Two summaries, two questions
For a gene in one donor and one cell type, there are two tempting ways to make a sample-level expression summary: add the cells’ raw counts, or normalize each cell first and then average. These operations answer different questions. A sum describes the total observed gene counts in the captured cell population; an average describes the typical normalized expression of a sampled cell, with each cell given equal weight. Neither quantity is automatically a substitute for the other.
This distinction matters when a sample-level value will be used as the response in a statistical model. Standard count-based pseudobulk models use a sum of raw counts. Averaged normalized expression is generally continuous and is not a count observation, so it does not satisfy the response scale assumed by a count likelihood. Choosing the response also affects what the result represents: cell numbers and sequencing depth influence a raw sum, whereas an unweighted average gives each included cell the same influence.
A common cell-level normalization divides a gene’s raw count by a cell-specific library size \(L_{idk}=\sum_gY_{idkg}\), then multiplies by a scale constant \(K\). For example, \(Z_{idkg}=K Y_{idkg}/L_{idk}\) gives counts per million when \(K=1{,}000{,}000\). Other normalizations use cell-specific size factors or transformations, but the central distinction remains: normalized cell values are not the original integer counts.
What the sum and average represent
The sum retains the total count accumulated across the cells assigned to a donor and cell type. It will tend to be larger when more cells are included, when those cells have greater sequencing depth, or when the gene contributes more counts per cell. These influences are not interchangeable. In particular, a larger sum does not by itself establish that a typical cell expresses more of the gene.
The unweighted average of normalized expression gives each included cell equal weight. It can summarize typical normalized expression among the sampled cells, but it depends on which cells were sampled and how their expression values were normalized. It is not the same as normalizing the summed count by the total pseudobulk library size. The latter weights each cell in proportion to its library size.
The equality follows by multiplying each cell’s gene fraction \(Y_{idkg}/L_{idk}\) by its share of the total library, then summing: each term becomes \(Y_{idkg}/\sum_jL_{jdk}\). Thus the pooled gene fraction is a library-size-weighted average of cell-level gene fractions. By contrast, the simple mean of those fractions gives each cell weight \(1/n_{dk}\), regardless of library size. The two coincide when library sizes are equal, but need not otherwise.
Worked Example: Summing Counts and Averaging CPM
Suppose three hypothetical cells of one donor’s cell type have gene counts 20, 10, and 30. Their total raw library sizes are 10,000, 20,000, and 30,000 counts. Calculate the summed raw count, the mean cell-level CPM, and the CPM of the pooled counts.
The summed raw count is \(20+10+30=60\) counts. Cell-level CPM values are calculated using \(Z_i=1{,}000{,}000Y_i/L_i\):
The unweighted mean is \((2{,}000+500+1{,}000)/3=1{,}166.7\) CPM. The pooled library size is \(10{,}000+20{,}000+30{,}000=60{,}000\), so the pooled CPM is \(1{,}000{,}000(60/60{,}000)=1{,}000\) CPM. These differ because the pooled value weights the cells by their library sizes; the largest-library cell receives half the total weight. The value 60 is an integer count, while 1,166.7 is a continuous average of normalized values.
Why count-based pseudobulk models use summed raw counts
A count-based model is specified for count observations: its response is a nonnegative integer, and its probability model describes variation in counts conditional on model parameters. In pseudobulk analysis, summing the original counts within each donor and cell type produces that kind of response. Averaging normalized values generally does not. It may produce a fraction or decimal, and its variation has been changed by cell-level normalization and averaging. Treating that average as if it were a raw count would therefore mismatch the data with the likelihood.
A simple aggregation result helps explain why summation is natural. If, as an idealized model, the counts from cells are independent Poisson random variables with means \(\mu_i\), their sum is also Poisson, with mean equal to the sum of the means. This provides a direct count-scale justification for aggregation under those assumptions. Real pseudobulk data may have extra variation and dependence, so a model must still allow an appropriate variance structure; summing does not by itself guarantee that a particular count distribution fits.
The independence condition in that result is an assumption about the illustrative cell-level Poisson variables, not a claim that cells from one donor are independent biological replicates. Pseudobulk aggregation makes the donor-by-cell-type sample the observation used for comparative inference; it does not turn its constituent cells into separate donors. Donor-to-donor variation and the study design still determine the replication available for inference.
In practice, a count-based pseudobulk analysis pairs the summed counts with sample-level information about sequencing depth and other design features. A library-size adjustment is used so that differences in sequencing opportunity are not mistaken for expression differences. That adjustment changes how the sum is interpreted in the model; it does not change the response from summed counts into an average of normalized cell values. The sum alone is not a depth-adjusted comparison.
Worked Example: More Cells Can Mean a Larger Sum
Consider two hypothetical donor samples in the same cell type. In sample A, four cells each have 5 raw counts for a gene and a library size of 1,000. In sample B, ten cells each have 5 raw counts and a library size of 1,000. Calculate both summaries and state what the difference does and does not show.
For sample A, the summed count is \(4(5)=20\) counts, and each cell has \(1{,}000{,}000(5/1{,}000)=5{,}000\) CPM. The average is therefore 5,000 CPM. For sample B, the sum is \(10(5)=50\) counts, and the average is also 5,000 CPM because every cell has the same normalized value.
