Convolution Formulations for Distributed Delays

From a Delay Distribution to a Model Input

A fixed delay assigns the same transit time to every unit of material or signal. In many mechanistic settings, transit times vary across cells, molecules, or other units. A distributed-delay formulation represents that variation by weighting past input according to the distribution of elapsed times before output occurs. The result is an average over past input, rather than a value taken only at one lag.

Let \(u(t)\) be the input rate into a process at time \(t\), and let \(y(t)\) be the rate at which the corresponding output appears. Suppose each input unit experiences a random nonnegative delay with density \(g(\tau)\), where \(\tau\) is the elapsed time between input and output. If the delay distribution does not change with calendar time and the process combines input cohorts linearly, then output at time \(t\) is a convolution:

y(t) = \int_0^\infty g(\tau)u(t-\tau)\,d\tau.

Here \(g(\tau)\) is a probability density over delay, not a rate of input or a cumulative probability. The integral adds contributions from all past input times \(t-\tau\), weighted by the fraction of each cohort expected to emerge at delay \(\tau\). If the model begins at time zero, specify \(u(s)=0\) for \(s<0\) when there was no earlier input; if input before time zero matters, its history must instead be supplied.

Definition: A delay kernel \(g(\tau)\) is a nonnegative density for the delay \(\tau\geq 0\), normalized so that \(\int_0^\infty g(\tau)\,d\tau=1\). For a time-invariant, linear process, the delayed output rate is \(y(t)=\int_0^\infty g(\tau)u(t-\tau)\,d\tau\). The kernel has units of inverse time, so \(g(\tau)d\tau\) is dimensionless.

If \(u\) has units of amount per hour, for example, then \(g\) has units of inverse hours and \(u(t-\tau)g(\tau)d\tau\) has units of amount per hour. Normalization means that every input unit eventually emerges in this output stream. If some units are lost, diverted, or remain permanently in the process, a normalized delay density alone does not represent that loss; a yield or loss mechanism must be included separately.

Why the Convolution Has This Form

Consider the input delivered during a short interval around time \(s\). Its amount is approximately \(u(s)\,ds\). The fraction of that cohort emerging with delay between \(\tau\) and \(\tau+d\tau\) is approximately \(g(\tau)d\tau\). It therefore contributes approximately \(u(s)ds\,g(\tau)d\tau\) to output over the corresponding time interval around \(s+\tau\). For output at a fixed time \(t\), cohorts with delay \(\tau\) entered at \(s=t-\tau\). Summing their contributions and taking the continuous limit gives the integral above.

This argument requires a linear superposition of cohorts: doubling input at any past time doubles that cohort's contribution, and one cohort does not alter another cohort's delay distribution. The kernel must also be nonnegative and integrable. For the usual interpretation as a probability density, it must integrate to one. Together, these assumptions make the convolution a mechanistic description of a stationary delay distribution, not a universal form for every process with memory.

Interpretation: The cumulative distribution \(G(a)=\int_0^a g(\tau)\,d\tau\) is the fraction of an input cohort expected to have emerged by elapsed time \(a\). The density \(g(a)\), by contrast, describes how concentrated emergence is around delay \(a\). Do not substitute the cumulative distribution for the density in the convolution for an output rate.

Uniform Delay Kernel

A uniform kernel assigns equal density to all delays in a specified interval \([a,b]\), where \(0\leq a<b\). It rules out delays outside that interval and gives equal weight per unit of time inside it:

g(\tau)= \begin{cases} \dfrac{1}{b-a}, & a\leq \tau\leq b,\\ 0, & \text{otherwise}. \end{cases}

The density has units of inverse time, since \(b-a\) is a duration. The mean delay is \((a+b)/2\), and the variance is \((b-a)^2/12\). A uniform kernel is useful when a process has a plausible minimum and maximum delay and a simple approximation treats all delays between them as equally likely. That assumption is substantive: it is not implied merely by knowing the range of possible delays.

