Why Track Moments of a Stochastic Cell Population?
A deterministic model summarizes a population with a typical trajectory. A stochastic model instead describes a distribution over possible cell counts. Its mean gives the expected count, while its variance describes the spread across realizations. These summaries can be easier to interpret and calculate than the full distribution, but they do not always evolve independently of it. The central issue in this tutorial is to determine exactly when equations for the mean and variance are self-contained—and when nonlinear state-dependent rates make them depend on still higher moments.
Consider a continuous-time Markov chain \(X(t)\) whose state is a nonnegative integer cell count. Birth events increase the count by one, with propensity \(\lambda(x)\) in state \(x\); death events decrease it by one, with propensity \(\mu(x)\). Both propensities must be nonnegative, and \(\mu(0)=0\) prevents a transition to a negative count. Assume the process is non-explosive and has the finite moments needed for the calculations below. Rates may be specified in units such as events per hour; a propensity is a rate conditional on the current state, not a probability.
Exact Mean and Variance Equations
In state \(x\), a birth changes \(X\) by \(+1\) and occurs at rate \(\lambda(x)\); a death changes it by \(-1\) and occurs at rate \(\mu(x)\). For any suitable function \(f\), the expected rate of change is therefore \[ \frac{d}{dt}E[f(X)]=E\left[\lambda(X)\{f(X+1)-f(X)\}+\mu(X)\{f(X-1)-f(X)\}\right]. \] This follows by considering the possible single jumps over a short time interval: each jump's change in \(f\) is multiplied by its state-dependent rate. Applying the identity to \(f(x)=x\) gives the mean equation; applying it to \(f(x)=x^2\) gives the second raw moment equation.
To obtain the variance equation, differentiate \(v=E[X^2]-m^2\), so that \(dv/dt=dE[X^2]/dt-2m\,dm/dt\). Rearranging the resulting terms gives a compact exact form:
The first term reflects how the net growth rate varies with the current count; the second reflects the variability contributed by individual jumps. These identities require no linearity assumption. However, their right-hand sides are expectations under the full distribution of \(X\), and may not be expressible using only \(m\) and \(v\).
Worked Example: Immigration and Linear Death
Suppose cells enter at a constant rate \(b=3\) cells per hour and each existing cell dies independently at rate \(\delta=0.4\) per hour. Thus \(\lambda(x)=3\) cells per hour and \(\mu(x)=0.4x\) cells per hour. Assume the initial count has mean \(m(0)=5\) cells and variance \(v(0)=2\) cells².
The mean equation is \(dm/dt=3-0.4m\), because \(E[0.4X]=0.4m\). Solving this linear equation gives
For the variance, \(\lambda(X)-\mu(X)=3-0.4X\), so its covariance with \(X\) is \(-0.4v\). Also, \(E[\lambda(X)+\mu(X)]=3+0.4m\). Hence
At time zero, \(dm/dt=3-0.4(5)=1\) cell per hour, and \(dv/dt=-0.8(2)+3+0.4(5)=3.4\) cells² per hour. The equations are closed: their right-hand sides use only the mean and variance (in this case the variance equation also uses the mean). No assumption about a particular distribution is needed beyond the stated moment conditions.
When Linear Propensities Give Closure
If both propensities are affine functions of the count, then the mean equation depends only on the mean, and the variance equation depends only on the mean and variance. For example, let \(\lambda(x)=b+gx\) and \(\mu(x)=dx\), where \(b\) is an immigration rate and \(g,d\) are per-cell rates. These rates must remain nonnegative on the states considered; with \(b,g,d\geq0\), that holds for every nonnegative count.
Worked Example: Instantaneous Moment Changes with Linear Rates
Let \(\lambda(x)=2+0.3x\) cells per hour and \(\mu(x)=0.5x\) cells per hour. At a particular time, suppose \(m=10\) cells and \(v=4\) cells². The affine propensities are nonnegative for all nonnegative \(x\), so the linear-rate equations apply.
Substitution into the mean equation gives
For the variance,
The mean is momentarily stationary, but the variance is increasing. A stationary mean does not imply that individual realizations or their distribution are stationary. These are instantaneous derivatives at the stated moments, not predictions that either derivative will remain constant over time.
Nonlinear Propensities Create a Moment Hierarchy
A nonlinear rate makes an expected rate depend on how counts are distributed, not just on their average. To see the consequence explicitly, consider \(\lambda(x)=rx\) and \(\mu(x)=\kappa x(x-1)\), where \(r\) and \(\kappa\) have units of inverse time. These are valid nonnegative birth and death propensities on nonnegative integer states, with no death at zero. The death rate represents state-dependent competition in this one-step model: each event decreases the count by one.
