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Glen_b
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In general this will not be a suitable estimate of the covariance matrix. Indeed the copula alone cannot be used to get a reasonable estimate of covariance, since the copula is invariant to monotonic increasing transformation of the margins, but the covariance is certainly not.

However, in some particular situations it may be possible to get a good estimate of a covariance matrix.

For example, consider he situation where $(X,Y)$ is bivariate Gaussian but where the only information you have on the strength of the bivariate relationship is the value of the Kendall correlation.

Even in this case, it's not going to be suitable to use the Kendall correlation directly. Here's the result for 1000 simulated samples across the range of possible correlation values, each with sample size 100:

Sample Kendall correlation vs population Pearson correlation

We can see that the sample Kendall correlation is closer to 0 than the population Pearson correlation in general, and the difference is fairly subtantialsubstantial when $|\rho|$ is between about 0.5 and 0.975 or so. The relationship is fairly linear near 0 but nonlinear further away.

You could use the relationship between the population Kendall and Pearson correlations for the bivariate normal case ($\tau=\frac{2}{\pi}\arcsin({\rho})$) to estimate the Pearson correlation from the sample Kendall correlation - and from there, obtain an estimate of the covariance.

Here's a simulation to show this approach in action. In this case I just used the naive estimator $\hat{\rho}=\sin(\frac{\pi}{2}\hat{\tau})$, and samples of size 100:

1: plot of r vs rho. 2: plot of sin(pi t/2) vs rho.  3: plot of sin(pi t/2) vs r

As we see, the naive estimator of $\rho$ based off a transformed sample Kendall correlation is excellent (bottom left); its variance is only a little larger than using the usual sample Pearson correlation (top left). Indeed the transformed sample Kendall correlation is very close to the corresponding value of the sample Pearson correlation (bottom right). There may well be a better estimator than the one I used, but this indicates that an approach based on the sample Kendall correlation may well be feasible.

If you do this at other typical sample sizes the spread around the relationships changes (the variance of the sample correlations will be proportional to $\frac{1}{n}$) but the basic pattern of results is similar.

The relationship between the Kendall and Pearson correlations will depend on the bivariate distribution (not just the copula) so I think you'll need a new analysis for each such case.

In general this will not be a suitable estimate of the covariance matrix. Indeed the copula alone cannot be used to get a reasonable estimate of covariance, since the copula is invariant to monotonic increasing transformation of the margins, but the covariance is certainly not.

However, in some particular situations it may be possible to get a good estimate of a covariance matrix.

For example, consider he situation where $(X,Y)$ is bivariate Gaussian but where the only information you have on the strength of the bivariate relationship is the value of the Kendall correlation.

Even in this case, it's not going to be suitable to use the Kendall correlation directly. Here's the result for 1000 simulated samples across the range of possible correlation values, each with sample size 100:

Sample Kendall correlation vs population Pearson correlation

We can see that the sample Kendall correlation is closer to 0 than the population Pearson correlation in general, and the difference is fairly subtantial when $|\rho|$ is between about 0.5 and 0.975 or so. The relationship is fairly linear near 0 but nonlinear further away.

You could use the relationship between the population Kendall and Pearson correlations for the bivariate normal case ($\tau=\frac{2}{\pi}\arcsin({\rho})$) to estimate the Pearson correlation from the sample Kendall correlation - and from there, obtain an estimate of the covariance.

Here's a simulation to show this approach in action. In this case I just used the naive estimator $\hat{\rho}=\sin(\frac{\pi}{2}\hat{\tau})$, and samples of size 100:

1: plot of r vs rho. 2: plot of sin(pi t/2) vs rho.  3: plot of sin(pi t/2) vs r

As we see, the naive estimator of $\rho$ based off a transformed sample Kendall correlation is excellent (bottom left); its variance is only a little larger than using the usual sample Pearson correlation (top left). Indeed the transformed sample Kendall correlation is very close to the corresponding value of the sample Pearson correlation (bottom right). There may well be a better estimator than the one I used, but this indicates that an approach based on the sample Kendall correlation may well be feasible.

If you do this at other typical sample sizes the spread around the relationships changes (the variance of the sample correlations will be proportional to $\frac{1}{n}$) but the basic pattern of results is similar.

The relationship between the Kendall and Pearson correlations will depend on the bivariate distribution (not just the copula) so I think you'll need a new analysis for each such case.

In general this will not be a suitable estimate of the covariance matrix. Indeed the copula alone cannot be used to get a reasonable estimate of covariance, since the copula is invariant to monotonic increasing transformation of the margins, but the covariance is certainly not.

However, in some particular situations it may be possible to get a good estimate of a covariance matrix.

For example, consider he situation where $(X,Y)$ is bivariate Gaussian but where the only information you have on the strength of the bivariate relationship is the value of the Kendall correlation.

