I have seen properties named $\rho$-, $\beta$-, and $\alpha$-mixing conditions in papers related to Copulas and Markov processes like this one:

In this paper, we identify conditions on $C$ that suffice for geometrically fast mixing rates. Geometric $\beta$-mixing, equivalent to geometric ergodicity for stationary Markov chains, is established under a rather strong condition that excludes copulas that exhibit tail dependence or asymmetry. Geometric $\rho$-mixing, which implies geometric $\alpha$-mixing, is obtained under a much weaker condition. We verify this condition for various parametric copula functions that are popular in applied work. $\rho$-, $\beta$-, and $\alpha$-mixing conditions may be used as the basis for a range of inequalities and limit theorems that are useful in demonstrating the asymptotic validity of statistical methods;

Beare, B., 2010, Copulas and temporal dependence, Econometrica, 78(1).

I am an engineering student and I have trouble understanding these conditions in statistical texts. Can somebody explain them?


Before going into the formal definition of mixing coefficients, the main idea of this notion is to quantify the dependence of a sequence of random variables. We would like to have theorems which do not only apply in the independent case, but "if the time series is not too far away from an independent sequence". There are several way for a sequence of random variables to be dependent, and mixing (in the sense $\rho$, $\alpha$ or $\beta$) is one of them.

First of all, we compare the dependence between two $\sigma$-algebras, either directly by comparing the difference between the measure of the intersection and the product of measures ($\alpha$-mixing), or the correlation between what is measurable with respect to the first and second $\sigma$-algebra ($\rho$-mixing).

Then we compare the dependence in these sense between the past and future $\sigma$-algebras.


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.