Thread: Rocky table View Single Post
2007-03-21, 16:52   #11
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by hhh So, for the sake of understanding: is it like I described, space $R^2$, for any $x\inR^2$ you take a U[0,1] r.v., all iid? This floor would be as I said highly discontinuous. Or do you mean by "uniformly random" "arbitrary, but continuous". In this case, I can vaguely figure how it should work with what you cited (I think you meant Brouwer's fixed point theorem).
Highly discontinuous? Indeed!! It is nowhere continuous.