tag:blogger.com,1999:blog-5457514384557268934.post4333046268586169958..comments2020-02-14T06:48:09.358+00:00Comments on Kinetically Constrained: An unintuitive probability problemAnonymoushttp://www.blogger.com/profile/10673953249469231273noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-5457514384557268934.post-79852646764175935202009-12-03T19:03:32.960+00:002009-12-03T19:03:32.960+00:00Yes this is right. Your condition for equilibrium ...Yes this is right. Your condition for equilibrium is slightly stronger than the one I wrote down, it's known as "detailed balance" and enforces a stationary distribution with no <a href="http://en.wikipedia.org/wiki/Limit_cycle" rel="nofollow">limit cycles</a>.<br /><br />For the chain as I've described it it's totally appropriate as I don't see what limit cycle we could have.<br /><br />It was writing down something like this that eventually stopped me being confused. Whether or not it's intuitive I suppose is somewhat subjective!Anonymoushttps://www.blogger.com/profile/10673953249469231273noreply@blogger.comtag:blogger.com,1999:blog-5457514384557268934.post-50404179028412644842009-12-03T17:57:57.789+00:002009-12-03T17:57:57.789+00:00I provided another intuition (3?)
If we sit in th...I provided another intuition (3?)<br /><br />If we sit in the middle of x and x+1, we'll see the flux coming from the right (x+1) and flux coming from the left (x). In equilibrium they should be the same. So<br /><br /> [P(x)][p_r(x)] = [P(x+1)][p_l(x+1)].<br /><br />Since p_r(x) = p_l(x+1). P(x) = P(x+1).<br /><br /> Quite ituitive, but I don't know if this make sense rigorously though. =)Unknownhttps://www.blogger.com/profile/11920880208900388170noreply@blogger.com