How long does it take for a two-site wait activation? Consider two sites, linked as sketched

Initially, both sites are off (red). However, each activates (turns green) at a constant rate $f$. Once activated, a site remains activated. If one activates, it generates an activation wave (green line) which propagates at speed $v$ (given as the number of sites per time unit), as shown

Then, one of two things can happen: either the wave activates the neighboring site, provided it has not been activated yet, or the site activates on its own

How do we calculate the expected time it takes for one site to activate? In other words, what is the expected waiting time on one site? If there was only one site, the answer would simply be $1/f$, but I am struggling to frame the two (and possible $n$-case) scenario. Any suggestions?
My attempt: Given a time interval $dt$, the probability $P_f$ that one site is activated, within $dt$, on its own is simply $fdt$. However, the probability $P$ that a site is activated within $dt$ is given by
$$
\begin{align}
P(dt)=&P_1(\text{"site activates autonomously, given the wave didn't reach it within }dt\text{"})\\
&+P_2(\text{"wave activates the site, given it wasn't autonomously activated within }dt\text{"})
\end{align}
$$
However, it seems relatively tricky to write these expressions. Intuitively, I would expect something like
$$
P_1(t)=1-e^{-ft/v}
$$
for example, but I have no particular motivation for this choice. It simply satisfies some expected asymptotics. Any ideas? Perhaps an approach based on first-hitting time analysis?
 A: This analysis in the top part of the answer is correct but inefficient; I include it so that comments continue to make sense. See below the line for a better approach.
In the below I use the convention that Exp($\lambda$) means the exponential distribution with rate $\lambda$.
The two node case can be analyzed like this: first nothing is happening, then after a time with Exp($2f$) distribution, one node is activated. With probability $1/2$, that node was the one of interest, so we stop. With probability $1/2$, the other node was the one of interest, so now after an additional time $1/v$ we will stop. However, we can stop before that if the node just activates itself, which will happen after a time with Exp($f)$ distribution after the first activation. (Here I have exploited the memoryless property of the exponential distribution.)
So using the total expectation formula, conditioning on which state activates first, gives the following.
$$E[T]=\frac{1}{2} \frac{1}{2f} + \frac{1}{2} \left ( \frac{1}{2f} + \int_0^\infty \min \{ t,1/v \} f e^{-ft} dt \right ) \\
= \frac{1}{2f} + \frac{1}{2} \left ( \int_0^{1/v} t f e^{-ft} dt + \int_{1/v}^\infty (f/v) e^{-ft} dt \right ) \\
= \frac{1}{f} \left ( 1 - \frac{1}{2} e^{-f/v} \right ).$$
In the three nodes equally spaced on a ring case, everything is almost the same because a second wave will always reach our node after the first wave does. So you get
$$E[T]=\frac{1}{3} \frac{1}{3f} + \frac{2}{3} \left ( \frac{1}{3f} + \int_0^\infty \min \{ t,1/v \} f e^{-ft} dt \right ) \\
= \frac{1}{3f} + \frac{2}{3} \left ( \int_0^{1/v} t f e^{-ft} dt + \int_{1/v}^\infty (f/v) e^{-ft} dt \right ) \\
= \frac{1}{f} \left ( 1 - \frac{2}{3} e^{-f/v} \right ).$$
We see a pattern in these formulae, but unfortunately this pattern corresponds to a star-shaped network of nodes ($n-1$ nodes connected to our node of interest in the center) rather than to a ring of nodes. A larger ring of nodes is more complicated because there are more categories of nodes. For four, there are three categories of nodes: the node of interest, the two neighboring nodes, and the node on the opposite side. So you get a schematic like
$$E[T]=\frac{1}{4} E[T \mid \text{ node of interest activates first }] + \frac{1}{2} E[T \mid \text{ neighboring node activates first }] + \frac{1}{4} E[T \mid \text{ opposite node activates first }].$$
That last term is now more complicated, because it could happen that our node activates itself, or that a neighboring node activates itself before time $1/v$ and then the wave reaches our node in time $1/v$ after that, or that no further self-activations occur and our node activates in time $2/v$ after the first activation.

First let me write out the nondimensionalization in this method. Let $T$ be the time to activation of our node of interest, let $t$ denote time as a variable, then $\tilde{T}=fT$ and $\tilde{t}=ft$, then abuse notation by dropping tildes in the analysis. Next introduce $\tilde{v}=v/f$, $s=1/\tilde{v}$, then again abuse notation by dropping tildes in the analysis.
We can think of $T$ as an explicit function of the node self-activation times $A_i$. For ease of notation, I will index the nodes so that their distance from the node of interest is $|i|$. In the ring network you then have
$$T=\min_i \{ A_i + |i| s \}$$
since it takes time $|i| s$ for a wave from node $i$ to reach our node of interest. Note that in the ring network, the fact that the wave activates other nodes along the way does not matter.
This minimum is greater than some $t$ if all its components are, which occurs with probability
$$P(T>t)=\prod_i P(A_i>t-|i| s)=\prod_i \min \{ 1,\exp(-(t-|i| s)) \}.$$
So the expectation of the activation time for any one node is given as $E[T]=\int_0^\infty P(T>t) dt=\int_0^\infty \prod_i \min \{ 1,\exp(-(t-|i| s)) \} dt$. This integral can be split up at $s,2s,\dots,\lceil (n-1)/2 \rceil s$ for $n$ nodes, and each integrand is of the form $ae^{-bt}$, so you can evaluate them analytically and sum them.
A few particular cases:

*

*Two nodes:

$$E[T]=\int_0^s e^{-t} dt + \int_s^\infty e^{-2t+s} dt.$$

*

*Three nodes:

$$E[T]=\int_0^s e^{-t} dt + \int_s^\infty e^{-3t+2s} dt.$$

*

*Four nodes:

$$E[T]=\int_0^s e^{-t} dt + \int_s^{2s} e^{-3t+2s} dt + \int_{2s}^\infty e^{-4t+4s} dt.$$
Now for the general case. The situation is slightly different between $n$ odd and $n$ even. When $n$ is odd, for each $k$, there are $2$ nodes at a distance of $k=1,2,\dots,(n-1)/2$ from the node of interest. In addition, we need to add up the distances up to $k$, each twice, which add up to $k(k+1)$. So we get
$$E[T]=\int_0^s e^{-t} dt + \sum_{k=1}^{(n-3)/2} \int_{ks}^{(k+1)s} e^{-(2k+1)t + k(k+1)s} dt + \int_{(n-1) s/2}^\infty e^{-nt+(n-1)(n+1) s /4} dt.$$
(The last term is just the $k=(n-1)/2$ term of the sum but with the upper limit replaced by $\infty$.)
Doing the integrals yields:
$$E[T]=1-e^{-s} + \sum_{k=1}^{(n-3)/2} \frac{e^{-k^2 s}-e^{-(k+1)^2 s}}{2k+1} + \frac{e^{-(n-1)^2 s/4}}{n}.$$
When $n$ is even, for each $k$ there are $2$ nodes at a distance of $k=1,2,\dots,(n-2)/2$, and then there is $1$ node at a distance of $n/2$. Again we add up the distances, each twice, which add up to $k(k+1)$, but since there is only one node at a distance of $n/2$, the very last distance sum is $2+4+\dots+n-2+n/2=n(n-2)/4+n/2=n^2/4$. So we get
$$E[T]=\int_0^s e^{-t} dt + \sum_{k=1}^{(n-2)/2} \int_{ks}^{(k+1)s} e^{-(2k+1)t + k(k+1)s} dt + \int_{ns/2}^\infty e^{-nt+n^2s/4} dt.$$
