Why a Sequence of Compartments Creates a Distribution of Transit Times
A cell population may be represented as moving through ordered maturation states. Even if every cell follows the same sequence, cells need not spend exactly the same time in each state. When the exit from each compartment is a constant-rate process, the resulting total time is distributed rather than fixed. This distinction matters when predicting the timing of downstream cell appearance: a chain can represent a gradual delay and spread of response, not just a single average delay.
Here, residence time means the elapsed time from a cell’s entry into the first compartment until it exits the last. We consider \(n\) sequential compartments. Each has the same constant exit rate \(k\), measured in inverse time, and exit from one compartment leads directly to the next. The focus is the transit-time distribution for one cell moving along this route, not the later consequences of cell division or lineage branching.
The assumptions are substantive. The exit rate \(k\) is constant over time and the same in every compartment. Every cell traverses the compartments in order, with no bypass, branching, return to an earlier state, or competing loss. Conditional on entering a compartment, its dwell time follows the exponential waiting-time model; the successive dwell times are independent. The model describes transit for a cell that completes the route. If cells can die or leave by another route, the completed-transit distribution alone does not describe all outcomes.
Distribution and Moments of Residence Time
Let \(T_i\) be the dwell time in compartment \(i\), for \(i=1,\ldots,n\). A constant exit rate gives \(T_i\sim\operatorname{Exponential}(k)\), with density \(k e^{-kt}\) for \(t\geq 0\), mean \(1/k\), and variance \(1/k^2\). Since total residence time is the sum \(T=T_1+\cdots+T_n\), the distribution of \(T\) is an Erlang distribution: the gamma distribution with integer shape \(n\) and rate \(k\).
The density has units of inverse time, as a probability density for a time variable should. Its cumulative distribution function is \(P(T\leq t)\). The complementary probability, \(P(T>t)\), is especially useful for asking what fraction of cells have not yet completed the chain by time \(t\).
A compact derivation of the moments uses the moment-generating function. For a single exponential dwell time, the moment-generating function is \(k/(k-s)\), for \(s<k\). Independence means that the moment-generating function of a sum is the product of the individual moment-generating functions.
Taking the logarithm gives \(\log M_T(s)=n[\log(k)-\log(k-s)]\). Its first derivative at \(s=0\) is \(n/k\), the mean. Its second derivative at \(s=0\) is \(n/k^2\), the variance of \(T\), because the second derivative of the log moment-generating function at zero is the variance.
The mean adds across compartments, as does the variance of independent dwell times. The coefficient of variation (standard deviation divided by mean) describes spread relative to the mean. It falls as \(n\) increases, but this statement must be interpreted carefully: it applies to a comparison in which the compartment exit rate is fixed. In that comparison, adding compartments also increases the mean transit time.
Worked Calculations
Worked Example: Mean and Spread With Three Compartments
Suppose a hypothetical cell passes through \(n=3\) sequential compartments, each with exit rate \(k=0.40\) per day. Assume the equal-rate, independent exponential dwell-time model applies. Calculate the mean residence time, variance, standard deviation, and coefficient of variation.
The mean is the number of stages divided by the rate:
The variance and standard deviation are:
The coefficient of variation is \(SD(T)/E[T]=4.33/7.50=0.577\), which also equals \(1/\sqrt{3}\). Thus the model predicts a mean transit time of 7.50 days, with substantial cell-to-cell variation in transit time despite identical compartment rates.
Worked Example: Comparing Chain Lengths at the Same Mean
Consider hypothetical chains with mean residence time fixed at 8 days. Compare chains with \(n=1\), \(n=2\), and \(n=4\) compartments. To keep the mean fixed, choose \(k=n/8\) per day for each chain. This is a different comparison from holding the exit rate fixed.
For \(n=1\), \(k=0.125\) per day and the variance is \(1/(0.125)^2=64\ \text{days}^2\). For \(n=2\), \(k=0.25\) per day and the variance is \(2/(0.25)^2=32\ \text{days}^2\). For \(n=4\), \(k=0.50\) per day and the variance is \(4/(0.50)^2=16\ \text{days}^2\). The standard deviations are 8.00, 5.66, and 4.00 days, respectively.
The chains have the same mean, but the longer chain is less variable relative to that mean. For example, the probability of still being in the chain after 8 days for the four-compartment model is:
For the one-compartment model, \(P(T>8)=e^{-1}\approx0.3679\). Thus the four-compartment model has a larger fraction of transit times beyond the common mean in this particular comparison, while its overall spread is narrower. This is not a contradiction: the distributions have different shapes, and a single tail probability does not summarize all aspects of spread.
Worked Example: Inferring Chain Parameters From Theoretical Moments
Suppose hypothetical transit-time measurements are summarized by a mean of 6 days and a variance of 9 days squared. If an equal-rate chain is assumed to describe the population exactly, what chain length and rate correspond to those moments?
From \(E[T]=n/k\) and \(\operatorname{Var}(T)=n/k^2\), dividing the squared mean by the variance gives \(n\):
Then \(k=n/E[T]=4/(6\ \text{days})=0.667\) per day. Equivalently, \(k=E[T]/\operatorname{Var}(T)=6/9=0.667\) per day. These calculations recover the parameters implied by the assumed model and exact moments; in actual data, sampling uncertainty and departures from the model mean that these moment-based values should not be treated as certain.
