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I was going to go with a title based on to the maths of the problem, but I think that this one will better get the information to others who have the same query, particularly as I struggled to think of apt (titular) description of the problem.

As the title suggests, I am working with the SIR model for infectious diseases, and am planning on extending and refining it. The component I am currently working on is the rate of transfer from $S$ to $I$: the rate of infection.

The problem

Here are my premises and how I have constructed my result:

  • Let there be $m$ of object $I$ and $n$ of object $S$, with $N_0 \geq 0$ other objects (neither $I$ nor $S$) for a total of $N$ total objects.
  • Over the timestep, each $I$ interacts with $i < N$ objects other than itself
  • Let $e$ denote the infection (event) of an $S$, which occurs with probability $\beta \leq 1$ when an $I$ interacts with $S$, and converts $S$ to $I$.

For any given $S$, $P(e)$ is therefore the complement of the probability that every $I$-$S$ interaction does not result in infection, i.e. the probability of at least one infection event. As the probability of a single-interaction infection is $\beta$, and the probability of any given $I$ interacting with a particular $S$ over the timestep is $\frac{i}{N-1}$, then the infection probability for a particular $S$ over the timestep is $$P(e)=1-\left(1-\beta\frac{i}{N-1}\right)^{m}$$

Over all $n$ of $S$, the average (expected) number of infections should then simply be $$E = n\left[1-\left(1-\beta\frac{i}{N-1}\right)^m\right]$$

This result satisfies one of the criteria I expect, which is that $$\text{as } n \to \infty, E \to \beta i m$$ This is because $$\left(1-\beta\frac{i}{N-1}\right)^m = 1 - m\beta\frac{i}{N-1}+{m \choose 2}\left(\beta\frac{i}{N-1}\right)^2+\dots\\\text{but } N \gg \beta i\\\text{so }\left(1-\beta\frac{i}{N-1}\right)^m \approx 1 - m\beta\frac{i}{N-1}\\\implies E = n\beta\frac{i m}{N-1}\\\text{but as }n \to \infty, N \to n \text{ and }n \gg 1\\\implies E \to \beta\frac{i m n}{n}=\beta i m$$

This is because as $n$ becomes much larger than $m$, the number of 'overlapping' interactions between $I$ and $S$ tends to zero (as well as inconsequential $I$-$I$ interactions) and the number of infections in a timestep should simply be the expected-to-be-infected proportion $\beta$ of the number of interactions, $i m$, which will be effectively all $I$-$S$ interactions.

Questions

  1. Is any of my maths/Are any of my premises wrong?
  2. Am I missing anything that will affect the result? What else can I add to make the model more (mathematically) accurate?
  3. In a system with $i \geq N$, how might I smooth the transition from $i$ to $N-1$ other than taking the $\text{min}$? i.e., how could I consider multiple interactions per $I$ with the same $S$?
  4. How is this for a first post? I'll try to get better as I go (^:
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  • $\begingroup$ Shouldn't your first equation be $P(e)=1-\left(1-\beta\frac{m}{N-1}\right)^{i}$? $\endgroup$
    – Dmitry
    Nov 22, 2018 at 14:13
  • $\begingroup$ I don't think so. My reasoning is that each of $m$ $I$s interacts with $i$ individuals; the chance of a given S being interacted with by each $I$ is $\frac{i}{N-1}$ (and hence infection chance is $\beta \frac{i}{N-1}$). This process happens $m$ times (once for each $I$) and so the total infection chance is the complement of the chance that every infection event fails. $\endgroup$
    – Fie
    Nov 25, 2018 at 19:32

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