Sojourn times in different compartments

How long does an individual spend in an infected state or compartment? This duration, also known as the sojourn time, isn’t a fixed value, so it’s important to know the distribution of these sojourn times. In Markov-Chain-Monte-Carlo simulations like ours, the sojourn times are exponentially distributed. First we will explain why the distribution is exponential, which can allow us to change our algorithm for non-exponential cases.

We discretise time into small intervals or time-steps of $\Delta t$, so that the total time $T = N \Delta t$, where $N$ is the total number of time-steps. In every small step $\Delta t$ an individual in our disease model has a small probability, $p$, of exiting the compartment. The value of $p$ depends on factors such as the number of individuals in that compartment at that time, and so on.

To simulate $p$, we draw a random number $r$ from a uniform distribution. If $r < p$, the individual exits the compartment. This is equivalent to saying:

\[\begin{equation}\texttt{The probability of exiting the compartment is p}.\end{equation}\]

This is the only role of the uniform distribution here.

Now, we ask ourself the following question: what is the probability that an individual will leave a compartment at some time $t$? In our scheme, this is equivalent to asking what the probability is of the event occurring between time $t$ and $t + \Delta t$. We will call this probability $P(t)\Delta t$. We’ve also discretized time into units of $\Delta t$, so we can define $t = n\Delta t$, where $n$ is the number of steps of $\Delta t$.

This allows us to reframe the problem as this: what is the probability that the individual will leave the compartment after exactly $n$ steps? This is given by:

\[\text{Prob. that event occurs exactly after $n$ steps} = (1-p)^n p.\]

Using the fact that $p = \lambda \Delta t$, and that $\Delta t = T/N$, we can show that:

\[\begin{split} \begin{align} \mathcal{P}_{\Delta t}(t)\dd t &= (1 - \lambda \Delta t)^{t/\Delta t} \lambda \Delta t\\ &= \left(1 - \frac{\lambda T}{N} \right)^{t N/T} \lambda \Delta t \end{align} \end{split}\]

To find the true probability density, we need to take the limit $N\to\infty$, i.e. $N/T \to \infty$, and so

\[\mathcal{P}(t)\dd t = \lim_{N/T \to\infty} \left(1 - \frac{\lambda T}{N} \right)^{t N/T} \lambda \Delta t = \lambda e^{-\lambda t} \dd t, \]

so that the probability distribution is:

\[\mathcal{P}(t) = \lambda e^{-\lambda t}.\]

Exercise

Show that the probability that an individual exits the compartment $I$ exactly after some time $t = n \Delta t$ is given by: $$\mathcal{P}(t)\,\Delta t = p_\text{IR} (1-p_\text{IR})^n.$$

Using the fact that $p=\lambda_I\,\Delta t$, and $\Delta t = T/N_\text{steps}$ (Where $T$ is the total simulation time and $N_\text{steps}$ the total number of steps): $$\mathcal{P}(t)\,\Delta t = \left(1 - \frac{\lambda_I T}{N_\text{steps}}\right)^{t N_\text{steps}/T}\, \lambda_I \, \Delta t.$$

Next, take the limit $N_\text{steps}\to\infty$, and $\Delta t\to 0$, keeping $N_\text{steps}\,\Delta t = T$, and argue that $$\mathcal{P}(t) = \lambda_I \, e^{-\lambda_I t}.$$ The sojourn times in the infected compartment are exponentially distributed, with mean $\tau_I = 1/\lambda_I$! In other words, this is a succession of Poisson processes – characteristic of Markov Chain Monte Carlo processes: happens every time the probability of a transition is independent of the history.

Can the same argument be used to say that the sojourn time in the susceptible compartment is $\tau_S = 1/\lambda_S$? Explain.