Timeline for distance and correlations
Current License: CC BY-SA 3.0
9 events
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
replaced http://stats.stackexchange.com/ with https://stats.stackexchange.com/
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| Sep 1, 2016 at 9:07 | comment | added | ttnphns | I agree with you. | |
| Sep 1, 2016 at 9:03 | comment | added | Petreius | I understand ! So in the end, proving that this distance is a Euclidean distance boils down to prove that covariance is a scalar product because we build d in such a way that the distance and the scalar product are tied by the cosine theorem. Also I think I solved my second problem with the axiom: up to a rescaling + centering, two perfectly correlated variables are identical. | |
| Sep 1, 2016 at 8:56 | comment | added | ttnphns | Both euclidean distance and scalar product are properties of euclidean space. The law of cosine establishes their relation. If you show that $\sum XY$ is a scalar product in euclidean space and you believe in the law of cosines then automatically follows that the d is euclidean. | |
| Sep 1, 2016 at 8:38 | comment | added | Petreius | ok, I would perfectly agree with you if I could find the result: " By the cosine theorem formula, the distance which is tied with the scalar product is Euclidean". I know that if the distance is Euclidean, then the cosine law holds but I'm not sure of the converse. | |
| Sep 1, 2016 at 8:07 | comment | added | ttnphns | What you say is correct, but I wonder why you still need a proof. 1) By the cosine theorem formula, the distance which is tied with the scalar product is euclidean (if vectors are considered in euclidean space). 2) r is cosine of the scalar product when the two variables are centered; also, if both variables are standardized to unit variance, then the cosine theorem becomes $d=\sqrt{2(1-r)}$. That distance is still euclidean because centering/scaling don't make the space spanned by the vectors non-euclidean. So this d is simply the euclidean d between the transformed, standardized variables. | |
| Sep 1, 2016 at 0:30 | comment | added | Petreius | Yes I have seen the link. But is the fact that Pearson correlation obeys the law of cosines (because the covariance is indeed a scalar product) enough to claim that the distance is Euclidean? Also, a Euclidean distance is a metric but it seems to me that the distance I consider does not verify the axiom "d(x,y) = 0 iff x = y". If X is proportional to Y for instance, the distance is zero because X and Y are perfectly correlated. | |
| Sep 1, 2016 at 0:08 | comment | added | ttnphns |
a proof of the fact that...d is Euclidean. But did you see there the link to this? The distance is euclidean because Pearson correlation is the cosine between centered vectors (variables), its formula is the formula of cosine similarity.
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| Aug 31, 2016 at 23:23 | history | asked | Petreius | CC BY-SA 3.0 |