2

I'm reasonably certain the absolute value of the distance covariance satisfies

  1. $d(x, y) \ge 0$ (non-negativity, or separation axiom)
  2. $d(x, y) = 0$ if and only if $x = y$ (identity of indiscernibles, or coincidence axiom)
  3. $d(x, y) = d(y, x)$ (symmetry)

But I'm not sure about:

4) $d(x, z) \le d(x, y) + d(y, z)$ (subadditivity / triangle inequality).

I'm thinking in particular of a space of timeseries, but I don't think that matters too much.

amoeba
  • 93,463
  • 28
  • 275
  • 317
naught101
  • 4,973
  • 1
  • 51
  • 85

4 Answers4

4

If by distance covariancen you mean this

http://en.wikipedia.org/wiki/Distance_correlation#Distance_covariance

then the point 2. is false. There are many possibilities for this distance to be 0.

Münchhausen
  • 51
  • 2
2

Distance covariance and distance Correlation do not satisfy the triangular inequality. Pearson's correlation itself on centered data or otherwise, does not satisfy the inequality, and hence is known as a semi-metric.

hearse
  • 2,355
  • 1
  • 17
  • 30
  • This appears to conflict with Münchhausen answer. – naught101 Mar 08 '12 at 04:17
  • It does not. Münchhausen deals with point 2. Mine deals with point 4. Münchhausen says that if x,y are statistically dependent, yet unequal. dCov(x,y) can be zero. – hearse Mar 09 '12 at 12:27
  • I meant the "semi-metric" comment. A semi-metric DOES satisfy point 2. – naught101 Mar 10 '12 at 02:24
  • 1
    Also, http://en.wikipedia.org/wiki/Distance_correlation says "[distance correlation] is zero if and only if the random variables are statistically independent" – naught101 Mar 10 '12 at 04:44
  • If you read again on what I said-The semi-metric, was about the Pearson's correlation. Distance Correlation and Pearson's Correlation are different. – hearse Mar 10 '12 at 22:42
  • I know you were talking about Pearson's, but Pearson's correlation coefficient can clearly be 0 when x!=y.. – naught101 Mar 12 '12 at 01:11
  • 1
    Ok-that was a misuse of terminology. Pearson's correlation cannot be a semimetric because of the iff condition over 2. – hearse Mar 12 '12 at 14:47
1

You got it: covariance, of course, induces a metric: $$ d(x,y) = \sqrt{ \text{Cov}(x-y,x-y) } $$ but NOT: $$ d(x,y) = \text{Cov}(x,y) $$ Actually I'm thinking about the property of the triangle inequality. Or, what effect does this property induce?

gung - Reinstate Monica
  • 132,789
  • 81
  • 357
  • 650
misssprite
  • 136
  • 2
0

Ok, from doing some more reading, the assumption in the question is wrong. Covariance actually defines an inner product, if (as I understand it) you first remove the mean of each variable. That is, covariance satisfies the following:

  1. Conjugate symmetry: $Cov(X,Y) = Cov(Y,X)$
  2. Linearity in the first argument: $Cov(aX+bY,Z) = a\cdot Cov(X,Z) + b\cdot Cov(Y,Z)$
  3. Positive semi-definiteness: $Cov(X,X) = 0$ with equality only for X constant (This is where the removal of the mean comes in: $Cov(X,Y) = \langle X-\bar{X},Y-\bar{Y}\rangle$, such that $Var(X) = 0 \iff X_i = \bar{X}\ \forall\ i$ [X is constant, so removing the mean gives the zero vector])

This inner product induces a norm, also with the mean of the random variable removed: $\|X-\bar{X}\| = Var(X)$

And the norm in turn induces a metric: $d(X,Y) = \|(X-\bar{X})-(Y-\bar{Y})\| = Var(X-Y)$

So in a sense, Covariance does define a metric, just not in the way I first thought. It's been a whole 6 months since I studies any linear algebra, so I might have some of those slightly screwed up. Please correct me if I'm wrong!

The same applies to distance covariance, as is pretty clear from the wikipedia article, although it's a different inner product.

naught101
  • 4,973
  • 1
  • 51
  • 85
  • Although a covariance defines an inner product (Parzen, 1961), it is only SEMI-definite. To be a metric, you need strict definiteness. (e.g. zero covariance doesn't imply identically distributed random variables.) I agree with @Münchhausen that this is not the case for distance covariances, unless you consider a equivalence class of zero (distance) covariance random variables. – Memming Mar 26 '12 at 19:06
  • The "positive semi-definite" aspect is defined as the inner product of X *with itself*. How can it *not* be identically distributed? – naught101 Mar 27 '12 at 00:01