A simpler condition for consistency of a kernel independence test
Abstract
A statistical test of independence may be constructed using the HilbertSchmidt Independence Criterion (HSIC) as a test statistic. The HSIC is defined as the distance between the embedding of the joint distribution, and the embedding of the product of the marginals, in a Reproducing Kernel Hilbert Space (RKHS). It has previously been shown that when the kernel used in defining the joint embedding is characteristic (that is, the embedding of the joint distribution to the feature space is injective), then the HSICbased test is consistent. In particular, it is sufficient for the product of kernels on the individual domains to be characteristic on the joint domain. In this note, it is established via a result of Lyons (2013) that HSICbased independence tests are consistent when kernels on the marginals are characteristic on their respective domains, even when the product of kernels is not characteristic on the joint domain.
1 Introduction
The HilbertSchmidt Independence Criterion [4] provides a measure of dependence between random variables on domain , and on domain , with joint probability measure on . This dependence measure may be used in statistical tests of dependence [5, 6]. The simplest way to understand HSIC is as the distance between an embedding of the joint distribution and the product of the marginals, to an appropriate feature space [9, 3], which is in our case a reproducing kernel Hilbert space. The distance covariance of [12] is a special case, for a particular choice of kernel [8]. We say the feature space is characteristic when the embedding is injective, and uniquely identifies probability measures [11, 1, 10]. A test based on HSIC is consistent when product of kernels on the domains being compared is characteristic to the joint domain [1, Theorem 3]. This is shown to be the case e.g. when Gaussian kernels are used on each of the domains.
We propose a simpler condition: namely, that the kernels on each of the individual domains and should be characteristic to those domains. The result is a direct consequence of [7, Lemma 3.8]. The result is of particular interest since it may be easier to define characteristic kernels on individual domains than on the joint domain. For example, characteristic kernels may be defined on the group of orthogonal matrices [2, Section 4], and on the semigroup of vectors of nonnegative reals [2, Section 5], however a kernel jointly characteristic to both domains (i.e., to orthogonal matrix/nonnegative vector pairs) is harder to define.
2 Results
We begin with a result from [10] that characteristic, translation invariant kernels provide injective embeddings of finite signed measures.
Proposition 1 (Injective embeddings of finite signed measures).
Let be a Polish, locally compact Hausdorff space. Let be a kernel, i.e. a bounded kernel for which , where is the class of continuous functions on that vanish at infinity.^{1}^{1}1Continuous functions vanishing at infinity are members of such that for all the set is compact. Assume , i.e. the kernel is translation invariant. Define as the RKHS induced by . The following statements are equivalent:

is characteristic

The embedding of a finite signed Borel measure , defined as
(1) is injective.
This result may be obtained by combining [10, Proposition 2], which states that an RKHS is universal iff the embedding in (1) is injective, with the result in [10, Section 3.2] that translation invariant kernels are universal iff they are characteristic.
This being the case, a minor adaptation of the proof of [7, Lemma 3.8] leads to the following result.
Theorem 2 (Characteristic kernels and independence measures).
Let and be kernels for the respective RKHSs on and on , with respective feature maps and . Assume both and are characteristic, translation invariant kernels, satisfying the conditions of Proposition 1. Define the finite signed measure
Define the covariance operator as the embedding of this signed measure into the tensor space^{2}^{2}2The tensor product is defined such that , . ,
Then iff .
Proof.
The result is straightforward. We now prove the other direction. For every and , we define the finite signed Borel measure
where is the indicator of the set The embedding of this measure to is injective, and is written
where we have used the linearity of the tensor product
Since the embedding is injective, we have that . Since this is true for all , we have that
Define the finite signed measure on , . The above equation can be interpreted as the embedding of this measure to ,
hence , given that the embedding is injective. We conclude that for all Borel sets , and hence . ∎
An important point to note is that the embedding of need not be characteristic to all probability measures: only the embeddings of each of the individual dimensions and need be characteristic. A second point is that a consistent test still requires characteristic kernels on both domains; it is not sufficient for one domain alone to have a characteristic kernel. A simple example can be used to illustrate the resulting failure mode: with a characteristic kernel, with the linear kernel , and points are distributed uniformly on a circular ring centered at the origin. The data are dependent, but HSIC with these kernels will not detect this dependence.
Acknowledgements: Thanks to Joris Mooij, Jonas Peters, Dino Sejdinovic, and Bharath Sriperumbudur for helpful discussions.
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