Worked Example: A Uniform Window of Delays

Suppose input enters a hypothetical process at a constant rate of \(6\) units per hour from time zero onward. The delay is uniform between \(2\) and \(4\) hours. Find the output rate at \(t=3\) hours.

The kernel is \(g(\tau)=1/(4-2)=0.5\) per hour for \(2\leq\tau\leq4\), and zero otherwise. In the convolution, only delays with \(t-\tau\geq0\) have nonzero input. At \(t=3\), the contributing delay interval is therefore \(2\leq\tau\leq3\):

y(3)=\int_2^3 (0.5\ \mathrm{hour}^{-1})(6\ \mathrm{units/hour})\,d\tau =3\ \mathrm{units/hour}.

The integration interval is one hour, so the result is \(0.5\times6\times1=3\) units per hour. At this time, delays between three and four hours cannot yet contribute because their associated input times are before time zero. Once the input has been constant for at least four hours, the full kernel contributes and the output rate equals the input rate.

Exponential Delay Kernel

An exponential kernel has a single positive rate parameter \(\lambda\), measured in inverse time:

g(\tau)=\lambda e^{-\lambda\tau},\qquad \tau\geq0.

It is normalized because \(\int_0^\infty\lambda e^{-\lambda\tau}\,d\tau=1\). Its mean delay is \(1/\lambda\), and its variance is \(1/\lambda^2\). The density is largest at zero and declines thereafter. This kernel has the memoryless property: conditional on not yet emerging, the remaining delay has the same exponential distribution. This may be a reasonable assumption for some mechanisms, but it is not a generic feature of delay processes.

Worked Example: Step Input With an Exponential Kernel

Consider a hypothetical input that switches on at time zero and remains at \(8\) units per hour. Let delays follow an exponential distribution with \(\lambda=0.5\) per hour. Calculate the output rate at \(t=2\) hours.

For \(\tau>2\), the input time \(2-\tau\) is negative and the input is zero. Thus:

\begin{aligned} y(2) &=\int_0^2 (0.5\ \mathrm{hour}^{-1})e^{-(0.5\ \mathrm{hour}^{-1})\tau} (8\ \mathrm{units/hour})\,d\tau\\ &=8\left(1-e^{-0.5(2)}\right)\ \mathrm{units/hour}\\ &=8(1-e^{-1})\ \mathrm{units/hour}\\ &\approx 5.057\ \mathrm{units/hour}. \end{aligned}

The value is below the eventual output rate of \(8\) units per hour because some input cohorts have not yet emerged. The mean delay is \(1/0.5=2\) hours, but this does not mean every unit emerges at two hours; the kernel spreads emergence across all nonnegative delays.

Gamma Delay Kernels

A gamma kernel can represent a delay distribution with a range of shapes. With shape \(k>0\) and rate \(\lambda>0\), it is:

g(\tau)=\frac{\lambda^k\tau^{k-1}e^{-\lambda\tau}}{\Gamma(k)},\qquad \tau\geq0,

where \(\Gamma(k)\) is the gamma function. The rate \(\lambda\) has units of inverse time, and the density has units of inverse time. The mean delay is \(k/\lambda\), and its variance is \(k/\lambda^2\). When \(k=1\), this reduces to the exponential kernel. For \(k>1\), the density is zero at zero and typically concentrates around positive delays, unlike the exponential density, which is highest at zero.

Worked Example: Output From a Brief Input Under a Gamma Kernel

Suppose a hypothetical process receives a brief input of \(12\) units at time zero, represented as an impulse with total amount \(12\) units. Let the delay have a gamma distribution with shape \(k=2\) and rate \(\lambda=1\) per day. Find the output rate at one day.

For shape two, \(\Gamma(2)=1\), so the kernel is \(g(\tau)=\lambda^2\tau e^{-\lambda\tau}\). With \(\lambda=1\) per day, and \(\tau\) measured in days, its numerical value is \(g(\tau)=\tau e^{-\tau}\) per day. In particular, at one day:

g(1\ \mathrm{day})=(1\ \mathrm{day}^{-1})^2(1\ \mathrm{day})e^{-1} =\frac{e^{-1}}{\mathrm{day}}.

