The Poisson Model For Molecular Counts

Why Model Molecular Counts?

A single-cell assay records nonnegative integer counts: for example, the number of unique molecular identifiers (UMIs) assigned to a gene in a cell. Even when the underlying molecular abundance is stable, the number captured and recorded can vary from one sampling process to another. A probability model makes that sampling variation explicit. The Poisson model is a useful starting point because it describes counts generated by many independent, relatively rare capture events.

The model concerns the count produced by sampling, conditional on the cell’s abundance and sequencing exposure. It does not claim that cells have identical expression, or that all observed variation is caused by sampling. Those distinctions matter: a count can be Poisson-like under fixed conditions even when the conditions differ between cells.

Definition: A random count $Y_{ig}$ for gene $g$ in cell $i$ has a Poisson distribution with rate $\lambda_{ig}$ if $$ P(Y_{ig}=k\mid \lambda_{ig})= \frac{e^{-\lambda_{ig}}\lambda_{ig}^{k}}{k!}, \qquad k=0,1,2,\ldots, $$ where $\lambda_{ig}\geq 0$ is the expected captured count under the conditions being held fixed. In particular, $P(Y_{ig}=k\mid\lambda_{ig})$ is the probability of observing exactly $k$ molecules for that gene in that cell.

What the Sampling Assumptions Mean

A simple account of the Poisson model starts with a pool of $N$ molecules that could be captured. Suppose each molecule is captured independently with probability $p$, and a captured molecule is counted once. The captured count is then binomial with mean $Np$. When the pool is large, $p$ is small, and $Np$ stays near a fixed value, this count is well approximated by a Poisson random variable. The rate $\lambda$ represents the expected number captured, not the total number of molecules in the cell.

For a gene in a cell, the rate can depend on both molecular abundance and sampling exposure. One convenient representation is $\lambda_{ig}=s_iq_{ig}$, where $s_i$ is a nonnegative exposure or depth factor for cell $i$, and $q_{ig}$ is the expected count per unit exposure for gene $g$ in that cell. This representation is a way to express how deeper sampling can increase expected counts; it does not by itself specify how $q_{ig}$ is estimated.

The model is conditional: the stated rate is treated as fixed when calculating count probabilities. For the interpretation to be appropriate, the conditions represented by that rate must be sufficiently stable over the sampling process. A useful idealization is that individual molecules are captured independently, each contributes at most one counted molecule, and the gene and cell assignments are correct. Any exposure or cell-specific abundance that changes the expected count should be reflected in the rate or held fixed in the conditioning.

Conditional independence is a separate assumption, not a consequence of knowing the marginal distribution of each count. A common idealization is that, conditional on cell-specific rates and exposures, the capture outcomes for different genes are independent, and that capture outcomes across cells are independent if the cells’ sampling processes are independent. The assumption can be inappropriate when observations share unaccounted-for influences. Also, if one conditions on a fixed total number of reads or molecules allocated across genes, the gene counts are constrained to sum to that total and are not independent. Thus, independent Poisson counts are a sampling model, not a universal description of every sequencing design.

A Short Derivation: Rare-Event Sampling and the Moments

The connection to rare-event sampling follows by taking a limit of the binomial probability. Let $B$ be the number captured from $N$ molecules, each independently captured with probability $p=\lambda/N$. For fixed nonnegative integer $k$,

\begin{aligned} P(B=k) &=\binom{N}{k}\left(\frac{\lambda}{N}\right)^k \left(1-\frac{\lambda}{N}\right)^{N-k}\\ &=\frac{N(N-1)\cdots(N-k+1)}{N^k} \frac{\lambda^k}{k!} \left(1-\frac{\lambda}{N}\right)^N \left(1-\frac{\lambda}{N}\right)^{-k}\\ &\longrightarrow \frac{\lambda^k}{k!}e^{-\lambda} \qquad\text{as }N\to\infty. \end{aligned}

The first factor in the second line tends to 1, the factor $(1-\lambda/N)^N$ tends to $e^{-\lambda}$, and the final factor tends to 1. The limit is the Poisson probability formula.

The same formula gives the model’s mean and variance. For $\lambda>0$, summing over the possible counts,

\begin{aligned} E(Y) &=\sum_{k=1}^{\infty}k\,e^{-\lambda}\frac{\lambda^k}{k!} =\lambda,\\ E\{Y(Y-1)\} &=\sum_{k=2}^{\infty}k(k-1)e^{-\lambda}\frac{\lambda^k}{k!} =\lambda^2,\\ \operatorname{Var}(Y) &=E(Y^2)-[E(Y)]^2\\ &=E\{Y(Y-1)\}+E(Y)-[E(Y)]^2 =\lambda. \end{aligned}

When $\lambda=0$, the count is zero with probability 1, so its mean and variance are both zero as well. Therefore, conditional on the rate, the Poisson model has mean $\lambda$ molecules and variance $\lambda$ squared molecules. The equality is a property of the conditional model; it should not be read as a claim that all genes or cells have the same rate.

Worked Example: A Low Expected Count

Suppose a hypothetical gene in a hypothetical cell has expected captured count $\lambda=0.25$ UMIs under the sampling conditions. Find the probabilities of zero, one, and at least two UMIs.

Using the Poisson formula, the probability of zero is $$ P(Y=0)=e^{-0.25}\frac{0.25^0}{0!}=e^{-0.25}\approx 0.7788. $$

The probability of one is $$ P(Y=1)=e^{-0.25}\frac{0.25^1}{1!} =0.7788(0.25)\approx 0.1947. $$

Since the count must be a nonnegative integer, the probability of at least two is the complement of the probabilities of zero and one: $$ P(Y\geq2)=1-P(Y=0)-P(Y=1) =1-0.7788-0.1947\approx 0.0265. $$

Thus, under this hypothetical rate, a zero count is common even though the expected count is positive: its probability is about 77.88%. An observed zero alone does not establish that the gene is biologically absent; it is compatible with low expected abundance and sampling. The expected count is an average over repetitions under the model, not a promise that a particular cell will yield a fractional count.

