Tutorials › Statistical Methods for Single-Cell RNA Sequencing › Expression-Dependent Dropout Probability

Noise Models and Dropout · Tutorial 66 of 700

Expression-Dependent Dropout Probability

Learn why a gene with a positive Poisson expression rate can still be undetected, and how its probability of detection changes with that rate.

Intermediate 10 min read

What You'll Learn

  • Derive the probability of a zero count from the Poisson probability mass function.
  • Distinguish an observed zero from a true absence of expression.
  • Calculate detection probabilities and expression rates needed to reach specified detection levels.
  • Understand how cell-to-cell variation in expression changes the expected fraction of zeros.
  • Identify when a Poisson abundance-dependent detection curve is an inadequate description.

Why Detection Depends on Expression

A gene can be expressed in a cell and still produce a recorded count of zero. When molecules are sampled to generate counts, a low underlying expression rate makes it more likely that none are sampled. Thus, even without a special “dropout” mechanism, the chance of observing a zero depends on abundance. This tutorial derives that relationship under Poisson sampling and shows what it does—and does not—allow us to conclude from an observed zero.

Here, detection means recording a count greater than zero for a gene in a cell. The term non-detection refers to recording zero. Non-detection is an observation about the count, not by itself a claim that the gene was biologically off. The model below describes sampling zeros conditional on an expression rate; it does not assert that every observed zero has this cause.

The Poisson Model and Its Detection Probability

Definition: Conditional on a fixed expression rate \(\lambda \geq 0\), a count \(X\) follows a Poisson distribution if $$ P(X=k\mid \lambda)=\frac{e^{-\lambda}\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ The rate \(\lambda\) is also the conditional expected count, measured in expected sampled counts per cell for the specified gene and sampling opportunity. Under this model, the probability of non-detection is \(P(X=0\mid\lambda)\), and the probability of detection is \(P(X>0\mid\lambda)\).

The Poisson model assumes that the rate is fixed for the cell and gene being considered, and that counts arise from events sampled independently at that rate over the specified sampling opportunity. In particular, the model does not include additional cell-to-cell rate variation, a distinct biological gene-off state, or another zero-generating mechanism. Those assumptions matter: the formula gives a conditional probability under this model, not a universal law for all single-cell count data.

Set \(k=0\) in the Poisson probability mass function. Since \(0!=1\) and \(\lambda^0=1\), the probability of a zero count is \(e^{-\lambda}\). Detection is the complementary event, so its probability is one minus that quantity.

$$ P(X=0\mid\lambda)=e^{-\lambda}, \qquad P(X>0\mid\lambda)=1-e^{-\lambda}. $$

These probabilities change smoothly with the rate. The derivative of the zero probability with respect to \(\lambda\) is \(-e^{-\lambda}\), which is negative for every finite \(\lambda\). The derivative of the detection probability is \(e^{-\lambda}>0\). Therefore, as the rate rises, non-detection becomes less likely and detection becomes more likely. Neither probability changes in equal steps for equal increases in rate: the slope of the detection curve becomes smaller as \(\lambda\) grows.

For example, at \(\lambda=0\), the model gives zero detection probability. At \(\lambda=1\), the detection probability is about 0.632. At \(\lambda=3\), it is about 0.950. A positive rate does not guarantee a positive observed count; rather, it determines the chance of one under the model.

Worked Example: Comparing Detection at Two Expression Rates

Suppose two hypothetical gene-cell combinations have fixed Poisson rates of \(\lambda=0.5\) and \(\lambda=2\) expected counts per cell. Calculate each probability of non-detection and detection.

For \(\lambda=0.5\), the zero probability is \(P(X=0\mid\lambda=0.5)=e^{-0.5}\approx0.6065\). Therefore, the detection probability is \(1-e^{-0.5}\approx1-0.6065=0.3935\).

For \(\lambda=2\), the zero probability is \(P(X=0\mid\lambda=2)=e^{-2}\approx0.1353\). The detection probability is \(1-e^{-2}\approx1-0.1353=0.8647\).

