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


*

*$-1 \leq scov(X_{1},X_{2}) \leq 1$

*if $X_{1},X_{2}$ are independent, then $scov(X_{1},X_{2}) = 0$;

*$|scov(X_{1},X_{2})| = 1$ if and only if $X_{2} = \lambda X_{1}$ for some $\lambda \in \mathbb{R}, \, \lambda \neq 0$;

*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
A: 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.
