7
$\begingroup$

Given two random variables $X_1$ and $X_2$, the Kendall-tau correlation coefficient could be defined as $$ \tau(X_{1},X_{2})=\mathbb{P}\Big((X_{1}-\tilde{X}_{1})(X_{2}-\tilde{X}_{2})>0\Big)-\mathbb{P}\Big((X_{1}-\tilde{X}_{1})(X_{2}-\tilde{X}_{2})<0\Big) $$ where $(\tilde{X}_1, \tilde{X}_2)$ are independent copies of $(X_1,X_2)$.

I was wondering if it is possible to provide an interpretation of Kendall-tau correlation as a kernel corresponding to RKHS, explicitly or implicitly ?

$\endgroup$
2
$\begingroup$

By the Moore-Aronszajn theorem, $\tau$ is the kernel for some RKHS iff it's symmetric and positive semidefinite. (The link uses the term "positive definite" to mean the equivalent of psd for matrices, unfortunately; that terminology isn't standardized.)

Update: What I had here before was based on a mistaken understanding of the framework (as well as a mistaken definition of $\tau$ in the original question); see the comments.

The new $\tau$ is clearly symmetric. I'm not sure yet whether it's psd. As @cardinal pointed out, it does at least satisfy $\tau(X, X) = 1$ and $-1 \le \tau(X, Y) \le 1$ for continuous RVs, which is a good start.

$\endgroup$
  • 1
    $\begingroup$ You lost me at the update. What does it mean to be "a standard normal with value $-1$"? And why is it not always the case (at least for continuous variables) that $\Pr(X\gt \tilde{X})=1/2$? $\endgroup$ – whuber Sep 18 '13 at 21:11
  • 1
    $\begingroup$ @whuber By that I mean that $X$ is assumed to be distributed as $N(0, 1)$ and observed to be -1. $\mathbb{P}(X > \tilde{X}) \ne 1/2$ because we have both an observation and an assumed distribution for $X$, at least in my understanding of the question, and so that's basically the value of the CDF. Based on the update from after my answer, though, it looks like the original definition of $\tau$ (on which my answer is based) may have been incorrect.... $\endgroup$ – Dougal Sep 18 '13 at 21:16
  • 2
    $\begingroup$ The original definition of $\tau$ takes as a domain an appropriate space of random variables, or, more properly, induced measures from a probability space, not a point in $\mathbb R \times \mathbb R$. Indeed, when interpreted in this way, I believe the answer is affirmative, i.e., there is a RKHS under an appropriate restriction on the random variables. $\endgroup$ – cardinal Sep 18 '13 at 21:19
  • $\begingroup$ @cardinal It makes more sense that $\tau$ is a function on pairs of random variables. But: since $X_1$ and $\tilde{X}_1$ have the same distribution, I don't see how $\mathbb{P}\left( (X_1 - \tilde{X}_1) (X_2 - \tilde{X}_2) > 0 \right) \ne \mathbb{P}\left( (\tilde{X}_1 - X_1) (X_2 - \tilde{X}_2) > 0 \right) = \mathbb{P}\left( (X_1 - \tilde{X}_1) (X_2 - \tilde{X}_2) < 0 \right)$, in which case the new definition of $\tau$ is always 0...as should be the old one, as whuber said, since $\mathbb{P}( X > \tilde{X} ) = 1/2$ when $X$ and $\tilde{X}$ have the same distribution. $\endgroup$ – Dougal Sep 18 '13 at 21:27
  • 1
    $\begingroup$ I just realized that $X_1$ and $Y_1$ are jointly-distributed, as are $\tilde{X}_1$ and $\tilde{X}_2$, though the tilded and the non-tilded are independent of one another. It makes sense to me now. $\endgroup$ – Dougal Sep 18 '13 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.