# Pearson correlation and metric properties

Assuming that the data set was $z$-standardized to zero mean and unit variance (also assuming that it does not contain constant vectors).

Then Pearson's r reduces to Covariance: $$\rho(X,Y) := \frac{Cov(X,Y)}{\sigma(X)\sigma(Y)} = Cov(X,Y)$$

Now I'm investigating the dissimilarity function $$d(X,Y):=\sqrt{1 - \rho(X,Y)}$$ which is the square root of a common transformation of $\rho$ for use as a dissimilarity function.

It can be shown that given above preconditions, it is in fact a linear multiple of Euclidean distance, and thus trivially metric:

$$\sqrt{\sum_i (x_i - y_i)^2} = \sqrt{\sum_i x_i^2+\sum_i y_i^2 - 2 \sum_i x_i\cdot y_i} \\ = \sqrt{n + n - 2n \cdot Cov(X,Y)} = \sqrt{2n} \cdot d(X,Y)$$ i.e. $$d(X,Y) = euclidean(X,Y) / \sqrt{2n}$$ Now I'm wondering if the properties of this function $d$ have been further explored. Is it metrical under a broader set of conditions than z-standardized data sets? Do you know related literature or proofs?

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Hmm, the only thing which comes to my mind is that this means: changing the view into the data. It means that each observed person defines a dimension in the euclidean space and each item, measured at this person, defines a vector from the origin into that space. Assume only two items now: then d is the distance between the arrowheads of the vectors. The beforehand centering of the items means to look at the difference-vectors (from the vectors to the means-vectors for each item) and standardizing means adapting lengthes. I've fiddled a bit with this but have no yet ready article about this. –  Gottfried Helms Jun 6 '14 at 10:20
Interesting idea. z-standardization is a linear transformation in the transposed space. If metric properties survive transposition, this would pretty much yield a proof. –  Anony-Mousse Jun 6 '14 at 10:58