The impulse contributes its total amount times the delay density:

y(1\ \mathrm{day})=12\ \mathrm{units}\times\frac{e^{-1}}{\mathrm{day}} \approx4.415\ \mathrm{units/day}.

The result is a rate, not an amount. The gamma density integrates to one, so the total output over all time is \(12\) units. Its mean delay is \(2\) days and its variance is \(2\) days squared. The density’s units are inverse days, as required for an impulse amount multiplied by the kernel to produce units per day.

Assumptions, Checks, and Alternatives

The convolution is most appropriate when the delay distribution is stable over time, input cohorts combine linearly, and the output of interest is the delayed version of the input. If the chance or timing of emergence depends on a changing state, concentration, or calendar time, a fixed kernel \(g(\tau)\) may be inadequate. Such mechanisms may require a time-varying kernel or additional state equations. Similarly, if output is only a fraction of input, include that yield explicitly rather than silently interpreting a sub-normalized function as a probability density.

Several checks help catch formulation errors:

Check normalization: verify \(g(\tau)\geq0\) and \(\int_0^\infty g(\tau)d\tau=1\) for a conserved process.
Check units: \(g\) must have inverse-time units, and \(g(\tau)d\tau\) must be dimensionless.
Check causality: input at \(t-\tau\) is past input for \(\tau\geq0\); do not include future input.
Check limiting behavior: for sustained constant input long enough to cover the kernel, a normalized kernel should return the same steady output rate.
Check support and history: a kernel with no mass before a minimum delay cannot produce output earlier than that delay, and missing pre-start input must not be mistaken for zero unless that is a model assumption.

A fixed-delay formulation is the main alternative when the mechanism is well represented by one common lag; it samples input at a single past time rather than averaging over past times. A distributed kernel is preferable when spread in delay is mechanistically important. Explicit state models are another option when the internal process or state-dependent rates need to be represented directly. The convolution is compact and interpretable, but it can require storing input history and does not by itself describe the internal state of the delayed process.

A frequent mistake is to multiply past input by the cumulative delay probability \(G(\tau)\) and integrate as though it were a density. That weights each past time by the fraction already emerged by that delay, rather than by the probability of emerging near that delay. The correction is to use the density \(g(\tau)\) for output rate, or to derive a cumulative-output quantity deliberately when that is the target.

Specify the input and target.
State whether the input and output are rates or amounts, and define the time origin and any pre-existing input history.

Choose and check the kernel.
Give its support and parameters, confirm nonnegativity and normalization, and verify the parameter units.

Evaluate the convolution.
Integrate over delays that have corresponding input history, then check the output units and limiting behavior.

Check Your Understanding

Try each question, then compare your work with the solutions.

A delay is uniform from \(1\) to \(3\) hours. What is the kernel density inside its support, and what are its mean delay and variance?
An input of \(5\) units per day switches on at time zero. Delays are exponential with rate \(1\) per day. What is the output rate at one day?
For a gamma kernel with shape \(2\) and rate \(2\) per hour, state the mean delay and the units of the density.
Why can a normalized kernel give an output rate below a constant input rate shortly after the input begins?

Solutions. (1) The density is \(1/(3-1)=0.5\) per hour on \([1,3]\). The mean is \((1+3)/2=2\) hours, and the variance is \((3-1)^2/12=1/3\) hour squared. (2) Only delays from zero to one day have nonzero input history, so \(y(1)=5\int_0^1e^{-\tau}\,d\tau=5(1-e^{-1})\approx3.161\) units per day. (3) The mean is \(k/\lambda=2/(2\ \mathrm{hour}^{-1})=1\) hour; the density has units of inverse hours. (4) Some of the probability mass is assigned to delays that have not elapsed yet, so those input cohorts have not contributed to output at that time.