Under the stated model, the rate-two combination is much more likely to be detected than the rate-0.5 combination. This comparison is about model-based probabilities for the two rates; it does not establish why either gene has its rate or why any particular cell had a zero.

Rates for Target Detection Levels

The formula can also be rearranged to find the rate at which a chosen detection probability is reached. If the target probability of detection is \(q\), where \(0<q<1\), then \(q=1-e^{-\lambda}\). Rearranging gives \(e^{-\lambda}=1-q\), and taking natural logarithms gives the required rate.

$$ \lambda=-\ln(1-q). $$

For a 50% detection probability, the required rate is \(-\ln(0.5)\approx0.6931\) expected counts per cell. For a 95% detection probability, it is \(-\ln(0.05)\approx2.996\). These are not universal expression cutoffs: they are thresholds implied by the Poisson sampling model and the chosen target probability.

Worked Example: Finding a Rate for 90% Detection

A hypothetical analysis asks what fixed Poisson rate would yield a 90% chance of detecting a gene in a cell. Set \(q=0.90\) in the rearranged equation: \(\lambda=-\ln(1-0.90)=-\ln(0.10)\).

Since \(\ln(0.10)\approx-2.3026\), the required rate is \(\lambda\approx2.303\) expected counts per cell. Checking by substitution, \(P(X>0\mid\lambda=2.303)=1-e^{-2.303}\approx1-0.1000=0.9000\).

Thus, if the fixed-rate Poisson assumptions hold, a rate of about 2.303 expected counts per cell corresponds to a 90% detection probability. The rate is an expectation, not a guarantee that each cell will yield that many observed counts.

When Rates Vary Across Cells

The fixed-rate formula is conditional. If different cells have different rates for the same gene, the zero probability for a randomly selected cell is the average of the cell-specific zero probabilities. Let \(\Lambda\) denote the rate across cells. Conditional on each cell’s rate, assume \(X\mid\Lambda\) is Poisson. Then the overall non-detection probability is \(E[e^{-\Lambda}]\), not generally \(e^{-E[\Lambda]}\).

This distinction follows because the exponential function \(e^{-\lambda}\) is convex: its second derivative is \(e^{-\lambda}>0\). Jensen’s inequality therefore gives \(E[e^{-\Lambda}]\geq e^{-E[\Lambda]}\). In words, among cells with the same average rate, variation in rates can produce a larger average zero probability than a single fixed rate equal to that average. This statement concerns rate heterogeneity under conditional Poisson sampling; it does not identify the source of the heterogeneity.

Worked Example: Averaging Cell-Specific Detection Probabilities

Consider a hypothetical population in which half the cells have rate \(\lambda=0.5\) and half have rate \(\lambda=2\). Calculate the average non-detection probability and compare it with the probability obtained by treating the average rate as fixed.

The cell-specific zero probabilities are \(e^{-0.5}\approx0.6065\) and \(e^{-2}\approx0.1353\). Because the two rate groups are equally common, the average non-detection probability is \((0.6065+0.1353)/2=0.3709\). The average rate is \((0.5+2)/2=1.25\). A fixed-rate Poisson model at that average rate gives \(e^{-1.25}\approx0.2865\).

The heterogeneous population’s average zero probability, about 0.3709, exceeds 0.2865. Its average detection probability is \(1-0.3709=0.6291\), compared with \(1-0.2865=0.7135\) under the fixed rate of 1.25. Plugging the mean rate into the fixed-rate formula would therefore understate the average zero probability in this example.

Interpreting the Abundance-Dependent Pattern

The curve \(e^{-\lambda}\) describes a basic abundance-dependent pattern: low-rate genes or gene-cell combinations are more likely to have zero counts, while high-rate ones are less likely. If a gene has a positive rate, its zero is compatible with Poisson sampling. Conversely, a sequence of zeros at low estimated abundance does not, by itself, demonstrate an extra dropout process or a biological off state.

