Exponential waiting time

Why do we assume that the waiting time in a compartmental model is exponentially distributed?
Tutorial
Modeling
Beginner
Published

Monday, July 1, 2024

Introduction

When you are diving in to a mathematical modeling paper with compartmental modeling, the authors may state that the waiting time within a compartment is exponentially distributed. This only comes up when the paper is pointing out how biologically unrealistic it is (another blog post). When I was an undergraduate, I was always puzzled by this! This concept of exponentially distributed dwelling-time is not intuitive without understanding some ordinary differential equation! Imagine a single compartment which we simply call A. The outflow out of the compartment is dependent on a constant rate \(k\) and the amount of individuals in A. So the easiest way to write this is:

\[\frac{dA}{dt} = -kA. \]

This equation means that the change in \(A\) over time is equal to \(kA\). This is a differential equation that we can actually solve analytically!

This is fairly simple to solve and let me show you the steps (honestly, I really recommend doing this yourself as practice!)

Imagine you have a stage A and there are individuals leaving it (note the hats and briefcases) and on the left, that is the solution of our differential equations. This is an exponential decay!

So now okay let’s look closer at the solution \(A= A_0 exp(-kt)\). First, we’re interested in how the proportion of individuals that stay in stage A change over time. To do this, we can divide the equation by A0 and get this:

\[\frac{A}{A_0} = exp(-kt). \]

This means that the proportion of individuals who are still in stage \(A\) can be represented by this exponential decay. Okay, what about the proportion leaving stage A? If 30 percent of the individuals are in stage A that means 70% left so the proportion leaving so \(1 - \frac{A}{A_0}\) thus we get:

\[ 1- \frac{A}{A0} = 1 -exp(-kt).\]

Hm, that’s interesting! That looks exactly like the cumulative distribution function (CDF) of the exponential distribution! Remember that to get the probability distribution function (PDF), we just have find the derivative of the CDF:

\[\frac{d}{dt} (1- \frac{A}{A0}) = \frac{d}{dt}(1 -exp(-kt)). \]

Which means that the PDF is \(k exp(-kt)\)

And if you remembered the mean of the exponential PDF is \(\frac{1}{k}\) (We could have found the mean of the CDF, but I just thought it was easier).

That means that the mean time that individuals dwell in A is \(\frac{1}{k}\).

Therefore, for example if \(k = \frac{1}{2}\) per day, then the mean time that an individual stay in \(A\) is \(\frac{1}{(1/2)}\) is 2 days.

Hopefully, I hope this makes it clear using differential equations that there is an exponential distribution. The exponentially distributed waiting time is an assumption! It can be biologically unrealistic because it assumes that that most individuals leave the compartment almost instantly.