I was reading
the random projection paper – written by Teija Seitola et al. – the other day, and I
think I have now quite a good understanding of how the method works.

The whole
thing is based on Johnson–Lindenstrauss lemma from 1984. The lemma says that
random projection (nearly) preserves the distances.

Let X be a
n×d-dimensional data matrix, where n is the number of data (or ensemble) points
and d is the actual dimension. Let f be d×k-dimensional random projection
matrix. Let x

_{i}and x_{j}be d-dimensional data vectors and ε>0. Now we have that
(1-ε)×ǁx

_{i}-x_{j}ǁ^{2}≤ ǁf(x_{i})-f(x_{j})ǁ^{2}≤ (1+ε)×ǁx_{i}-x_{j}ǁ^{2}.
In the paper
there are also to explicit formulas for k that surprisingly do not depend on
the dimension d, but only on the number of observations n and the error ε. For
example, k > 4×(ε

^{2}/2-ε^{3}/3)^{-1}×log(n).
So if, for
example in correlation dimension, we set n = 2000 and ε = 0.1, we need k that is
higher than 4*(.1^2/2-.1^3/3)^(-1)*log(2000) = 6515.1. I tested this with
320-dimensional Lorenz and set k to 10000 and the thing works since the curves became so similar (see figure below).

As a
conclusion, I would say that random projection method works “better” in large than small dimensional problems, and that the dimension have to be really large before it is worth to use random projections in applications where the
preservation of distances is very very important. Note that there are other kinds of
applications too…

Below, there
is a little Matlab example.

Ciao,

Janne

%% Random
projection Matlab example

n = 2000; %
Number of time points

d = 20000; %
Dimension

k = 10000; %
Random dimension

X =
10+5*randn(n,d); % Data matrix

R = randn(d,k);
% Random projection matrix

for kkk = 1:d

R(kkk,:) = R(kkk,:)/norm(R(kkk,:)); %
Normalization

end

P = X*R; %
Projection

% Distances

ii = 10; jj =
20;

norm(X(ii,:)-X(jj,:))

norm(P(ii,:)-P(jj,:))

http://nuit-blanche.blogspot.com/2017/11/biologically-inspired-random-projections.html

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