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10

This may not be a consequence of biased estimators or sampling error. I don't think it is a coincidence that $$\frac{6}{\pi} \arcsin\left(\frac{0.9}{2} \right) = 0.891457\ldots \approx 0.891$$ Copula construction involves applying nonlinear transformations to random variables which need not preserve correlation. If random variables $X$ and $Y$ are ...


4

I would guess you are calculating the maximum likelihood estimator: $ \hat{\theta} = \frac{1}{N} \sum (x_i - \bar{x}) (y_i - \bar{y}) $ instead of the unbiased estimator: $ \hat{\theta} = \frac{1}{N-1} \sum (x_i - \bar{x}) (y_i - \bar{y}) $ The unbiased estimator has a bias of zero, i.e. : $ E_{x|\theta}[\hat{\theta}] - \theta = 0 $ The unbiased ...


3

This is an interesting observation that you have. The interesting part is "consistently smaller". The normal copula is based on a multivariate normal distribution. The correlation you get out is the correlation parameter you put in. Everything else is most probably due to an issue in your approach. If you did not say "consistently smaller", I would say it ...


3

Since I think this is of interest for other people, I will post the approach I found: First, let $C_n(u_1,\ldots,u_n)$ be a $n$ - dimensional Clayton copula with generator function $F$ and inverse $F^{-1}$. Then, Generate $n$ independent r.v. from $U (0,1)$ Calculate $n-1$ derivatives of $F$, where $F_{n-1}$ denotes the $n-1$-th - order derivative of $F$ ...


2

It's very difficult to find accessible material on copulas. I'm still struggling to understand them myself. While I haven't come across any videos that explain copulas well, I have found the following resources very helpful. A blog post: An intuitive, visual guide to copulas, by Thomas Wiecki, is truly introductory with absolute minimal math but it doesn't ...


2

Your reasoning for the first property does not look correct or at least I do not understand it. Your arguments for the second property seem sound. But your wording of the second property is a bit fuzzy. You should state this more clearly, for example: $C(1,\ldots,1,u_j,1,\ldots,1) = u_j$ for all $u_j\in [0,1]$ and $j\in 1,\ldots, d.$ You don't mention it ...


1

Clayton Copula-Matlab Code %% Simulations of Clayton copulas using conditional cdf %Example for theta=4 n=3000; theta=5; u=rand(1,n); y=rand(1,n); v=((y.^(1/(1+theta)).*u).^(-theta)+1-u.^(-theta)).^(-1/theta); x1=norminv(u); x2=norminv(v); plot(x1,x2,'.') Though for me, Gaussian seems uneasy, can you share a code, on how to do Gaussian Copula in ...


1

Fix u, obtain derivative of v. And do it again for fixing v. To get marginal density of v, one has to do the integration w.r.t. u


1

I am sharing what I found most helpful for understanding the concept 'copula'. It is an academic paper written by an undergraduate student in Netherland, for her college graduation in 2007. Copulas: modeling dependencies in Financial Risk Management This paper started with why measuring dependencies among financial assets is critical and continued with ...


1

In the line: sim17 = cgarchsim(fit17, n.sim = 1, m.sim = 1000, startMethod = "sample", preR = preR, preQ = preQ, preZ = preZ, prereturns = prereturns, presigma = presigma, preresiduals = preresiduals, cluster = NULL) , you should add n.start=0: sim17 = cgarchsim(fit17, n.sim = 1, n.start=0, m.sim = 1000, startMethod = "sample", preR = preR, ...


1

I assume with "spread of a copula" you mean the spread/dispersion of points in a bivariate scatter plot of the copula. This spread is related to the density of the copula. In an area where the copula is near uniform, scatter will be somewhat uniform. Obviously in areas where the density is concentrated, scatter will be concentrated. You can verify this by ...


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