Worked Example: A Higher Expected Count

For a different hypothetical gene-cell combination, suppose $\lambda=16$ captured UMIs. Calculate the model’s standard deviation and the probability of observing exactly 16 UMIs.

The Poisson variance is $\lambda$, so the standard deviation is $$ \sqrt{\operatorname{Var}(Y)}=\sqrt{16}=4\text{ UMIs}. $$

For exactly 16 UMIs, $$ P(Y=16)=e^{-16}\frac{16^{16}}{16!} \approx 0.0992. $$

The expected count is 16 UMIs, but the probability of observing exactly that count is only about 9.92%. Counts on either side of 16 are possible; the model’s standard deviation of 4 UMIs describes the scale of its sampling variation, not a bound on possible outcomes. A high expected count can therefore still yield a range of integer observations.

Worked Example: Changing Exposure Changes the Rate

Suppose a hypothetical gene has $q_g=0.03$ expected UMIs per exposure unit, and the rate is $\lambda_i=s_iq_g$. Compare cells with exposure factors $s_i=10$ and $s_i=100$, assuming the same $q_g$.

At exposure 10, the rate is $$ \lambda_{10}=10(0.03)=0.30\text{ UMIs}, $$ so the probability of a zero is $$ P(Y=0\mid\lambda_{10})=e^{-0.30}\approx0.7408. $$

At exposure 100, the rate is $$ \lambda_{100}=100(0.03)=3.00\text{ UMIs}, $$ and the zero probability is $$ P(Y=0\mid\lambda_{100})=e^{-3.00}\approx0.04979. $$

Under this model, the tenfold increase in exposure produces a tenfold increase in expected count, from 0.30 to 3.00 UMIs. The zero probability falls from about 74.08% to about 4.979%. This comparison holds the per-unit rate fixed; it does not show that the cells have identical biological states in general.

Independence and the Main Alternatives

For a single gene and cell, the Poisson model specifies a distribution for one count. To model a matrix of counts, one must additionally say which rates are conditioned on and whether counts are independent after conditioning. Under a conditional independence model, the joint probability for a collection of counts is the product of the separate Poisson probabilities. For two genes $g$ and $h$ in cell $i$, for example,

P(Y_{ig}=a,Y_{ih}=b\mid\lambda_{ig},\lambda_{ih}) = P(Y_{ig}=a\mid\lambda_{ig}) P(Y_{ih}=b\mid\lambda_{ih}).

This factorization is justified only under the stated conditional independence assumption. It should not be inferred merely because each count separately has a Poisson distribution. Nor does it say counts from different cells are independent if they share a source of variation that has not been represented in the conditioning.

A binomial model is a natural alternative when the number of opportunities is finite and known and each opportunity has a specified capture probability. The Poisson distribution is a limiting approximation in the large-pool, small-probability setting described above. If the total number allocated across genes is fixed, a multinomial model expresses that fixed-total constraint; its component counts are dependent. Independent Poisson counts may be a convenient approximation when the total is not fixed in that way, but the sampling design determines which description is appropriate.

Where the Model Can Mislead

The Poisson model is an explicit account of conditional sampling noise, not a complete biological model of all single-cell count variation. If abundance differs across cells, then a single common rate does not describe those cells unless that variation is represented in the rates. Likewise, uncertainty about exposure, changes in capture conditions, assignment errors, or a fixed-total design can make the simple assumptions unsuitable. The conditional mean-variance equality derived above applies at a fixed rate; it need not describe a collection of counts whose rates vary.

A practical first check is to ask whether the rate definition and conditioning match the data-generating process: Is the count a molecule count or a read count? Is exposure treated consistently? Are the relevant cell- and gene-specific conditions held fixed? Is there a fixed total that couples the gene counts? Such questions clarify what the model claims before any formal assessment of fit. Evidence that observed variation departs from the conditional Poisson model calls for further assessment rather than automatic interpretation as biological absence or a particular biological mechanism.

Common pitfall: “The Poisson model says the mean equals the variance, so every gene should have the same mean and variance across all cells.” This confuses a conditional statement with a pooled one. The model says that for a count with a specified fixed rate $\lambda$, its conditional mean and variance are both $\lambda$. Different genes or cells can have different rates, and pooling counts with different rates changes what the overall mean and variance represent.

Check Your Understanding

Answer each question using the conditional Poisson model. Solutions follow the questions.

For a count with rate $\lambda=0.5$, calculate the probability of zero counts to three significant figures.
State the mean, variance, and standard deviation for a Poisson count with rate $\lambda=9$, including the count units where appropriate.
Why does a fixed total number of reads allocated across genes generally conflict with conditional independence of the gene counts?
A count is zero in one cell. Does that prove the gene is absent from the cell? Explain using the model.
In one sentence, state the sampling conditions that motivate a Poisson approximation to a binomial count.

Solutions. (1) $P(Y=0)=e^{-0.5}\approx0.607$. (2) The mean is 9 counts, the variance is 9 counts squared, and the standard deviation is $\sqrt{9}=3$ counts. (3) If the total is fixed, an increase in one gene’s allocation leaves fewer reads available for other genes, so the counts are constrained and dependent. (4) No. With any positive rate, the Poisson model assigns a positive probability $e^{-\lambda}$ to zero; a zero observation is not proof of biological absence. (5) The approximation is motivated by many independent capture opportunities, each with a small probability of capture, while the product of the number of opportunities and capture probability remains approximately fixed.