Right, that's not quite the issue. It's that some comparisons are made in 2-space, some in 10-space, etc. It would be nice to have some idea that a distance is 2-space is "about as meaningfully far" as some other distance in 10-space. I'm trying to find the order of that correcting factor and it seems to be sqrt(n). Within 2- or 10-space indeed those distances aren't randomly distributed... but would they be so differently distributed as to change this factor? Gut says no, but I have no more justification than that.
On Wed, Oct 19, 2011 at 10:15 PM, Ted Dunning <[email protected]> wrote: > None of this actually applies because real data are not uniformly > distributed (not even close). Do the sampling on your own data and pick a > good guess from that. > > On Wed, Oct 19, 2011 at 11:40 AM, Sean Owen <[email protected]> wrote: > >> Ah, I'm looking for the distance between points within, rather than >> on, the hypercube. (Think of it as random rating vectors, in the range >> 0..1, across all movies. They're not binary ratings but ratings from 0 >> to 1.) >> >> On Wed, Oct 19, 2011 at 6:30 PM, Justin Cranshaw <[email protected]> >> wrote: >> > I think the analytic answer should be sqrt(n/2). >> > >> > So let's suppose X and Y are random points in the n dimensional hypercube >> {0,1}^n. Let Z_i be an indicator variable that is 1 if X_i != Y_i and 0 >> otherwise. Then d(X,Y)^2 =sum (X_i - Y_i)^2 = sum( Z_i). Then the expected >> squared distance is E d(X,Y)^2 = sum( E Z_i) = sum( Pr[ X_i != Y_i]) = n/2. >> > >> > >> >