Even in this case, it's not going to be suitable to use the Kendall correlation directly. Here's the result for 1000 simulated samples across the range of possible correlation values, each with sample size 100:

Sample Kendall correlation vs population Pearson correlation

We can see that the sample Kendall correlation is closer to 0 than the population Pearson correlation in general, and the difference is fairly substantial when $|\rho|$ is between about 0.5 and 0.975 or so. The relationship is fairly linear near 0 but nonlinear further away.

You could use the relationship between the population Kendall and Pearson correlations for the bivariate normal case ($\tau=\frac{2}{\pi}\arcsin({\rho})$) to estimate the Pearson correlation from the sample Kendall correlation - and from there, obtain an estimate of the covariance.

Here's a simulation to show this approach in action. In this case I just used the naive estimator $\hat{\rho}=\sin(\frac{\pi}{2}\hat{\tau})$, and samples of size 100:

1: plot of r vs rho. 2: plot of sin(pi t/2) vs rho.  3: plot of sin(pi t/2) vs r

As we see, the naive estimator of $\rho$ based off a transformed sample Kendall correlation is excellent (bottom left); its variance is only a little larger than using the usual sample Pearson correlation (top left). Indeed the transformed sample Kendall correlation is very close to the corresponding value of the sample Pearson correlation (bottom right). There may well be a better estimator than the one I used, but this indicates that an approach based on the sample Kendall correlation may well be feasible.

If you do this at other typical sample sizes the spread around the relationships changes (the variance of the sample correlations will be proportional to $\frac{1}{n}$) but the basic pattern of results is similar.

The relationship between the Kendall and Pearson correlations will depend on the bivariate distribution (not just the copula) so I think you'll need a new analysis for each such case.

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Glen_b
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In general this will not be a suitable estimate of the covariance matrix. Indeed the copula alone cannot be used to get a reasonable estimate of covariance, since the copula is invariant to monotonic increasing transformation of the margins, but the covariance is certainly not.

However, in some particular situations it may be possible to get a good estimate of a covariance matrix.

For example, consider he situation where $(X,Y)$ is bivariate Gaussian but where the only information you have on the strength of the bivariate relationship is the value of the Kendall correlation.

Even in this case, it's not going to be suitable to use the Kendall correlation directly. Here's the result for 1000 simulated samples across the range of possible correlation values, each with sample size 100:

Sample Kendall correlation vs population Pearson correlation

We can see that the sample Kendall correlation is closer to 0 than the population Pearson correlation in general, and the difference is fairly subtantial when $|\rho|$ is between about 0.5 and 0.975 or so. The relationship is fairly linear near 0 but nonlinear further away.

You could use the relationship between the population Kendall and the Pearson correlationcorrelations for the bivariate normal case ($\tau=\frac{2}{\pi}\arcsin({\rho})$) to estimate the Pearson correlation from the sample Kendall correlation - and from there, obtain an estimate of the covariance.

Here's a simulation to show this approach in action. In this case I just used the naive estimator $\hat{\rho}=\sin(\frac{\pi}{2}\hat{\tau})$, and samples of size 100:

1: plot of r vs rho. 2: plot of sin(pi t/2) vs rho.  3: plot of sin(pi t/2) vs r

As we see, the naive estimator of $\rho$ based off a transformed sample Kendall correlation is excellent (bottom left); its variance is only a little larger than using the usual sample Pearson correlation (top left). Indeed the transformed sample Kendall correlation is very close to the corresponding value of the sample Pearson correlation (bottom right). There may well be a better estimator than the one I used, but this indicates that an approach based on the sample Kendall correlation may well be feasible.

If you do this at other typical sample sizes the spread around the relationships changes (the variance of the sample correlations will be proportional to $\frac{1}{n}$) but the basic pattern of results is similar.

The relationship between the Kendall and Pearson correlations will depend on the bivariate distribution (not just the copula) so I think you'll need a new analysis for each such case.

In general this will not be a suitable estimate of the covariance matrix. Indeed the copula alone cannot be used to get a reasonable estimate of covariance, since the copula is invariant to monotonic increasing transformation of the margins, but the covariance is certainly not.

However, in some particular situations it may be possible to get a good estimate of a covariance matrix.

For example, consider he situation where $(X,Y)$ is bivariate Gaussian but where the only information you have on the strength of the bivariate relationship is the value of the Kendall correlation.

Even in this case, it's not going to be suitable to use the Kendall correlation directly. Here's the result for 1000 simulated samples across the range of possible correlation values, each with sample size 100:

Sample Kendall correlation vs population Pearson correlation

We can see that the sample Kendall correlation is closer to 0 than the population Pearson correlation in general, and the difference is fairly subtantial when $|\rho|$ is between about 0.5 and 0.975 or so. The relationship is fairly linear near 0 but nonlinear further away.

