Generalisation of the notion of correlation for $\alpha$-stable distributions

Question

Pearson correlation is defined via variance and covariance, so will not work when applied to $\alpha$-stable distributions with $\alpha \neq 2$. Is there a way to generalise the notion of correlation to such distributions, e.g. by doing some form of renormalisation?

Example:

$$ \rho_{generalised}(X, Y) := \lim_{k \to \infty} \rho(X_k, Y_k) $$

where $X_k := \min(\max(X, -k), k)$ and $Y_k$ similarly.

Answer 1

I have found something that could be useful. An alternative to the traditional correlation for $\alpha$-stable distributions with $\alpha > 1$ is the signed symmetric covariation coefficient.

Definition. Let $(X_{1},X_{2})$ be a bivariate symmetric $\alpha$-stable random vector with $\alpha > 1$. The signed symmetric covariation coefficient between $X_{1}$ and $X_{2}$ is the quantity:

$$ scov(X_{1},X_{2}) = \kappa_{(X_{1},X_{2})} | \frac{[X_{1},X_{2}]_{\alpha}[X_{2},X_{1}]_{\alpha}}{|| X_{1}||_{\alpha}^{\alpha} || X_{2}||_{\alpha}^{\alpha}} |^{\frac{1}{2}}, $$

where

• $[X_{1},X_{2}]_{\alpha} = \int_{S_{2}} s_{1}s_{2}^{\langle\alpha -1\rangle} \mathbf{\Gamma}(d\mathbf{s})$, where $\mathbf{\Gamma}$ is the sprectral measure of the random vector $(X_{1},X_{2})$;

• $||X_{1}||_{\alpha} = ([X_{1},X_{1}]_{\alpha})^{\frac{1}{\alpha}}$;

• $ \kappa_{(X_{1},X_{2})} = sign([X_{1},X_{2}]_{\alpha}) \quad if\quad sign([X_{1},X_{2}]_{\alpha}) = sign([X_{2},X_{1}]_{\alpha})$;

• $ \kappa_{(X_{1},X_{2})} = - 1 \quad if\quad sign([X_{1},X_{2}]_{\alpha}) = - sign([X_{2},X_{1}]_{\alpha})$.

The following proposition shows that the signed symmetric covariation coefficient has desirable properties as does the ordinary correlation coefficient of a bivariate Gaussian random vector.

Proposition. Let $(X_{1},X_{2})$ be a bivariate symmetric $\alpha$-stable random vector with $\alpha > 1$. The signed symmetric covariation coefficient has the following properties:

1. $-1 \leq scov(X_{1},X_{2}) \leq 1$
2. if $X_{1},X_{2}$ are independent, then $scov(X_{1},X_{2}) = 0$;
3. $|scov(X_{1},X_{2})| = 1$ if and only if $X_{2} = \lambda X_{1}$ for some $\lambda \in \mathbb{R}, \, \lambda \neq 0$;
4. for $\alpha = 2$, $scov(X_{1},X_{2})$ coincides with the usual correlation coefficient.

For further details refer to: Estimation and comparison of signed symmetric covariation coefficient and generalized association parameter for alpha-stable dependence by Bernédy Kodia and Bernard Garel url: https://hal.archives-ouvertes.fr/hal-00951885/document

Answer 2

I found this idea in the book Gilchrist: "Statistical Modelling with Quantile Functions"; this are based on medians:

The comedian (don't laugh) of $X$ and $Y$ is defined by $$ \text{coMED}(X,Y) = M[(X-M(X))(Y-M(Y))] $$ where $M(X)$ is the median of $X$. Then one would have to standardize this by some measures of variability, that book gives MedAD as the sample median of the deviations $d_i$ from the median. How well that works for the stable distribution I do not know; you could investigate it by simulation.

Answer to additional question in the comments: "Do you think comedian is bilinear, like covariance?": First, the median itself satisfies $M(aX+b)=aM(X)+b$ (assuming that in the even $n$ case we use the mid-median, that is, taking the median of $X_1, X_2, X_3, X_4$ as $\frac{X_{(2)}+X_{(3)}}{2}$ where $X_{(i)}$ denotes the order statistics). This is because a linear transformation $ax+b$ of the data will not change the order ($a>0$) or will reverse the order ($a<0$). It is the last case which forces the use of the mid-median! But this is not enough to conclude that the median is linear, we would need that $M(X+Y)=M(X)+M(Y)$ and that is manifestly false. So without linearity of the median itself, bilinearity of the comedian is too much to ask. It cannot be true.