What steps could be taken to check for bivariate Gaussianity without using regression based check? Can we somehow employ the use of definition of variogram measure for assessing spatial variability?
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$\begingroup$ It is difficult to see how the variogram applies. It summarizes a single multivariate observation and requires strong additional assumptions in order to estimate variances and covariances. $\endgroup$– whuber ♦Commented Mar 31, 2011 at 22:49
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1$\begingroup$ @whuber: Thanks for the link. However, that question talks about univariate gaussianity, and nothing about bivariate gaussianity. $\endgroup$– user1102Commented Mar 31, 2011 at 22:53
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$\begingroup$ Good point. But it seems like those replies generalize in some ways. For instance, they suggest that one way to detect bivariate normality is to check various linear combinations of the two variables separately for normality. This reduces it to a set of univariate tests. $\endgroup$– whuber ♦Commented Mar 31, 2011 at 23:50
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$\begingroup$ @whuber: I guess what you are suggesting is regression, and I have stated in my question that without the use of regression $\endgroup$– user1102Commented Mar 31, 2011 at 23:53
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$\begingroup$ No, that's the whole point: univariate checks are manifestly not regression methods. Neither is taking a linear combination of variables. $\endgroup$– whuber ♦Commented Mar 31, 2011 at 23:54
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I have recently come across this method that was displayed in Johnson and Wichern.
Let the data points that you want to test for bivariate normality be designated as $\{ x_{i} \}$. Next, compute the sample covariance matrix and deisgnate it as $S$.
For each observed point calculate $d_{j}^{2} = (x_{j} - \bar{x})^{T} S^{-1} (x_{j} - \bar{x})$. Order the values of the $d_{j}^{2}$ from low to high. The last mathematical step is to plot the pair $(q_{c,p}((j- \frac{1}{2}), d_{j}^{2})$, where $q_{c,p}((j- \frac{1}{2})/n)$ is the $100(j- \frac{1}{2})$ quantile of the chi-squared distribution. The plot should be a straight line if the data has a bivariate normal distribution.
