# Tag Info

7

The best introduction to copulas I know, i.e. with rigour and intuition, is the following. THE QUANT CLASSROOM BY ATTILIO MEUCCI A Short, Comprehensive, Practical Guide to Copulas Visually introducing a powerful risk management tool to generalize and stress-test correlations

6

I found Coping With Copulas by Thorsten Schmidt really helped me to get a more basic understanding of copulas. As well as looking at some simple examples in R and thinking about different directions the transformations can happen. To answer your actual question I'll attempt to describe the steps involved as simply as I can. Let's say you use the copula ...

6

In the theory of copulas you want to model a multivariate (often bivariate) distribution and keep the marginals fixed. Thus you have random variables $X$ and $Y$ with cdf $F_X(x) = P[X \le x]$ and $F_Y(y) = P[Y\le y]$ and you want to find some $F_{X,Y}(x,y) = P[X \le x, Y\le y]$ such that when you look at marginals you get $F_{X,Y}(x,\infty) = F_X(x)$ and ...

5

Here is a working paper that you may be interested in.

5

It is hard to find a stable non-trivial dependence structure in financial data. Usually when such is found it is hard to rationalize. One of my favorite (although I am sure there are others) is the so called "Presidential Puzzle". This is an old finding by Santa-Clara and Valkanov (2003) They find that " Excess return in the stock market is higher under ...

4

Since your standard for meaningful risk measurement is pretty low ("Meaningful in this context simply means that we really capturing a sort of risk of the overall portfolio.") There are limited cases where such an approach might make sense. From a broader perspective I think it makes very little sense. If your matrix of pairwise risk numbers is symmetric ...

3

The answer of user27915816 led me into the right direction, yet I think I found an even better generalization: Distance Correlation (dCor) There are several reasons for that: It generalizes classical (i.e. linear) correlation in the sense that linearity is a special case. It gives identical readings for linear dependence. There are analogs for variance, ...

3

Mutual information measures how much knowing one variable reduces uncertainty about another variable. It considers any type of dependency (linear or non-linear), it's measured in bits, and it is widely used in machine learning, computer vision NLP and other fields.

3

Do you refer with 'negative tail dependence' to the case that one variable has a extremely low value and the other random variable has an extremely large value, i.e., $$\tau=\lim_{p \rightarrow 0} \frac{Pr[x>Q_x(1-p),y<Q_y(p)]}{p},$$ where $Q_x(1-p)$ and $Q_y(p)$ refer to the $(1-p)$-th quantile of the random variable $x$ and the $p$-th quantile of ...

2

There is a brief and not overly technical introduction here: http://prescientmuse.blogspot.co.uk/2015/01/a-brief-introduction-to-copula.html And an application of use in a trading system with full R code here: http://prescientmuse.blogspot.co.uk/2015/02/vanilla-trading-algorithm.html Hope that helps!

2

I think you fail to understand Multivariate Garch model such as DCC models since they do take into account non linearity. They are interested in jointly modeling the time series behavior of multiple conditional variance processes. Each couple of series has its own particular conditional correlation process evolving trough time in a non-linear way. In fact ...

1

Assume that $H_1 < S_0 < H_2$. let \begin{align*} \tau_1 = \inf\{t: \, t>0 \text{ and } S_t \le H_1 \}, \end{align*} and \begin{align*} \tau_2 = \inf\{t: \, t>0 \text{ and } S_t \ge H_2 \}. \end{align*} Then, the option payoff is defined by \begin{align*} X\, \mathbb{I}_{\{\tau_1 >T\}} \mathbb{I}_{\{\tau_2 >T\}} &= X\, \mathbb{I}_{\{\...

1

Let $X$ be an $n \times 1$ random vector and $w$ a vector of coefficients. Then we know that $$Var(w^TX)=w^T \;Var(X) \;w=w^T\Sigma w$$ where $\Sigma$ is the covariance matrix of $X$. Now suppose that $C$ is the correlation matrix of $X$ and $e$ is the $n \times 1$ vector of all ones. We see that $$w^T C w = w^Te +\sum_{i \neq j}w_iw_jc_{ij}$$ where $c_{ij}$ ...

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