Since \(E[X^2]=m^2+v\), the exact mean equation becomes
Unlike the linear example, the mean equation already needs the variance. The variance equation needs a third central moment. Define \(\mu_3=E[(X-m)^3]\). Since \(E[X^3]=m^3+3mv+\mu_3\), the covariance term in the exact variance equation is \(\operatorname{Cov}(X,\lambda(X)-\mu(X))=(r+\kappa)v-\kappa(2mv+\mu_3)\). Also, \(E[\lambda(X)+\mu(X)]=(r-\kappa)m+\kappa(m^2+v)\). Combining these expressions yields
Thus the exact variance equation is not determined by \(m\) and \(v\) alone: it also contains \(\mu_3\). An equation for that third moment generally involves still higher moments. This sequence of dependencies is the moment hierarchy. It is not a defect in the derivation; it is a consequence of nonlinear propensities and the fact that a few moments do not, in general, determine the full state distribution.
Worked Example: Nonlinear Rates and the Third Moment
Use the nonlinear propensities above with \(r=1\) per hour and \(\kappa=0.2\) per hour. Suppose the count distribution at a particular time is \(P(X=0)=0.25\), \(P(X=1)=0.50\), and \(P(X=2)=0.25\). The mean is \(m=0(0.25)+1(0.50)+2(0.25)=1\) cell. The variance is \(0.25(0-1)^2+0.50(1-1)^2+0.25(2-1)^2=0.5\) cells², and symmetry about 1 gives \(\mu_3=0\) cells³.
The mean derivative is
For the variance, substitute the same moments into the exact equation:
The third central moment happens to be zero for this distribution, but that does not make the variance equation generally closed: a different distribution with the same mean and variance can have a different \(\mu_3\), and therefore a different variance derivative.
Closure, Diagnostics, and Alternatives
A moment closure replaces untracked higher moments with expressions involving tracked moments. For example, setting \(\mu_3=0\) in the nonlinear variance equation imposes a zero-third-central-moment approximation. It is an additional distributional assumption, not a consequence of the birth-death model. Other closure assumptions posit a particular distribution family and express higher moments through its parameters. Their usefulness depends on whether that family represents the actual distribution over the times and states of interest.
The moment equations are valuable because they can summarize average behavior and variability without tracking every state probability. For affine rates, the equations above provide exact moment dynamics. With nonlinear rates, possible alternatives include solving the master equation for state probabilities when the relevant state space is manageable, or estimating moments from stochastic simulations. A simulation estimate has sampling error; it does not make an approximate closure exact. The earlier material on deterministic limits explains how deterministic trajectories can arise under appropriate scaling, but a deterministic trajectory alone does not supply the finite-population variance or higher moments.
Several checks can reveal problems with a proposed closure or calculation:
- Verify that propensities are nonnegative and respect the boundary at \(X=0\). A formula that permits a positive death rate at zero is not a valid birth-death specification.
- Check units: \(dm/dt\) has units of cells per time, while \(dv/dt\) has units of cells² per time.
- Check that a computed variance remains nonnegative. A negative variance signals an invalid approximation or numerical problem, not a possible stochastic spread.
- Compare closure predictions with distribution-level calculations or simulations where feasible. Large disagreement, strong skewness, or substantial probability near the zero boundary can indicate that the assumed distribution shape is inadequate.
A common mistake is to replace \(E[X^2]\) by \(m^2\) because the mean is known, and then treat the resulting nonlinear mean equation as exact. In fact, \(E[X^2]=m^2+v\); the replacement silently assumes \(v=0\). A population can have a nonzero variance even when its mean is known precisely. The correction is to retain the variance in the exact equation or state and validate an explicit approximation for it.
Check Your Understanding
Work through these questions, then check the solutions below.
- For \(\lambda(x)=a\) and \(\mu(x)=\delta x\), write the exact mean equation. Does it require the variance?
- For a birth-death process with \(m=6\) cells, \(v=3\) cells², and \(\lambda(x)=0.2x\), \(\mu(x)=0.1x\), calculate \(dm/dt\).
- For the nonlinear propensities \(\lambda(x)=rx\), \(\mu(x)=\kappa x(x-1)\), which extra moment appears in the exact variance equation derived above?
- Why is replacing \(E[X^2]\) with \(m^2\) not an exact step for a stochastic population?
Solutions. 1. \(dm/dt=a-\delta m\); it is closed and does not require the variance. 2. \(dm/dt=(0.2-0.1)(6)=0.6\) cells per time unit. The variance is not needed for this affine-rate mean equation. 3. The third central moment \(\mu_3=E[(X-m)^3]\) appears. 4. The identity is \(E[X^2]=m^2+v\), so the replacement discards the variance and is exact only when \(v=0\).