Thread: Rocky table
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Old 2007-03-21, 16:52   #11
R.D. Silverman
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Nov 2003

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Originally Posted by hhh View Post
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.
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