1
$\begingroup$

Introduction. I post here the same question that I asked on math.stackexchange, since it might be more relevant to this forum: Could it make sense to calculate the "distance correlation" among two probability measures (probability distributions)?.

Question. Would it make sense to calculate the "distance correlation" introduced by Székely et al.(2007) (also here on Wikipedia) among two probability measures $P$ and $Q$, i.e. $\mathcal{R}(P,Q)$, rather than between two random vectors $X$ and $Y$, as stated on the Székely et al.(2007) paper?

Edited after the Christian Hennig's Answer. @Christian Hennig, thanks! Do you mean something like this?

  1. If $X \sim P$ and $Y \sim Q$ and
  2. If (from @Artem Mavrin's reply in Joint probability measure - note that I used your notation for the joint probability, i.e. $J$)

the joint distribution of random variables $X$ and $Y$, defined on a common probability space $(\Omega, \mathcal{A}, J)$ and taking values in measurable spaces $(\mathcal{X}, \mathcal{B})$ and $(\mathcal{Y}, \mathcal{C})$, respectively, is the probability measure defined on $(\mathcal{X} \times \mathcal{Y}, \mathcal{B} \otimes\mathcal{C})$ by $$ J_{X, Y}(E) = J((X, Y) \in E) $$ for all $E \in \mathcal{B} \otimes \mathcal{C}$.

  1. Then, we can calculate the correlation distance among two probability measures $P$ and $Q$, i.e. $\mathcal{R}(P,Q)=\mathcal{R}\left({J_{X, Y}(E)}\right)$.

But then, I would not know how to connect $\mathcal{R}(P,Q)=\mathcal{R}\left({J_{X, Y}(E)}\right)$ to $\mathcal{R}(X,Y)$, which is (the latter) what stated on the Székely et al.(2007) paper... I think I did not get this connection: $\mathcal{R}(P,Q) = \mathcal{R}(X,Y)$.

$\endgroup$

1 Answer 1

1
$\begingroup$

If you just specify the distributions $P$ and $Q$ (of the two random variables $X$ and $Y$), you don't specify the dependence structure between them, which is essential for distance correlation. You'd need to define it depending on the joint distribution (let's call it $J$) of $X$ and $Y$, so you can have ${\cal R}(J)$, which would be the same thing as defining it based on $X$ and $Y$ (involving the dependence structure between them).

$\endgroup$
11
  • 2
    $\begingroup$ +1 - The same is true for the ordinary Pearson correlation $\endgroup$
    – Henry
    Commented Sep 20, 2023 at 9:56
  • $\begingroup$ Thanks @Christian Hennig! I wrote my comment in the main question :-) By chance, do you have a reference to look up? $\endgroup$
    – Ommo
    Commented Sep 20, 2023 at 10:43
  • $\begingroup$ Thanks @Henry! I try to ask the same as to Christian Henning.. By chance, do you have a reference to suggest about the comment you wrote? $\endgroup$
    – Ommo
    Commented Sep 20, 2023 at 10:46
  • $\begingroup$ @Ommo Saying "Pearson correlation is affected by the dependence structure" is not controversial: it is zero when the two random variables are independent (easily proved since the covariance will then be zero) so when it is not zero you can say they are not independent. The assertion with "distance correlation" is that you can also say that if it is zero then they are independent, which is not generally automatically the case with Pearson correlation. $\endgroup$
    – Henry
    Commented Sep 20, 2023 at 10:51
  • 1
    $\begingroup$ @Ommo The two are just the same thing by definition. ${\cal R}(J)$ is just defined by what goes on between random variables $(X,Y)$ that are jointly distributed according to $J$. So you wouldn't define $ {\cal R}(J)$ in a different way and then somehow "connect" this to ${\cal R}(X,Y)$, rather you define ${\cal R}(X,Y)$ as you know it, and then have the right to call this thing also $ {\cal R}(J)$. The thing that you may get wrong here is the expectation that these should somehow look very different, yet still be the same. They're not different in any sense. $\endgroup$ Commented Sep 20, 2023 at 13:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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