How Chain Length Changes the Distribution
At a fixed \(k\), increasing \(n\) raises both mean and variance: \(E[T]=n/k\) and \(\operatorname{Var}(T)=n/k^2\). It also lowers the coefficient of variation, \(1/\sqrt{n}\). The total time becomes less dispersed relative to its longer mean, but its absolute variance does not shrink.
At a fixed mean \(m\), the rate must instead be set to \(k=n/m\). Substitution into the variance gives:
Under this matched-mean comparison, increasing the number of compartments reduces both the absolute variance and relative spread. The distribution becomes more concentrated around the common mean. In the limit of many stages with the mean held fixed, the relative spread approaches zero. A one-stage chain, by contrast, is exponential and has standard deviation equal to its mean.
This comparison helps distinguish a chain’s two roles. A chain length can represent the number of sequential maturation steps, while its rates determine the time scale. Changing \(n\) without adjusting \(k\) changes both the number of steps and the mean delay. Adjusting \(k\) to preserve the mean isolates the effect of adding stages on the shape and spread of transit times.
Diagnostics, Alternatives, and Limitations
The equal-rate Erlang model is useful when transit is sequential and the available information supports a common first-order rate. It is not a universal description of maturation. Before interpreting its moments, check whether the proposed stages are genuinely ordered, whether exit rates can reasonably be treated as constant, and whether the observed outcome reflects completed transit rather than a mixture of completion and competing cell loss.
- Unequal stage rates: If different compartments have different exponential exit rates, the dwell times are not identically distributed. The total is generally not Erlang, and the equal-rate formulas do not apply.
- Non-exponential dwell times: If dwell times have a more concentrated or otherwise different shape than exponential waiting times, a constant-rate compartment is a poor representation of that stage.
- Alternative routes or loss: Bypass, branching, reversal, or death changes which paths are possible and which cells contribute to the observed transit-time distribution.
- Rates that vary over time: A time-varying exit process does not have the constant-rate exponential dwell-time distribution assumed here.
A useful diagnostic is to compare observed transit-time summaries with model implications. Under the equal-rate model, the coefficient of variation is \(1/\sqrt{n}\), and the moment ratio \(E[T]^2/\operatorname{Var}(T)\) equals the integer \(n\). A substantially different ratio, or a fitted non-integer value when using moments as an approximation, can signal sampling noise or model mismatch; it does not by itself identify the biological cause. When individual transit-time data are available, the distribution’s shape and tail behavior should also be checked, rather than relying only on its mean and variance.
A fixed-delay model is an alternative when transit is assumed to take nearly the same time for every cell; it has no residence-time spread. A single exponential stage is appropriate when one constant-rate waiting time is a reasonable description, but it has a coefficient of variation of one. An equal-rate multi-stage chain produces a more concentrated distribution relative to its mean. Unequal-rate stages or other dwell-time distributions may be more appropriate when stage-specific evidence supports them. No choice should be made solely because a model is mathematically convenient: it should reflect the intended biology and be assessed against relevant data.
A common mistake is to say, “Adding compartments reduces the mean residence time because the chain becomes more precise.” That conclusion confuses two comparisons. At fixed \(k\), the mean is \(n/k\) and increases with chain length. Only when \(k\) is increased along with \(n\) to hold the mean fixed does the variance decrease as \(m^2/n\). Always state which quantity is held constant before comparing chains.
Check Your Understanding
Answer each question, then compare your work with the solutions.
- A hypothetical five-stage chain has a common exit rate of 0.25 per day. Calculate its mean, variance, standard deviation, and coefficient of variation.
- A four-stage chain has a mean residence time of 10 days. What rate gives that mean, and what are its variance and standard deviation?
- Two chains both have a mean residence time of 10 days, but one has two stages and the other has eight. Which has the larger variance, and what are the two variances?
- Name two assumptions that would be violated if cells could bypass a stage or die before completing the chain. Explain why the equal-rate transit-time result would then need reconsideration.
Solutions. (1) \(E[T]=5/0.25=20\) days; \(\operatorname{Var}(T)=5/(0.25)^2=80\ \text{days}^2\); \(SD(T)=\sqrt{80}=8.94\) days; \(CV=1/\sqrt{5}=0.447\). (2) \(k=n/E[T]=4/10=0.400\) per day; the variance is \(4/(0.400)^2=25.0\ \text{days}^2\), and the standard deviation is 5.00 days. (3) At fixed mean \(m=10\) days, variance is \(m^2/n\). The two-stage chain has variance \(100/2=50\ \text{days}^2\); the eight-stage chain has variance \(100/8=12.5\ \text{days}^2\). The two-stage chain is more variable. (4) Bypass violates the assumption that every cell traverses the same ordered sequence; death violates the assumption that the modeled transit path has no competing loss. The distribution of completed transit times may then describe only a selected subset of cells, and the equal-rate Erlang result need not represent all cells or outcomes.