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!


3 Answers 3


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

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


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

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

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


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


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.

  • $\begingroup$ Good point! This subtle point with the denominator ... The above holds straight forward for the correlation (which is the only parameter of the copula). If the op performed some more scaling with standard deviations different from zero, then estimating the variances and the covariance should cancel the denominator-right? But very good point for the bias! $\endgroup$
    – Richi Wa
    Commented Apr 7, 2019 at 17:30
  • $\begingroup$ Yes in retrospect my estimator is of covariance rather than correlation, and if you used a consistent estimator for each (sample) standard deviation then the the denominator does indeed cancel. So I am going to shift my position and hypothesise, instead, that the OP is estimating correlation based on an MLE covariance estimator divided by the known (population) standard deviations. And then I still think he is using 100 samples! Time will tell... $\endgroup$
    – Attack68
    Commented Apr 7, 2019 at 17:41

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?


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