Consider 2 independent, parallel $M/M/1$ queues $Q_1, Q_2$ with identical arrival rate $\lambda$ (corresponding to an exponential random variable $A \sim \text{Exp}(\lambda)$) and service rate $\mu$ (corresponding to an exponential random variable $S \sim \text{Exp}(\mu)$). For each $M/M/1$ queue, we use the FCFS (First Come First Served) discipline and if there is some customer in service, no more customers can enter it.
For each customer $c$, its start time, service time, and service interval are denoted by $c_{st}, c_{ft},$ and $[c_{st}, c_{ft}]$, respectively.
For two customers $w$ served by $Q_1$ and $r$ served by $Q_2$, suppose that $r$ starts between $w$'s service interval (i.e., $r_{st} \in [{w}_{st}, {w}_{ft}]$) and the difference between their start times ${r}_{st} - {w}_{st}$ follows an exponential random variable $D$ with rate $\lambda$ (see the figure below).
Let $r'$ be the last previous customer served by $Q_2$ just before the one $r$.
Problem: What is the expectation of $D' = r'_{st} - {w}_{st}$, the time difference between the starts times of $r'$ and $w$?
My trial: $\mathbb{E}(D') = \mathbb{E}(r'_{ft} - r'_{st}) + \mathbb{E}(r_{st} - r'_{ft}) - \mathbb{E}(D) = \frac{1}{\mu} + \mathbb{E}(A_1 \mid A_1 < A_2) - \frac{1}{\lambda}) = \frac{1}{\mu} + \frac{3}{2\lambda} - \frac{1}{\lambda} = \frac{1}{2\lambda} + \frac{1}{\mu},$ where $A_1$ and $A_2$ are i.i.d. random variables with common distribution as $A$.
Does this make sense?
We have $$\begin{align*} D' &= r'_{st} - w_{st}\\ &= (r'_{st} - w_{st}) + (r_{st}-r_{st}) + (r'_{ft}-r'_{ft})\\ &= (r_{st} - w_{st}) - (r'_{ft}-r'_{st}) - (r_{st}-r'_{ft})\\ &= D - (r'_{ft}-r'_{st}) - (r_{st}-r'_{ft}). \end{align*}$$ Hence $$\begin{align*}\mathbb E[D'] &= \mathbb E[D] - \mathbb E[r'_{ft}-r'_{st}] - \mathbb E[r_{st}-r'_{ft}]\\ &= \frac1\lambda -\frac1\mu -\frac1{2\lambda}\\ &= \frac{\mu-2\lambda}{2\lambda\mu}. \end{align*}$$