In practice, \(\lambda\) is usually a model-based expected count, not a quantity directly observed for each cell. A single observed count cannot reveal both a cell’s rate and whether its zero arose from ordinary sampling. To assess whether a Poisson abundance-dependent curve is useful, compare observed detection frequencies across groups with rates estimated independently or specified by a model, and check whether the pattern is broadly compatible with \(1-e^{-\lambda}\). Uncertainty in the rate estimates and variation among cells also affect that comparison.

A systematic mismatch can indicate that the fixed-rate Poisson assumptions are insufficient, but the mismatch does not name its cause. A negative binomial model allows more count variation than a Poisson model, while zero-inflated and hurdle models represent zeros with different structures. These are alternatives with different assumptions, not automatic explanations for a zero. The earlier material on excess zeros emphasizes why an observed surplus of zeros alone is not proof of zero inflation.

Diagnostics, Limitations, and a Common Mistake

A useful diagnostic is to examine detection frequency across ranges of modeled expression rate. The Poisson curve predicts a rising detection probability that approaches one as the rate increases. Persistent non-detection at rates where \(e^{-\lambda}\) predicts very few zeros, or a pattern inconsistent with the curve across rate ranges, is a warning that the fixed-rate model may not describe the data well. Before assigning a biological interpretation, check that rates and counts refer to the same sampling opportunity and that cell-to-cell heterogeneity has not been ignored.

The curve has important limits. It is conditional on the specified Poisson rate and does not separate biological expression from technical sampling. It does not account for uncertain rates, variable rates across cells, or distinct gene-off states. Nor does the formula establish that all observed zero counts should be labeled “dropout” in a mechanistic sense. It gives a precise baseline against which more complex explanations can be considered.

Common mistake: “The gene’s rate is positive, so a zero count proves technical dropout.” This reasoning confuses a positive expected count with a guaranteed observed count. Under Poisson sampling, a positive rate still gives \(P(X=0\mid\lambda)=e^{-\lambda}>0\). The correction is to treat the zero as compatible with sampling at that rate, then assess whether the model and its assumptions fit the broader pattern of counts.

A related mistake is calculating the zero probability at the average rate when rates vary across cells. The correct population probability is \(E[e^{-\Lambda}]\); substituting the mean rate gives \(e^{-E[\Lambda]}\), which can be smaller. The two expressions coincide when all cells have the same rate, but not generally when rates vary.

1
Specify the rate and sampling unit.
State what \(\lambda\) measures and the cell and sampling opportunity to which it applies.
2
Calculate the conditional probabilities.
Use \(e^{-\lambda}\) for non-detection and \(1-e^{-\lambda}\) for detection under the fixed-rate Poisson model.
3
Check the interpretation.
Ask whether rates are plausibly fixed within the comparison and whether the observed detection pattern is consistent with the model before attributing zeros to another process.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. For a fixed rate of \(\lambda=1.5\) expected counts per cell, calculate the probability of non-detection and the probability of detection.
  2. What rate corresponds to a 50% probability of detection under the fixed-rate Poisson model?
  3. In a population of cells with varying rates, why is the overall zero probability generally not the zero probability calculated at the average rate?
  4. A gene is recorded as zero in one cell even though its modeled rate is positive. Does this observation alone establish that the gene was biologically off? Explain.

Solutions. (1) \(P(X=0\mid\lambda=1.5)=e^{-1.5}\approx0.2231\); \(P(X>0\mid\lambda=1.5)=1-0.2231=0.7769\). (2) Solve \(0.50=1-e^{-\lambda}\), giving \(\lambda=-\ln(0.50)\approx0.6931\) expected counts per cell. (3) The overall probability is \(E[e^{-\Lambda}]\), the average of the cell-specific probabilities. Because \(e^{-\lambda}\) is convex, this is at least \(e^{-E[\Lambda]}\), and the inequality is generally strict when rates vary. (4) No. At any finite positive rate, \(e^{-\lambda}>0\), so a zero is compatible with Poisson sampling; one observation does not identify a biological state.

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