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I have a question from epidemiology that I'm struggling 1) to write down mathematically and 2) determine if it has a closed form solution.

First here's the epidemiological question. Say I have $C$ children. They are all independently infected by a recoverable disease at a constant rate $\lambda$. After a fixed time interval $t$, the number of infections each child has had, $Y_i$, is Poisson distributed with parameter $\lambda t$:

$$Y_i\sim\text{Poi}(\lambda t)$$

The total number of infections across all children is $M=\sum_{i=1}^{C} Y_i$.

Let $N$ be a child's infection order, i.e, $N=1$ is a child's first infection, $N=2$ is a child's second infection, and so on.

My question is, what is the probability that an infection chosen at random from the set of $M$ infections an $N^\text{th}$ infection?

I think the correct mathematical formulation of the probability mass function is this:

$$p(n) = \text{Pr}(N=n) = \frac{\sum_{i=1}^{C} I(Y_i\ge n)}{M}$$

where $I(\cdot)$ is the indicator function (1 if $Y_i\ge n$, 0 otherwise).

If correct, can this function be written in terms of $\lambda t$ and $C$? Any thoughts most welcome.

RobPratt
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Nick Savill
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  • When you say "an infection chosen at random from the set of $M$ infections is an $N$th infection", do you mean that if the particular infection in question affects child $i$ then this is the $N$th time child $i$ has been infected, or that this is the $N$th infection overall? – Henry Jan 14 '25 at 12:54
  • Sorry for the confusion. I mean if the particular infection in question affects child i then this is the Nth time child i has been infected – Nick Savill Jan 14 '25 at 13:32
  • In the case $C=1$ we get $\mathbf P(N=n)=\sum_{k=n}^{+\infty}\frac{(\lambda t)^k}{k.k!}\mathrm e^{-\lambda t}$ which can be expressed with the exponential integral for $n=1$. – Christophe Boilley Jan 14 '25 at 17:24
  • The limit of $C\rightarrow\infty$ would be useful, and perhaps more tractable than finite $C$? – Nick Savill Jan 15 '25 at 09:06

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