Well, it is missing a matrix to define the projections, but it is the heart of the issue.
It also lacks a proof which is kind of important among some circles. The clever proof is actually not that hard to grok. On Fri, Oct 21, 2011 at 5:07 AM, Federico Castanedo <[email protected] > wrote: > 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] > > > > > >
