Epidemic modelling: expectation of time of infection given the distribution of transmission and recovery Can I express the expected value of
\begin{equation}
\langle \tau\rangle_\text{total}=\int_0^\infty \tau  \psi_\text{inf}(\tau)\Psi_\text{rec}(\tau)\mathrm{d}\tau
\end{equation}
in terms of the moment(s) of the two other distributions $\psi_\text{inf},\psi_\text{rec}$ where $\Psi_\text{rec}(\tau)$ is the survival function of the recover distribution.
If not, is it at least possible to express:
\begin{equation}
I=\int_0^\infty   \psi_\text{inf}(\tau)\Psi_\text{rec}(\tau)\mathrm{d}\tau
\end{equation}
in terms of the moments of  $ \psi_\text{rec}$ and $\psi_\text{inf}$?

Consider the following model: An infected individual will attempt to transmit the disease to a healthy neighbour at a random time whose distribution is $\psi_\text{inf}(\tau)$ such that $\langle \tau\rangle_\text{inf}$ is the expected time of transmission after initially contracting the disease. The probability that an individual stays infected for a duration $\tau$ without recovering and transmits the disease to the healthy neighbour in the next infinitesimal time interval $\mathrm{d}\tau$ is given by:
\begin{equation}
\psi_\text{total}(\tau)\mathrm{d}\tau=   \psi_\text{inf}(\tau)\Psi_\text{rec}(\tau)\mathrm{d}\tau
\end{equation}
Can I express the expected time of a successful transmission $\langle \tau\rangle_\text{total}$ in terms of the moments of the times of (attempted) infection and recovery: $\langle \tau\rangle_\text{inf},\langle \tau\rangle_\text{rec}$?
 A: I am not certain whether this is what you are looking for, but you could consider expanding $\Psi_{\sf rec}$ in power series (if it is regular enough) and the moments would emerge from this expansion. I am not sure you could find a much closer form given the quite general framework set in your question.
For example: If i) $\Psi_{{\sf rec}}$ is analytic about the origin; ii) $\langle \tau^i\rangle_{\sf inf}<\infty$ for all $i$; and
iii) $$\sum_{i\geq 0}\frac{\left|\overset{(i)}{\Psi}_{{\sf rec}}(0)\right|}{i!}\langle\tau^i\rangle_{{\sf inf}}<\infty,$$
where $\overset{(i)}{\Psi}_{{\sf rec}}(0)$ is the $i$th derivative of $\Psi_{{\sf rec}}$ at the origin, then
$$I=\sum_{i\geq 0}\frac{\overset{(i)}{\Psi}_{{\sf rec}}(0)}{i!}\int_0^{\infty} \psi_{{\sf inf}}(\tau) \tau^i d\tau=\sum_{i\geq 0}\frac{\overset{(i)}{\Psi}_{{\sf rec}}(0)}{i!}\langle\tau^i\rangle_{{\sf inf}}.$$
In the particular case where infections and healings are driven by Poisson processes -- i.e., once an individual is infected, it will attempt to transmit the infection after an exponentially distributed random time and the same for healing -- then, $\Psi_{\sf rec}(t)=e^{-t/\langle \tau\rangle_{\sf rec}}$ and $\psi_{\sf inf}(t)=e^{-t/\langle \tau \rangle_{\sf inf}}/\langle \tau\rangle_{\sf inf}$, and condition iii) above is fulfilled whenever $\langle \tau \rangle_{\sf inf}/\langle \tau \rangle_{\sf rec}<1$. In this case, we have
$$I= \frac{\langle \tau \rangle_{\sf rec}}{\langle \tau \rangle_{\inf}+\langle \tau \rangle_{\sf rec}},$$
as $\langle \tau^i\rangle_{\sf inf}=i!\langle \tau \rangle_{\sf inf}^i$ and
$$I=\sum_{i\geq 0}\overset{(i)}{\Psi}_{{\sf rec}}(0)\langle\tau\rangle_{{\sf inf}}^i=\sum_{i\geq 0} (-1)^i \left(\frac{\langle \tau \rangle_{\sf inf}}{\langle \tau \rangle_{\sf rec}}\right)^i=\sum_{i\geq 0} \left(\frac{\langle \tau \rangle_{\sf inf}}{\langle \tau \rangle_{\sf rec}}\right)^{2i}-\sum_{i\geq 0}\left(\frac{\langle \tau \rangle_{\sf inf}}{\langle \tau \rangle_{\sf rec}}\right)^{2i+1}=\frac{\langle \tau \rangle_{\sf rec}}{\langle \tau \rangle_{\inf}+\langle \tau \rangle_{\sf rec}}.$$
That is, for Poisson driven clocks, you only need the first moments.
In the same vain,
$$\langle \tau \rangle_{\sf total}=\sum_{i\geq 0}\frac{\overset{(i)}{\Psi}_{{\sf rec}}(0)}{i!}\langle\tau^{i+1}\rangle_{{\sf inf}}.$$
For the Poisson example, this reduces to
$$\langle \tau \rangle_{\sf total}=\frac{\lambda_{\sf inf}}{\left(\lambda_{\sf inf}+\lambda_{\sf rec}\right)^2}$$
where $\lambda_{\sf inf}:=1/\langle \tau\rangle_{\sf inf}$ and $\lambda_{\sf rec}:=1/\langle \tau\rangle_{\sf rec}$.
I believe that there are more relaxed conditions than the sufficient ones I cooked up now, namely, i), ii) and iii).
