Deriving Delay Moments From Transit-Chain Rates

Why Transit-Chain Delay Has Moments

A biomarker need not appear as soon as its formation is affected. Material may pass through several maturation stages before reaching the measured pool. A transit chain represents this sequence with first-order transfer between compartments. Its stage rates determine not only how long appearance is delayed on average, but also how broadly appearance is distributed over time.

The earlier material on transit compartments as biomarker maturation chains establishes the compartment interpretation. Here the focus is the delay distribution implied by that chain. We treat an input pulse as the starting point and ask when its material reaches the final stage. This gives precise definitions of mean delay and variance that can also describe the timing kernel for a time-varying input.

The moment formulas below assume an irreversible, linear chain with constant positive rates, no branching or loss within the chain, and initially empty stages. Under these assumptions, each stage transfers all material onward, and the chain has a normalized appearance-time density. If an input varies over time rather than arriving as an impulse, the output is the input convolved with this density. Thus the chain acts as a delay-and-spread kernel.

Deriving the Delay Moments

For a unit pulse, each stage contributes a first-order waiting time. The mathematical representation is a sum \(T=T_1+\cdots+T_m\), where \(T_i\) is exponentially distributed with rate \(k_i\), and the stage waiting times are independent. Equivalently, the Laplace transform of the appearance-time density is the product of the transforms of the individual stages:

Because \(L(0)=1\), differentiating the Laplace transform at zero gives \(E[T]=-L'(0)\) and \(E[T^2]=L''(0)\). To obtain the variance cleanly, take the logarithm of the product:

At \(s=0\), \(L'(0)=L(0)\left.\frac{d}{ds}\log L(s)\right|_{s=0}=-\sum_i 1/k_i\), so the mean is the sum of the stage means. Differentiating once more gives \(\left.\frac{d^2}{ds^2}\log L(s)\right|_{s=0}=\sum_i 1/k_i^2\). Since \(L''/L=(\frac{d}{ds}\log L)^2+\frac{d^2}{ds^2}\log L\), it follows that \(E[T^2]=(\sum_i 1/k_i)^2+\sum_i 1/k_i^2\). Subtracting \(E[T]^2\) yields the variance.

Rates have units of inverse time. Therefore, \(E[T]\) has units of time, while \(\operatorname{Var}(T)\) has units of time squared. The standard deviation, \(\sqrt{\operatorname{Var}(T)}\), returns to time units and describes the scale of spread around the mean.

Equal-Rate Chains

When every stage has the same rate \(k\), the chain has an Erlang appearance-time density. For \(m\) stages this density is \(f_T(t)=k^m t^{m-1}e^{-kt}/(m-1)!\) for \(t\geq 0\). Applying the general moment formulas gives:

The coefficient of variation \(CV(T)=SD(T)/E[T]\) is dimensionless. For a fixed number of equal-rate stages, increasing \(k\) shortens both the mean and standard deviation in inverse proportion. For fixed mean delay and fixed \(m\), the equal-rate chain has the smallest variance among chains with different positive rates. Indeed, setting \(a_i=1/k_i\), the Cauchy–Schwarz inequality gives \((\sum_i a_i)^2\leq m\sum_i a_i^2\), or \(\sum_i a_i^2\geq E[T]^2/m\), with equality only when all \(a_i\) are equal.

Unequal Rates: Timing and Spread

With unequal rates, each stage contributes a different amount to both moments. A slow stage has a large mean waiting time \(1/k_i\) and an even larger variance contribution relative to its rate, \(1/k_i^2\). Consequently, a slow step can dominate the total delay and widen its distribution. The order of the rates does not alter these two moments: the formulas depend on the rates as a set, not on their sequence. For a linear chain, the transfer-function factors also multiply, so reordering the stages leaves the overall impulse-response kernel unchanged.

Moments summarize a distribution but do not completely describe its shape. For two unequal stages, the density can be written explicitly. If \(k_1\neq k_2\), then for \(t\geq0\),

Here \(F_T(t)=P(T\leq t)\) is the fraction of a unit pulse that has appeared by time \(t\). The expressions are written with \(k_2>k_1\); swapping the stage labels gives the same distribution. When the rates approach equality, the density approaches the two-stage equal-rate Erlang density rather than becoming singular.

Interpretation, Diagnostics, and Alternatives

The moment formulas are useful for translating rates into interpretable timing summaries and for checking whether a proposed rate combination implies a plausible delay. They also show why a single average delay can be incomplete: two chains may have the same mean but different variances and differently shaped appearance profiles. For equal rates, the Erlang form is convenient and compact. For unequal rates, the product form of the transfer function remains simple, but the density is generally not the same Erlang curve.

A pure time delay is a different approximation. It shifts an input by one fixed interval and has zero variance in delay; a transit chain spreads an input over time and has positive variance for finite positive rates. A pure delay may be adequate when the spread is negligible for the application, but it cannot represent gradual appearance. Conversely, a transit chain imposes a particular family of distributions and should not be assumed to capture every biological maturation process.

Potential warning signs include an observed appearance profile with a much longer tail, substantially different spread, or multiple separated peaks than the fitted chain predicts. Such discrepancies may reflect time-varying rates, branching, material loss, feedback, or more than one pathway, all of which violate the assumptions used in these moment formulas. Also check dimensional consistency: a reported mean in \(\text{h}^2\) or variance in hours signals an error. Moment calculations alone cannot establish that a chain is biologically correct; they describe the consequences of its assumed rates.