The raw sum in B is 2.5 times the sum in A, but there are also 2.5 times as many cells. These numbers do not show that a typical cell in B has higher normalized expression: the observed cell-level CPM values are identical. The pooled library sizes are 4,000 and 10,000, respectively, and the pooled CPM is 5,000 in both samples. If the scientific target concerns typical-cell expression, comparing the sums without accounting for the scale of the pooled libraries would be misleading. If the target concerns aggregate captured counts, the totals are relevant, but cell sampling and sequencing depth remain part of their interpretation.
Unequal library sizes change the weighting
Even if every cell is normalized to the same units, the mean across cells and the pooled normalized count can differ when library sizes vary. This is not an arithmetic error. The unweighted average describes the mean of the cell-level normalized values; pooling counts first gives more weight to cells with larger libraries. When library sizes are similar, the summaries may be close. When they differ substantially, the choice of summary can materially affect the result.
Worked Example: A Pooled Fraction Is Not an Unweighted Mean
Take two hypothetical cells with gene counts 10 and 10, and library sizes 100 and 1,000. Compare the unweighted mean of their gene fractions with the fraction obtained by pooling counts and libraries.
The cell-level fractions are \(10/100=0.10\) and \(10/1{,}000=0.01\). Their unweighted mean is \((0.10+0.01)/2=0.055\), or 5.5%. The pooled gene fraction is \((10+10)/(100+1{,}000)=20/1{,}100=0.01818\), or about 1.818%.
The pooled fraction can also be computed as a weighted average. The first cell’s library weight is \(100/1{,}100\), and the second cell’s is \(1{,}000/1{,}100\). Thus \((100/1{,}100)(0.10)+(1{,}000/1{,}100)(0.01)=0.01818\). The high-library cell has most of the pooled weight, so the pooled value is much closer to its 1% fraction than to the unweighted mean of 5.5%.
Assumptions, diagnostics, and alternatives
Before calculating either summary, the aggregation must be defined consistently. The included cells should belong to the intended donor and cell type, gene identifiers and count definitions must agree across cells, and cell-level normalization must be applied consistently if averages are compared. The number of included cells and the total library size should be recorded for each donor-by-cell-type sample. Without these quantities, it is difficult to tell whether a changed sum reflects expression, cell yield, or sequencing depth.
Useful checks include verifying that each pseudobulk count equals the sum of its contributing cell counts, that the result is a nonnegative integer, and that the associated library size is the sum of the same cells’ library sizes. Examine the number of cells and total library size across donors. Large imbalances, extreme library sizes, or a few highly influential cells warrant investigation. Also compare summaries only after confirming that normalization factors and units are the same. A disagreement between an unweighted average and a pooled normalized value may be expected when library sizes vary; it is a prompt to identify the estimand and weighting, not to choose whichever result looks more convenient.
| Summary | What it summarizes | Use and caution |
|---|---|---|
| Sum of raw counts | Total observed gene counts across included cells | Appropriate count-scale response for standard count-based pseudobulk models; affected by cell yield and sequencing depth and needs sample-level adjustment. |
| Mean of normalized cell values | Unweighted mean normalized expression among included cells | Answers a typical-sampled-cell question under the chosen normalization; is generally continuous, not a raw count response. |
| Pooled normalized count | Summed gene counts divided by a pooled library-size scale | Equivalent to a library-size-weighted mean of cell-level fractions, not usually the unweighted cell mean. |
These summaries are not interchangeable alternatives for the same model. A mean of normalized expression can be appropriate when the target is a cell-level average and the analysis uses a method suited to that continuous outcome. A count-based pseudobulk model instead uses summed counts and models them on the count scale, with sample-level adjustment for library size. The first choice makes the cell the unit contributing to an average; the second makes the donor-by-cell-type aggregate the count observation. In either case, donors—not the number of cells alone—govern biological replication.
Check Your Understanding
Answer each question, then compare with the solutions below.
- A donor has three cells with raw counts 4, 8, and 12 for a gene. What is the summed pseudobulk count, and why is it on the count scale?
- Two cells have normalized values 2 and 10. What is their unweighted mean? Does that mean qualify as a raw count response?
- When does the pooled gene fraction equal the unweighted average of the cell-level gene fractions?
- Why can a larger summed count in one donor sample fail to show higher typical-cell expression?
- Under what assumptions is a sum of Poisson counts also Poisson, and what does that result not establish about donor replication?
Solutions. (1) The sum is \(4+8+12=24\) counts, a nonnegative integer formed by adding the observed raw gene counts. (2) The mean is \((2+10)/2=6\). It is an average normalized value and need not be an integer, so it is not a raw count response. (3) The equality holds when the cell library sizes are equal, so the pooled calculation weights each cell equally. (4) The sum can increase because more cells were included or because the sample had greater sequencing depth; the sum alone does not isolate expression per typical cell. (5) The cell counts must be independent Poisson variables; their sum is then Poisson with mean equal to the sum of their means. This distributional result does not make cells independent biological replicates or increase the number of donors.