You could use the relationship between the population Kendall and the Pearson correlation for the bivariate normal case ($\tau=\frac{2}{\pi}\arcsin({\rho})$) to estimate the Pearson correlation from the sample Kendall correlation - and from there, obtain an estimate of the covariance.

Here's a simulation to show this approach in action. In this case I just used the naive estimator $\hat{\rho}=\sin(\frac{\pi}{2}\hat{\tau})$, and samples of size 100:

1: plot of r vs rho. 2: plot of sin(pi t/2) vs rho.  3: plot of sin(pi t/2) vs r

As we see, the naive estimator of $\rho$ based off a transformed sample Kendall correlation is excellent (bottom left); its variance is only a little larger than using the usual sample Pearson correlation (top left). Indeed the transformed sample Kendall correlation is very close to the corresponding value of the sample Pearson correlation (bottom right). There may well be a better estimator than the one I used, but this indicates that an approach based on the sample Kendall correlation may well be feasible.

The relationship between the Kendall and Pearson correlations will depend on the bivariate distribution (not just the copula) so I think you'll need a new analysis for each such case.

In general this will not be a suitable estimate of the covariance matrix. Indeed the copula alone cannot be used to get a reasonable estimate of covariance, since the copula is invariant to monotonic increasing transformation of the margins, but the covariance is certainly not.

However, in some particular situations it may be possible to get a good estimate of a covariance matrix.

For example, consider he situation where $(X,Y)$ is bivariate Gaussian but where the only information you have on the strength of the bivariate relationship is the value of the Kendall correlation.

Even in this case, it's not going to be suitable to use the Kendall correlation directly. Here's the result for 1000 simulated samples across the range of possible correlation values, each with sample size 100:

Sample Kendall correlation vs population Pearson correlation

We can see that the sample Kendall correlation is closer to 0 than the population Pearson correlation in general, and the difference is fairly subtantial when $|\rho|$ is between about 0.5 and 0.975 or so. The relationship is fairly linear near 0 but nonlinear further away.

You could use the relationship between the population Kendall and Pearson correlations for the bivariate normal case ($\tau=\frac{2}{\pi}\arcsin({\rho})$) to estimate the Pearson correlation from the sample Kendall correlation - and from there, obtain an estimate of the covariance.

Here's a simulation to show this approach in action. In this case I just used the naive estimator $\hat{\rho}=\sin(\frac{\pi}{2}\hat{\tau})$, and samples of size 100:

1: plot of r vs rho. 2: plot of sin(pi t/2) vs rho.  3: plot of sin(pi t/2) vs r

As we see, the naive estimator of $\rho$ based off a transformed sample Kendall correlation is excellent (bottom left); its variance is only a little larger than using the usual sample Pearson correlation (top left). Indeed the transformed sample Kendall correlation is very close to the corresponding value of the sample Pearson correlation (bottom right). There may well be a better estimator than the one I used, but this indicates that an approach based on the sample Kendall correlation may well be feasible.

If you do this at other typical sample sizes the spread around the relationships changes (the variance of the sample correlations will be proportional to $\frac{1}{n}$) but the basic pattern of results is similar.

The relationship between the Kendall and Pearson correlations will depend on the bivariate distribution (not just the copula) so I think you'll need a new analysis for each such case.

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Glen_b
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In general this will not be a suitable estimate of the covariance matrix. Indeed the copula alone cannot be used to get a reasonable estimate of covariance, since the copula is invariant to monotonic increasing transformation of the margins, but the covariance is certainly not.

However, in some particular situations it may be possible to get a good estimate of a covariance matrix.

For example, consider he situation where $(X,Y)$ is bivariate Gaussian but where the only information you have on the strength of the bivariate relationship is the value of the Kendall correlation. However, you do have marginal information on $X$ and $Y$ (such as their variances).

Even in this case, it's not going to be suitable to use the Kendall correlation directly. Here's the result for 1000 simulated samples across the range of possible correlation values, each with sample size 100:

Sample Kendall correlation vs population Pearson correlation

We can see that the sample Kendall correlation is closer to 0 than the population Pearson correlation in general, and the difference is fairly subtantial when $|\rho|$ is between about 0.5 and 0.975 or so. The relationship is fairly linear near 0 but nonlinear further away.

You could use the relationship between the population Kendall and the Pearson correlation for the bivariate normal case ($\tau=\frac{2}{\pi}\arcsin({\rho})$) to estimate the Pearson correlation from the sample Kendall correlation - and from there, obtain an estimate of the covariance.

