I think, that's a good explanation of the Johnson-Lindenstrauss Lemma, which is the basis of the manifold learning theory using random projections.
2011/10/21 Ted Dunning <[email protected]> > Sort of. > > I may be misunderstanding the question. > > If you take a random orthogonal projection, then distances will be > preserved > within a reasonably small epsilon to reasonably high probability. > > Mathematically, if you take a random matrix \Omega which is tall and skinny > and do a QR decomposition: > > QR = \Omega > > Then Q is tall and skinny and Q^T projects vectors into a much lower > dimensional space. > > If you take vectors x and y, then > > |x - y| \approx |Q' x - Q'y| > > or, more precisely, we have, with high probability, > > |x - y| - \epsilon \le |Q' x - Q'y| \le |x - y| + \epsilon > > This is very close to what you were saying, I think. > > On Thu, Oct 20, 2011 at 4:07 PM, Lance Norskog <[email protected]> wrote: > > > Does this all translate to doing high-dimensional distance with random > > projection? Project each vector to one dimension and subtract? This > sounds > > like a really useful distance measure. > > > > On Wed, Oct 19, 2011 at 7:32 PM, Ted Dunning <[email protected]> > > wrote: > > > > > The distribution of the dot product of two randomly chosen, uniformly > > > distributed unit vectors is roughly normally distributed with a > standard > > > deviation that declines with increasing dimension roughly with your > > > observed > > > sqrt scaling factor. > > > > > > In fact, it is just this scaling property that makes the stochastic SVD > > > work > > > with high probability of high accuracy. The general property that > random > > > unit vectors are nearly orthogonal is called "quasi-orthogonality" > > > > > > On Wed, Oct 19, 2011 at 4:32 PM, Sean Owen <[email protected]> wrote: > > > > > > > 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. > > > > >> > > > > > >> > > > > > >> > > > > > > > > > > > > > > > > > > > > -- > > Lance Norskog > > [email protected] > > >
