Consider two random processes $X, Y$. $Y$ is a Poisson process with rate $\lambda$. $X$ behaves as follows: whenever $X < Y$, it acts equivalent to a Poisson process with rate $\lambda$ (independent of $Y$). Whenever $X = Y$, it waits until $Y$ increments, so that $Y - X = 1$, before continuing to increment itself. (In other words, $X$ never allows itself to surpass $Y$).
My question is: what is known about the distribution of $X$? Specifically, can anyone offer any bounds regarding the expected time until $X > n$, for arbitrary $n$?
My simulations indicate that it's roughly Y's arrival time plus some very slowly growing "lag" when $n$ is very big. This lag is sort of intuitive: consider the case where you have a whole bunch of processes each bounding the one to its left; in that case you would intuitively have quite a big discrepancy between the leftmost process and the rightmost one.
Fix a positive integer $n$. If $T_{ij}$ is the expected time for $X$ to reach state $n$, given $(X(t),Y(t))=(i,j)$, then you can stop the $Y$ process once it reaches state $n$ and you get a state space of integers $(i,j)$ for $0\leq i\leq j \leq n$ and equations \begin{align} T_{ij} &=\frac{1}{2\lambda} + \frac{1}{2}T_{i+1,j} + \frac{1}{2}T_{i,j+1}, \quad 0\leq i<j<n\\ T_{jj}&= \frac{1}{\lambda} + T_{j,j+1} , \quad 0\leq j<n\\ T_{i,n} &= \frac{n-i}{\lambda} , \quad 0\leq i < n\\ T_{nn}&=0\end{align} You can solve them by hand for small integers $n$ to compute the desired value $T_{00}$.
$n=1$: $T_{00} = 2/\lambda$.
$n=2$: $T_{00} = 3.5/\lambda$.