Here's a simulation to show this approach in action. In this case I just used the naive estimator $\hat{\rho}=\sin(\frac{\pi}{2}\hat{\tau})$, and samples of size 100:

1: plot of r vs rho. 2: plot of sin(pi t/2) vs rho.  3: plot of sin(pi t/2) vs r

As we see, the naive estimator of $\rho$ based off a transformed sample Kendall correlation is excellent (bottom left); its variance is only a little larger than using the usual sample Pearson correlation (top left). Indeed the transformed sample Kendall correlation is very close to the corresponding value of the sample Pearson correlation (bottom right). There may well be a better estimator stillthan the one I used, but this indicates that an approach based on the sample Kendall correlation may well be feasible.

The relationship between the Kendall and Pearson correlations will depend on the bivariate distribution (not just the copula); so I think you'll need a new analysis for each such case.

In general this will not be a suitable estimate of the covariance matrix. Indeed the copula alone cannot be used to get a reasonable estimate of covariance, since the copula is invariant to monotonic transformation of the margins, but the covariance is certainly not.

However, in some particular situations it may be possible to get a good estimate of a covariance matrix.

For example, consider he situation where $(X,Y)$ is bivariate Gaussian but where the only information you have on the bivariate relationship is the value of the Kendall correlation. However, you do have marginal information on $X$ and $Y$ (such as their variances).

Even in this case, it's not going to be suitable to use the Kendall correlation directly. Here's the result for 1000 simulated samples across the range of possible correlation values, each with sample size 100:

Sample Kendall correlation vs population Pearson correlation

We can see that the sample Kendall correlation is closer to 0 than the population Pearson correlation in general, and the difference is fairly subtantial when $|\rho|$ is between about 0.5 and 0.975 or so. The relationship is fairly linear near 0 but nonlinear further away.

You could use the relationship between the population Kendall and the Pearson correlation for the bivariate normal case ($\tau=\frac{2}{\pi}\arcsin({\rho})$) to estimate the Pearson correlation from the sample Kendall correlation - and from there, obtain an estimate of the covariance.

Here's a simulation to show this approach in action. In this case I just used the naive estimator $\hat{\rho}=\sin(\frac{\pi}{2}\hat{\tau})$, and samples of size 100:

1: plot of r vs rho. 2: plot of sin(pi t/2) vs rho.  3: plot of sin(pi t/2) vs r

As we see, the naive estimator of $\rho$ based off a transformed sample Kendall correlation is excellent (bottom left); its variance is only a little larger than using the usual sample Pearson correlation (top left). Indeed the transformed sample Kendall correlation is very close to the corresponding value of the sample Pearson correlation (bottom right). There may well be a better estimator still, but this indicates that an approach based on the sample Kendall correlation may well be feasible.

The relationship between the Kendall and Pearson correlations will depend on the bivariate distribution (not the copula); I think you'll need a new analysis for each such case.

In general this will not be a suitable estimate of the covariance matrix. Indeed the copula alone cannot be used to get a reasonable estimate of covariance, since the copula is invariant to monotonic increasing transformation of the margins, but the covariance is certainly not.

However, in some particular situations it may be possible to get a good estimate of a covariance matrix.

For example, consider he situation where $(X,Y)$ is bivariate Gaussian but where the only information you have on the strength of the bivariate relationship is the value of the Kendall correlation.

Even in this case, it's not going to be suitable to use the Kendall correlation directly. Here's the result for 1000 simulated samples across the range of possible correlation values, each with sample size 100:

Sample Kendall correlation vs population Pearson correlation

We can see that the sample Kendall correlation is closer to 0 than the population Pearson correlation in general, and the difference is fairly subtantial when $|\rho|$ is between about 0.5 and 0.975 or so. The relationship is fairly linear near 0 but nonlinear further away.

You could use the relationship between the population Kendall and the Pearson correlation for the bivariate normal case ($\tau=\frac{2}{\pi}\arcsin({\rho})$) to estimate the Pearson correlation from the sample Kendall correlation - and from there, obtain an estimate of the covariance.

Here's a simulation to show this approach in action. In this case I just used the naive estimator $\hat{\rho}=\sin(\frac{\pi}{2}\hat{\tau})$, and samples of size 100:

1: plot of r vs rho. 2: plot of sin(pi t/2) vs rho.  3: plot of sin(pi t/2) vs r

As we see, the naive estimator of $\rho$ based off a transformed sample Kendall correlation is excellent (bottom left); its variance is only a little larger than using the usual sample Pearson correlation (top left). Indeed the transformed sample Kendall correlation is very close to the corresponding value of the sample Pearson correlation (bottom right). There may well be a better estimator than the one I used, but this indicates that an approach based on the sample Kendall correlation may well be feasible.

The relationship between the Kendall and Pearson correlations will depend on the bivariate distribution (not just the copula) so I think you'll need a new analysis for each such case.

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Glen_b
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