I have a system in which emails arrive at a Poisson process of parameter $\lambda$ (emails/day). Then, a secretary reads it, writes a reply and sends it at a $\mu$ (emails read/day) to a senior officer who authorizes the secretary's reply and sends it to the original email author at a $\phi$ rate (authorization+email to original author/day). Both $\mu$ and $\phi$ distribute Exponentially.
My question is: If I have 1 email to answer, the time it will take to answer it will be the sum of the two times (the time it takes for the secretary to read and then write a reply + the time it takes for the senior officer to authorize and send the reply). Both of these times distribute Exponentially. So, will the time to read, reply, authorize, and send the reply distribute as an Expontential of parameter (($\mu * \phi$)/($\mu + \phi$))?
If $X\sim\mathrm{Expo}(\mu)$ and $Y\sim\mathrm{Expo}(\phi)$ then $Z=X+Y$ does not have an exponential distribution. Indeed, the density of $Z$ is given by the convolution of the densities of $X$ and $Y$. Assuming $\mu\ne\phi$, this is \begin{align} f_Z(t) &= f_X\star f_Y(t)\\ &= \int_0^t f_X(t-s)f_Y(s)\ \mathsf ds\\ &= \int_0^z \mu e^{-\mu (t-s)}\phi e^{-\phi s}\ \mathsf ds\\ &= \mu\phi e^{-\mu t}\int_0^t e^{-(\phi-\mu)s}\ \mathsf ds\\ &= \frac{\mu\phi}{\phi-\mu} e^{-\mu t}\left(1 - e^{-(\phi-\mu)t}\right). \end{align} If $\mu=\phi$ then this is \begin{align} f_Z(t) &= f_X\star f_X(t)\\ &= \int_0^t f_X(t-s)f_X(s)\ \mathsf ds\\ &= \int_0^t \mu e^{-\mu(t-s)}\mu e^{-\mu s}\ \mathsf ds\\ &= \mu^2 e^{-\mu t}\int_0^t \mathsf ds\\ &= \mu^2 t e^{-\mu t}. \end{align}