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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.

aellab
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    Not sure about the answer, but try to give a formulation first. Let $Z_n, W_n$ be the $n$-th arrival time of the process $X, Y$ respectively. $W_n$ is well known to have a Gamma/Erlang distribution, while $Z_n = \max{Z_{n-1}, W_n} + V_n$ where $V_n$ is an exponential random variable with rate $\lambda$, independent. – BGM Jul 11 '18 at 03:17
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    You can view this as a symmetric random walk $Z=Y-X$ on $\mathbb Z$ in which steps are taken with rate $2\lambda$ and all negative steps at the origin are discarded. You want to know the expected number of steps until $n+1$ negative steps have been taken. I think you should be able to derive a generating function for this. Some of the techniques I used in this answer might be useful. If I find the time, I'll try to work it out in more detail. – joriki Jul 11 '18 at 06:11
  • @joriki That's a clever observation. I'd greatly appreciate if you find the time to post an answer; I am not too familiar with generating functions. – aellab Jul 11 '18 at 12:41

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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 $T_{00}$.

  • $n=1$: $T_{00} = 2/\lambda$.

  • $n=2$: $T_{00} = 3.5/\lambda$.

Michael
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  • A simple lower bound on the expected time for $X$ to reach state $n$ is the starting average delay of $1/\lambda$ (for $Y$ to make the first move) and then the expected time for a Poisson process to move $n$ steps unhindered by $Y$, so $$ \frac{n+1}{\lambda}\leq T_{00}$$ – Michael Jul 11 '18 at 06:10
  • I think you've shifted $n$ by $1$ relative to the question. It says $X\gt n$, not $X\ge n$. (Though your version seems more natural; it might be preferable to change the question instead.) – joriki Jul 11 '18 at 06:14
  • @joriki : So I have. But solving one solves the other. My answer specifies what I solve for so I don't think there is any confusion. – Michael Jul 11 '18 at 06:15
  • Never underestimate the potential for confusion :-) – joriki Jul 11 '18 at 06:32
  • Thank you for the answer. I am interested in estimating for arbitrary $n$; can we derive some sort of asymptotic upper bound on $T_{00}$ using your method? I'm not very familiar with this kind of recursive relation.

    I am mostly interested in showing that $X$ and $Y$ behave very similar as $n \to \infty$.

     – aellab Jul 11 '18 at 12:21
  • @mich: Exactly what sort of similarity do you want to prove? Given the random walk equivalent in my comment under the question, we know that $X$ will almost surely infinitely often coincide with $Y$. Do you want to show more than that? – joriki Jul 11 '18 at 12:45
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    @joriki Yep. Specifically, if $A_x, A_y$ are the arrival times of $X$ and $Y$ at $n$, I'm interested in the asymptotic growth (well, decrease) of $A_x / A_y$ as $n \to \infty$. Hopefully to be able to roughly estimate how close $X$ and $Y$ are going to "act" for some $n$. – aellab Jul 11 '18 at 12:55
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    To be more clear, I'd like to be able to find for example, some upper bound on for which $n$s is $A_x \leq 1.01A_y$. – aellab Jul 11 '18 at 13:00
  • @mich : The equations I give can indeed be written as a simple numerical recursion: We have states $(i,j)$ defined by a triangle and we know values $T_{ij}$ for the top row. From that we can compute the second-to-top row, then the third, and so on, using a simple equation. This will give an easy numerical method but won't give asymptotics. Right now, my only asymptotic observation is that the time average fraction of time that $X(t)<Y(t)$ is 1 (since we have renewals every meeting, and once $Y$ increases by 1 beyond $X$, the mean time for $X$ to catch up is $\infty$). – Michael Jul 11 '18 at 18:31