# Average number of infections in SIR(S) model

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 (^:
• Shouldn't your first equation be $P(e)=1-\left(1-\beta\frac{m}{N-1}\right)^{i}$? Nov 22, 2018 at 14:13
• 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.
– Fie
Nov 25, 2018 at 19:32