2
$\begingroup$

I am reading the paper multimodal image registration by maximization of the correlation ratio and maybe I am wrong but I am not sure to understand why this formula is correct

$$ \rho(X, Y)=\frac{\operatorname{Cov}(X, Y)^{2}}{\operatorname{Var}(X) \operatorname{Var}(Y)}=\frac{[E(X Y)-E(X) E(Y)]^{2}}{\left[E\left(X^{2}\right)-E(X)^{2}\right]\left[E\left(Y^{2}\right)-E(Y)^{2}\right]} $$

Then the author prove that

$$ \rho(X, Y)=\cos ^{2} \alpha $$

But is it not supposed to be ?

$$ \rho(X, Y)=\frac{\operatorname{Cov}(X, Y)}{\sqrt{\operatorname{Var}(X)} \sqrt{\operatorname{Var}(Y)}} $$

which leads to

$$ \rho(X, Y)=\cos \alpha $$

$\endgroup$
2
  • $\begingroup$ What is alpha here? $\endgroup$
    – jros
    Commented Jan 24, 2022 at 14:36
  • $\begingroup$ The angle between the two random variables X and Y in the L2 space $\endgroup$
    – glouis
    Commented Jan 24, 2022 at 14:39

1 Answer 1

4
$\begingroup$

Yes, I think so. Looking at section 3.3 of the paper, the notation and the terminology the authors use seem to be wrong. They are talking about correlation but writing down squared correlation.

$\endgroup$

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.