Why Normal Copula Correlation Differs from Gaussian Parameter

# Copula Correlations


# Copula Correlations












It seems that the sample linear correlation coefficient $\hat{\rho}$ of samples generated by a copula that is parametrized by $\rho$ is unequal to $\rho$. For example, I construct a Normal copula with parameter $\rho = 0.9$. I generated a bunch of samples from this copula, and calculate their sample correlation coefficient $\hat{\rho}$. I am finding that $\hat{\rho}$ is slightly, but consistently less than $\rho$ for large samples (around 0.891). This is confusing to me as shouldn't $E[\hat{\rho}] = \rho$, as the Normal copula is solely parametrized by $\rho$? Any help would be greatly appreciated. Thanks!

## Answer by RRL (score 14, accepted)

https://quant.stackexchange.com/a/44979

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 jointly normal with correlation $\rho$, then with $\Phi$ denoting the univariate standard normal distribution function, we have

$$corr(\Phi(X), \Phi(Y)) = \frac{6}{\pi}\arcsin \frac{\rho}{2}$$.

For a proof, we can make a suitable transformation so that $X$ and $Y$have standard normal marginal distributions and the same correlation.

We then have

$$\tag{*}E(\Phi(X) \Phi(Y)) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \left(\int_{-\infty}^x \phi(u)\, du\right) \left( \int_{-\infty}^y \phi(v)\, dv\right)f(x,y;\rho)\, dx \, dy,$$

where $\phi$ is the standard normal density function and $f$ is the standard bivariate normal density function.

Observe that (*) is equivalent to the joint probability that $U \leqslant X$ and $V \leqslant Y$ where $U$ and $V$ are standard normally distributed random variables that are uncorrelated with each other and where each is uncorrelated with $X$ and $Y$:

$$E(\Phi(X)\Phi(Y)) = P(U \leqslant X, V\leqslant Y) = P(X-U \geqslant 0, Y-V \geqslant 0).$$

Now $X_1 = X-U$ and $X_2 = Y-V$ are both normally distributed with mean $0$, standard deviation $\sqrt{2}$, and with correlation

$$\text{corr}(X_1,X_2) = \frac{E((X-U)(Y-V))}{\sqrt{\text{var}(X-U)}\sqrt{\text{var}(Y-V)}} \\= \frac{E(XY)-E(XV) - E(YU) + E(UV)}{\sqrt{2}\sqrt{2}} \\ = \frac{\rho}{2}$$

It is well known that if $Z_1$ and $Z_2$ have a standard bivariate normal distribution with correlation $\rho'$, then the orthant probability is

$$P(Z_1 \geqslant 0, Z_2 \geqslant 0) = \frac{1}{4} + \frac{1}{2 \pi}\arcsin \rho'.$$

Hence,

$$E(\Phi(X)\Phi(Y)) = P(X_1 \geqslant 0, X_2\geqslant 0) = P(X_1/\sqrt{2} \geqslant 0, X_2/\sqrt{2}\geqslant 0) =\frac{1}{4} + \frac{1}{2\pi} \arcsin \frac{\rho}{2}.$$

SInce $\Phi(X)$ and $\Phi(Y)$ are uniformly distributed in $[0,1]$ we have

$$E(\Phi(X)) = E(\Phi(Y)) = \frac{1}{2}, \\ var(\Phi(X)) = var\Phi(Y)) = \frac{1}{12}$$

Thus

$$corr(\Phi(X), \Phi(Y)) = \frac{E(\Phi(X)\Phi(Y)) - E(\Phi(X))E(\Phi(Y))}{\sqrt{var(\Phi(X))}\sqrt{var(\Phi(Y))}} =\frac{6}{\pi} \arcsin \frac{\rho}{2}$$

## Answer by Attack68 (score 4)

https://quant.stackexchange.com/a/44975

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 estimator is obviously slightly larger and fits your described scenario, based on my hypothesis it implies you used 100 samples, which is quite low by the way.

## Answer by Richi Wa (score 3)

https://quant.stackexchange.com/a/44972

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 is sampling error.

Let's try to find the issue: What are your marginal distributions? What is the dimension of your model? How many samples do you draw?