simplifying a summation related to an epidemic process

Question

Here is a question that I came across while trying to understand some fundamental mathematical concepts related to the spread of epidemics. The problem turns out to be a mathematical puzzle related to the binomial coefficient.

Consider an epidemic process such as the one described in [zhang2017contact]. Let $\lambda(N)$ be an exogenous infection rate. Let $\gamma^{N/2}$ be an endogenous infection rate. The total rate at which a node is contaminated (endogenously and exogenously) is the product of these two quantities. Let also $\mu$ be the recovery rate. The ratio of the contamination rate over the recovery rate plays an important role in the study of epidemics.

The probability that $i$ node are infected is $\tilde{\pi}(i;N)/Z$ where \begin{equation} \tilde{\pi}(i;N)={N \choose i} \left( \frac{\lambda(N)}{\mu} \gamma^{N/2} \right )^{i} \textrm{ , } i =0,\ldots, N \end{equation}

The fraction of infected nodes is \begin{equation} \rho(N) = \frac{E(I)}{N} = \sum_{i=0}^ N i \pi(i;N)/{N} \end{equation}

where \begin{equation} \pi(i;N) = \frac{\tilde{\pi}(i;N)}{Z} \end{equation} and \begin{equation} Z=\sum_{i=0}^N \tilde{\pi}(i;N) \end{equation}

Questions

Our goal is to analyze $\rho(N)$ and to show that, for certain shapes of $\lambda(N)$, we have that $\rho(N)$ first decreases and then increases. This means that, as $N$ increases, the system is first dominated by exogenous factors, and then by endogenous factors.

1) for which shapes of the $\lambda(N)$ function we are able to obtain simple closed form expressions for $\rho$?

2) this is the case of most interest: are we able to treat the case $\lambda(N)=1/N$?

Zhang et al., Contact process with exogenous infection and the scaled SIS process. Journal of Complex Networks, 2017

Answer 1

We obtain \begin{align*} \color{blue}{\rho(n)}&=\frac{1}{N}\sum_{j=0}^N j\pi(j;N)\\ &=\frac{1}{N}\sum_{j=0}^N j\binom{N}{j}\left(\frac{\lambda(N)}{\mu}\gamma^{\frac{N}{2}}\right)^j\\ &=\sum_{j=1}^N\binom{N-1}{j-1}\left(\frac{\lambda(N)}{\mu}\gamma^{\frac{N}{2}}\right)^j\tag{1}\\ &=\sum_{j=0}^{N-1}\binom{N-1}{j}\left(\frac{\lambda(N)}{\mu}\gamma^{\frac{N}{2}}\right)^{j+1}\tag{2}\\ &\color{blue}{=\left(\frac{\lambda(N)}{\mu}\gamma^{\frac{N}{2}}\right)\left(1+\frac{\lambda(N)}{\mu}\gamma^{\frac{N}{2}}\right)^{N-1}}\tag{3} \end{align*}

Comment:

• In (1) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.

• In (2) we shift the index to start with $j=0$.

• In (3) we apply the binomial theorem.