Wikipedia:Reference desk/Archives/Mathematics/2019 May 23

= May 23 =

How to find the stationary distribution of a random walk on real numbers?
Start at the point $$(1/2, 1/2)$$. At each time step, choose a point uniformly at random from $$(0,0), (0,1), (1,0), (1,1), (1/2, 1/2)$$, and then move half way from the current position to the chosen point.

In other words, at each time step transition from $$(x, y)$$ to one of $${(x/2, y/2), (x/2, (y+1)/2), ((x+1)/2, y/2), ((x+1)/2, (y+1)/2), ((x+1/2)/2, (y+1/2)/2)}$$.

After $$n$$ time steps measure the distance from the starting position.

What is the distribution of distances as $$n \rightarrow \infty$$? What is the expected distance?

I understand how to find stationary distributions for finite Markov chains, but this is a bit beyond me. What topics would be needed to answer this, or is there a general approach to answer questions of this form? 98.190.129.147 (talk) 23:41, 23 May 2019 (UTC)


 * What you're describing is similar to the chaos game. There are several closely related examples there, but not quite the one you're describing exactly.  I suspect it might be worth it to just write a program to see if there's anything interesting as a result (often you get something self-similar from such a process).  I'm not sure how to analyze a description like that systematically, but there might be some better refs at that article.  –Deacon Vorbis (carbon &bull; videos) 00:24, 25 May 2019 (UTC)


 * I did a plot of the cumulative distribution function for the x-variable alone; it's not that hard to compute for values of the form p/2n where n is small. Judging from the roughish appearance of the graph the cdf is fractal in nature, something like a smoothed version of the Cantor function. It would follow that the distribution has no density function in the normal sense. The average position is obviously (1/2, 1/2), and from that the average displacement is 0, but the average distance is a more difficult calculation even you just look at horizontal or vertical distance. Given a closed form for the distribution is unlikely, I think your best bet is to use a Monte Carlo method to get an estimate for something like the average distance. --RDBury (talk) 13:37, 25 May 2019 (UTC)