Diagnosing The Poisson Mean-Variance Assumption

Why Check the Mean–Variance Relationship?

A Poisson model makes a precise prediction: for counts generated independently with the same rate, the variance equals the mean. That equality is useful as a diagnostic because it links two features of the data that can be summarized across repeated observations. It is not, however, a rule that every collection of observed counts must satisfy exactly. A finite sample can fall above or below the model’s prediction just by chance.

The earlier material on the Poisson model for molecular counts establishes the distributional assumption used here. This tutorial focuses on assessing its mean–variance implication with replicate summaries and plots. The goal is to judge whether the observed counts look compatible with that implication, not to identify the cause of a discrepancy or select a more elaborate count model.

Definition: If \(X_1,\ldots,X_n\) are independent observations from the same Poisson distribution with rate \(\lambda\), then each observation has mean \(\lambda\) and variance \(\lambda\). The Poisson mean–variance assumption is the equality \(\operatorname{Var}(X_i)=E(X_i)=\lambda\). In a sample, compare the sample variance \(s^2\) with the sample mean \(\bar{x}\); their equality is a model prediction, not a requirement that the two computed numbers match exactly.

Conditions for a Meaningful Comparison

A direct comparison of sample mean and sample variance is most informative when the observations represent repeated counts for the same gene under comparable conditions. For example, counts might come from replicate observations with the same measurement exposure and a stable underlying count rate. The observations should be independent under the model, and each should be adequately represented by the same Poisson rate \(\lambda\).

These conditions matter in single-cell data. Counts from cells with substantially different library sizes or capture opportunities may have different expected counts even if each count is Poisson conditional on its own rate. Pooling those observations and comparing the pooled variance with the pooled mean does not directly check a single-rate Poisson model. Similarly, combining different genes into one set of counts is not a gene-wise check: genes can have different rates. A mean–variance plot should generally use one point per gene, with that gene’s counts summarized across a suitable set of comparable observations.

When replicate counts are available, use the ordinary sample summaries:

\bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_i, \qquad s^2=\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar{x})^2, \qquad R=\frac{s^2}{\bar{x}}\quad(\bar{x}>0).

The ratio \(R\) is a compact descriptive summary. A value near 1 is consistent with the equality at the level of these summaries; a value appreciably above 1 means the observed sample variance exceeds the sample mean, and a value below 1 means it is smaller. The ratio alone does not say whether a difference is larger than would be expected from sampling variability.

Why the Sample Variance Is Noisy

The sample variance is an estimate, so it fluctuates from one finite sample to another. In particular, using \(n-1\) in its denominator makes \(s^2\) unbiased for the population variance when observations are independent and identically distributed with a finite variance. Under the single-rate Poisson model, this means \(E(s^2)=\lambda\); also, \(E(\bar{x})=\lambda\). Thus the model predicts equality of the expectations of the two summaries. It does not assert that \(s^2=\bar{x}\) in every sample.

Derivation: For independent Poisson observations with rate \(\lambda\), \(E(X_i^2)=\lambda+\lambda^2\), since \(\operatorname{Var}(X_i)=E(X_i^2)-E(X_i)^2=\lambda\). Also, \(E(\bar{X}^2)=\lambda/n+\lambda^2\). Using \(\sum_i(X_i-\bar{X})^2=\sum_iX_i^2-n\bar{X}^2\), we obtain

E\!\left[\sum_{i=1}^{n}(X_i-\bar{X})^2\right] =n(\lambda+\lambda^2)-n\left(\frac{\lambda}{n}+\lambda^2\right) =(n-1)\lambda.

Dividing by \(n-1\) gives \(E(s^2)=\lambda\), the same as \(E(\bar{X})\). With only a few replicates, however, the realized variance can be far from its expectation. A variance-to-mean ratio moderately above or below 1 should therefore be treated as a clue to inspect, not as a verdict.

