# Random walk on infinite line - can it be stationary?

Suppose a random walk on an infinite line $[...-3,-2,-1,0,1,2,3,...]$, starting from 0. Probability to go right or left are equal. Does such a process stationary? I think that it is NOT, since the support of each step is different. I.e., $x_1\in{\{-1,1\}}, x_2\in{-2,0,2}, ...$. Thanks.

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The argument has nothing to do with the conclusion, since it would apply equally well to simple random walks on discrete circles with an even number of states, which DO have a stationary measure. // You should definitely explain precisely what it is you mean by Does such a process stationary? – Did May 3 '12 at 14:50
Thanks Didier. By stationary I mean the standard definition of a stationary process: $\Pr(X_{t_1+\tau},...,X_{t_k+\tau})=\Pr(X_{t_1},...,X_{t_k})$, for all $\tau,k$, and $t_1,...,t_k$. So, does such a random walk have this property, let's say from some finite point of time? From my argument in the question follows that: $\Pr(X_{n})\neq\Pr(X_{n+1})$ for all $n$, since $\Pr(X_{n}=n+1)=0$, while $\Pr(X_{n+1}=n+1)>0$. After all, such a random walk can be viewed as a Markov chain, so I wonder if it has a stationary distribution. – Michael May 4 '12 at 15:21
I think that a similar case is a Markov chain that represents an MM1 queue, which do have a stationary distribution under certain conditions (utilization \rho<1). – Michael May 4 '12 at 15:29
So... Does such a process stationary? means Does such a process admits a stationary distribution? The answer is NO but it has nothing to do with the fact that the distributions of $X_n$ and $X_{n+1}$ are not equal when $X_0=0$ almost surely, see the example in my first comment. My impression at this point is that finding some good introductory textbook on random processes and pondering it quietly would save you some time. Let me suggest that you try the (freely available) first chapter of Markov chains by James Norris. – Did May 4 '12 at 16:07
Thanks. I'll read this chapter. – Michael May 4 '12 at 19:46