Is the absolute value of distance covariance a metric? I'm reasonably certain the absolute value of the distance covariance satisfies


*

*$d(x, y) \ge 0$     (non-negativity, or separation axiom)

*$d(x, y) = 0$   if and only if   $x = y$     (identity of indiscernibles, or coincidence axiom)

*$d(x, y) = d(y, x)$     (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.
 A: 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.
A: 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.
A: 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?
A: 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:


*

*Conjugate symmetry: $Cov(X,Y) = Cov(Y,X)$

*Linearity in the first argument: $Cov(aX+bY,Z) = a\cdot Cov(X,Z) + b\cdot Cov(Y,Z)$

*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.