Worked Example: A Sample Close to the Poisson Prediction

Suppose a hypothetical gene has counts \(0,1,1,2,2,3,3,4\) across eight comparable, independent replicate observations. Calculate its mean, sample variance, and variance-to-mean ratio.

The sum of the counts is \(16\), so \(\bar{x}=16/8=2\) counts per observation. The sum of squared counts is \(44\). Using the computational form of the sample variance,

s^2=\frac{\sum_i x_i^2-n\bar{x}^2}{n-1} =\frac{44-8(2^2)}{7} =\frac{12}{7} \approx 1.714\ \text{counts}^2.

The variance-to-mean ratio is \(R=1.714/2\approx0.857\). This sample variance is somewhat below the sample mean, but the summaries are in the same general range. With only eight observations, this result alone does not establish that the Poisson assumption fails. The counts are nonnegative integers, and the calculation checks the equality for this gene; whether the observations truly have equal rates and are independent still depends on how they were obtained.

Reading a Mean–Variance Plot

For each gene, calculate the mean and sample variance across comparable replicate observations, then plot the variance on the vertical axis against the mean on the horizontal axis. The reference line \(y=x\) marks the Poisson equality. Points near the line have sample summaries of similar magnitude. Points well above it have larger sample variance than mean; points below it have smaller sample variance. The pattern across genes is often more informative than any one point.

A plot is diagnostic rather than a proof. Each point is based on a finite number of observations, and the variance estimate can be especially erratic when there are few replicates. Genes with very different average counts can also occupy very different parts of the plot, so inspect both the overall pattern and individual points. If the mean is zero, the ratio \(s^2/\bar{x}\) is undefined; show or report the mean and variance directly rather than assigning a ratio.

Worked Example: Comparing Three Genes on a Mean–Variance Plot

Consider hypothetical counts for three genes, each measured across four comparable replicate observations: Gene A has \(1,2,2,3\); Gene B has \(0,1,3,4\); and Gene C has \(2,3,4,5\). Calculate one point for each gene.

For Gene A, the sum is \(8\), so \(\bar{x}=2\). The sum of squares is \(18\), giving \(s^2=(18-4(2^2))/3=2/3\approx0.667\). Its point is approximately \((2,0.667)\), below the reference line.

For Gene B, the sum is also \(8\), so \(\bar{x}=2\). The sum of squares is \(26\), giving \(s^2=(26-4(2^2))/3=10/3\approx3.333\). Its point is approximately \((2,3.333)\), above the line.

For Gene C, the sum is \(14\), so \(\bar{x}=3.5\). The sum of squares is \(54\), giving \(s^2=(54-4(3.5^2))/3=5/3\approx1.667\). Its point is approximately \((3.5,1.667)\), below the line.

The three genes with mean counts around 2–3.5 fall on both sides of the Poisson reference line. That spread illustrates why isolated points are not conclusive, particularly with only four observations per gene. Before interpreting the pattern, confirm that counts are being compared across suitably comparable observations and that the summaries were calculated gene by gene.

Replicate Summaries and Practical Diagnostics

A replicate summary can make the assumptions behind the plot easier to inspect. For a particular gene, arrange the counts in observation order and check whether the observations represent comparable measurement opportunities. Then calculate \(\bar{x}\), \(s^2\), and, when \(\bar{x}>0\), \(R\). Keep the raw replicate counts available: two sets of counts can have the same mean and variance but very different arrangements or unusual observations.

As a rough diagnostic, one can also compare the total squared deviations with the fitted Poisson mean. Define \(Q=\sum_i(x_i-\bar{x})^2/\bar{x}\) when \(\bar{x}>0\). Because the numerator equals \((n-1)s^2\), \(Q=(n-1)R\). Under a well-fitting single-rate Poisson model, this quantity is often on the scale of \(n-1\), though it is not a universal cutoff. Formal calibration depends on the sample size and model conditions; especially with few observations or small expected counts, a numerical threshold should not be treated as definitive evidence.

