1$ queue w.r.t. the input rate

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when toggle format	what		by	license	comment
Sep 19, 2014 at 7:47	vote	accept	Indigo
Sep 19, 2014 at 7:47	vote	accept	Indigo
Sep 19, 2014 at 7:47
Sep 18, 2014 at 14:57	comment	added	James Martin		Indeed. For a general arrival process, you could say that the process is "stationary" (or, if you prefer, has stationary increments) if the distribution of $(W(u+r_1,u+r_2), r_1, r_2\in\mathbb{R}, r_1<r_2)$ is the same for all $u$, where $W(s,t)$ is the amount of work arriving in $[s,t)$.
Sep 18, 2014 at 13:04	comment	added	Indigo		What do you mean by stationary arrival process? A Poisson process has stationary increments, but it's not itself stationary (meaning: the distribution of $N(t)$ is different from the distribution of $N(t+h)$).
Sep 18, 2014 at 11:45	comment	added	James Martin		I edited the end of the explanation in the answer also to try to make it a bit clearer.
Sep 18, 2014 at 11:44	comment	added	James Martin		Now for $\lambda_i<\lambda_j$, couple the arrival processes so that $W_i[s,t]\leq W_j[s,t]$ for all $s$ and $t$. Then $X_i(t)\leq X_j(t)$ for all $t$. In particular if the $j$-system is empty at some time in $[−T,0]$, then so is the $i$-system. If, further, the two arrival processes are identical on $[−T,0]$, then also $X_i(0)=X_j(0)$. But the system is stationary so 0 is just a typical time. If we can show that with high probability two systems have the same state at time 0, then their stationary distributions are close in total variation distance.
Sep 18, 2014 at 11:42	history	edited	James Martin	CC BY-SA 3.0	added 76 characters in body
Sep 18, 2014 at 11:27	history	edited	James Martin	CC BY-SA 3.0	added 76 characters in body
Sep 18, 2014 at 11:20	comment	added	James Martin		From a stationary arrival process, you can construct a stationary evolution for the queue. This is sometimes called "Loynes' construction"; for example, you might want to look at the book by Baccelli and Bremaud. For $s<t$ let $W[s,t]$ be the amount of work arriving during the interval $[s,t]$. Then the amount of work in the queue at time t is given by $X(t)=\sup_{s<t}(W[s,t]-t)$.
Sep 17, 2014 at 15:06	comment	added	Indigo		Hi @James ! Thank you for your answer. The answer itself is very clear, but I am not familiar with the technique, so I have a a question. It looks like you are proving that for every $t$ there exists a coupling such that the queues with $\lambda_n$ input and a queue with $\lambda$ input agree from $t$ on with high probability.. How does stationarity enter the picture here?
Sep 17, 2014 at 8:44	history	edited	James Martin	CC BY-SA 3.0	added 14 characters in body
Sep 15, 2014 at 10:09	history	answered	James Martin	CC BY-SA 3.0