Euclidean distance scaled by the variance

Hello,

Let a and b are two n-dimensional vectors.

Also, let s2_i is the variance of the of the i-th element of all vectors and s2_mean is the average value of the variance for i=1,2,...,n

How the following version of Euclidean distance is implemented?

d(a,b)=sqrt(sum (s2_i/s2_mean)*(a_i-b_i)^2)

where a_i and b_i are the i-th elements of vectors a and b respectively.

The function

y=pdist2(a,b,'seuclidean') corresponds to the euclidean distance divided by the corresponding element of the standard deviation.

The one I am refer to is a slight variation to the 'seuclidean'.

Thank you very much.

Best,

Natasha

Answer 1

[Jan]

What exactly is "the variance of the of the i-th element of all vectors"? You have two [1 x n] vectors a and b, but what does "all" mean then?

"d(a,b)=sqrt(sum (s2_i/s2_mean)*(a_i-b_i)^2)" seems to be almost the full solution:

d = sqrt(sum(s2 ./ s2_mean) .* (a - b) .^ 2)