Worked Example: A Replicate Set with a Larger Variance

For a second hypothetical gene, suppose eight comparable replicate counts are \(0,0,1,1,2,2,4,6\). Assess the mean–variance discrepancy and calculate \(Q\).

The sum is \(16\), giving \(\bar{x}=16/8=2\) counts per observation. The sum of squares is \(62\), so

s^2=\frac{62-8(2^2)}{7} =\frac{30}{7} \approx4.286\ \text{counts}^2, \qquad R=\frac{4.286}{2}\approx2.143.

The sample variance is a little more than twice the sample mean. The scaled squared-deviation summary is \(Q=(n-1)R=7(2.143)\approx15.0\), compared with \(n-1=7\) as a rough reference scale. This is a marked descriptive discrepancy in this sample and warrants checking the observations and the assumptions. It does not, on its own, identify why the discrepancy occurred or prove which alternative model is appropriate. In particular, the calculation assumes that these eight observations are independent and have comparable rates; if that condition is not credible, the single-rate comparison is not the right diagnostic.

Common Mistakes, Alternatives, and Limits

Treating equality as an exact sample rule. The reasoning “the Poisson variance equals the mean, so every sample must have \(s^2=\bar{x}\)” confuses population properties with estimates. The correction is to consider sampling variation and replicate count when interpreting the difference.
Pooling observations with unequal exposure without checking. A raw count comparison can reflect different measurement opportunities rather than a common rate. The correction is to compare observations with comparable exposure or use a model that accounts for the exposure before applying a mean–variance diagnostic.
Combining genes into a single variance calculation. A pooled result can mix genes with different mean counts and does not provide a gene-wise check. The correction is to compute one mean and variance per gene, then inspect the resulting points.
Calling a point above the line proof of a particular mechanism. A diagnostic describes a mismatch with the simple Poisson equality; it does not identify its cause. The correction is to report the observed pattern and check the data and assumptions before considering alternatives.

A mean–variance plot is transparent and useful for spotting broad patterns, but it compresses each gene’s replicate counts into two numbers. A replicate table retains the individual observations and helps reveal data issues that a plot can hide, though it is less convenient for summarizing many genes. Neither replaces checking whether the observations are independent and comparable. When those assumptions are not plausible, a more flexible model or a design-aware analysis may be needed; choosing and interpreting such models requires additional assumptions beyond this diagnostic.

Key takeaway: Under independent, equal-rate Poisson sampling, the population mean and variance are both \(\lambda\). Use gene-wise replicate summaries and a \(y=x\) reference line to assess whether the observed counts are compatible with that equality, while treating finite-sample departures as diagnostic evidence rather than a final explanation.

Check Your Understanding

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

For counts \(1,1,2,2,3\), calculate \(\bar{x}\), \(s^2\), and \(R\). Is the sample variance exactly equal to the mean?
A gene has a point well above the \(y=x\) line. What does this say, and what does it not establish?
Why is it potentially misleading to pool counts from cells with substantially different library sizes for a single-rate Poisson mean–variance check?
What does the derivation of \(E(s^2)=\lambda\) establish, and what does it not guarantee for one finite sample?

Solutions. (1) The sum is \(9\), so \(\bar{x}=9/5=1.8\). The sum of squares is \(19\), so \(s^2=(19-5(1.8^2))/4=2.8/4=0.700\). Thus \(R=0.700/1.8\approx0.389\); the sample variance is not exactly equal to the mean. (2) It says the observed sample variance exceeds the sample mean for that gene. It does not prove a specific cause or establish that a particular alternative model is correct. (3) Different library sizes can correspond to different expected counts, so the pooled observations may not have one common Poisson rate; the pooled variance is then not a direct check of the single-rate equality. (4) It establishes that the sample variance is unbiased for \(\lambda\) under independent, equal-rate Poisson sampling. It does not guarantee that a realized sample variance equals the realized sample mean.