# Show that there does not exist a unique stationary distribution.

So I was doing some self study and came across a proposition in one of my chemical engineering course's prescribed textbooks. I can't quite get the proof out. It's to do with a particle moving through a medium such that when it makes contact with to either of two plates $$L$$ units apart (i.e. one at $$0$$ and one at $$L$$), it remains there.

Consider that the movement of a single particle follows a random walk which can be described by a Markov chain with states $$[0, L]$$ where $$P(X_n = -1) = p_{-1}$$,$$P(X_n = 0) = p_{0}$$ and $$P(X_n = 1) = p_{1}$$ with $$p_{-1} + p_{0} + p_{1} = 1$$. Show that if states $$0$$ and $$L$$ are completely absorbing, then there does not exist a stationary distribution. Hint: Start by considering $${\pi} = \pi P$$

This makes sense intuitively since we have two recurrent classes $$\{0\}$$ and $$\{L\}$$ and one transient class $$\{1, 2, ..., L - 2, L - 1\}$$. However, once I try and expand $${\pi} = \pi P$$, I don't know how to proceed next. Ideally I'd like a few more hints rather than an answer.

• There’s really not much more in the way of hints to give you. Once you have $\mathbf\pi=\mathbf\pi P$ you’re already almost at the solution.
– amd
Commented Oct 12, 2018 at 6:10

$$(1,0,0,...,0)$$ and $$(0,0,...0,1)$$ are two invariant distributions so uniqueness fails.

• @amWhy Your comment is totally unjustified. My answer is correct and anyone who understands the question will give the same answer. Please withdraw your comment or tell me why this is not an answer. You can see that user amd has posted the same answer. If you flagged this answer you have done something very unreasonable. Commented Oct 14, 2018 at 12:22
• I believe short answers are sometimes automatically sent to the LQ review queues. Maybe that's what happened here? Commented Oct 15, 2018 at 12:18
• @ArnaudD. Thank you. Even though I have been here for quite some time I get confused when something like this happens. As far as I am concerned my answer is compete, correct and useful. Commented Oct 15, 2018 at 12:22
• I buy ArnaudD.'s idea, but I don't exclude the possibility that someone flagged it as VLQ/NAA. A multitude of short (but excellent) answers like yours has passed the LQ review without problem. What's confusing are the reviewers' votes, which are their individual (and unjustified) decision. In light of the amount of work that they deleted and the editing wars in which they engaged themselves, I'm worried about the health of CRUDE. Commented Oct 15, 2018 at 13:09
• Related meta discussion. I don't think user amWhy flagged your answer because one can't flag and review one single answer at the same time. For example, I had flagged an answer as VLQ, so that it entered into the LQ review. I saw others' votes, but I couldn't review it myself since I had flagged this. Unluckily, the reviewers let go that answer. Commented Feb 14, 2019 at 13:59

Write your equation as $$\mathbf\pi(P-I)=0$$. The first and last rows of $$P-I$$ are zero, so $$(1,0,\dots,0)$$ and $$(0,\dots,0,1)$$ are obvious independent solutions of the equation.

• I posted the same answer and it was flagged for deletion with some absurd comments. Commented Oct 12, 2018 at 4:54
• @Kavi, it may have been flagged for deletion, but it says it was you who actually deleted it. Not clear to me why you did that. Commented Oct 12, 2018 at 11:50
• I deleted it after very adverse and meaningless comments and after it was flagged. I have no idea what those comments meant. Commented Oct 12, 2018 at 11:52
• @KaviRamaMurthy I've checked the related tag scores (and the contributing posts) of the two reviewers who actively participate in CRUDE and have deleted plenty of stuff. Your post is just one of those they voted to delete. Commented Oct 12, 2018 at 13:17
• I don't see any "very adverse" comment, only an automatic comment from Review queues. I'll admit that I don't understand the reason you got this comment, since it is the one usually adressed to new users that post questions as answers. I also admit that I know close to nothing about the tags, so I can't judge the quality of @KaviRamaMurthy's answer (or this one). Commented Oct 12, 2018 at 17:33