For count-based models, assess whether the assumed sampling variation is plausible; extra-binomial variation calls for a model that allows it rather than treating a simple binomial standard error as reliable.
The main alternatives answer related but not identical questions. An independent-samples comparison is appropriate for unrelated donors, but wastes valid matching when pairs exist. A paired t analysis is direct and transparent for a continuous donor-level outcome with a credible difference-scale approximation. A donor-intercept regression allows the same within-donor logic while accommodating covariates and a specified outcome distribution. Count-based paired regression can account for different denominators, but relies on its likelihood assumptions. More complex analyses may be needed for multiple measurements per donor, incomplete or irregular sampling, or several conditions; those settings are not reduced to the simple two-condition paired contrast described here.
Check Your Understanding
Answer each question, then compare with the brief solutions.
- Four donor-level differences in proportion are 1, 2, 3, and 6 percentage points. What is the mean paired change?
- Why can positive within-donor correlation make a paired comparison more precise?
- A fitted paired logit model has condition coefficient \(\hat{\beta}=0.30\). What is the estimated odds ratio, approximately, and what does it describe?
- What is the independent unit in a study with 12 donors, each contributing two condition samples and thousands of cells per sample?
- Name one reason the simple paired t procedure may be unreliable even when all pairs are correctly matched.
Solutions. 1. The mean is \((1+2+3+6)/4=3.00\) percentage points. 2. The variance of a difference subtracts twice the within-pair covariance, so positive covariance reduces difference variance. 3. \(\exp(0.30)=1.350\); it is the estimated condition 1 to condition 0 odds ratio conditional on donor intercepts, for the outcome and denominator used. 4. The independent units are the 12 donors, not the cells or 24 donor-condition samples. 5. Examples include a highly skewed or outlier-prone distribution of donor differences, especially with few pairs, or a condition effect confounded with batch.