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I have some bivariate data and I have calculated the error ellipse in the following way: I have first calculated the covariance matrix and then to obtain the radii of the ellipse I have taken the eigenvalues and then their squared root.

Does the length of each of these radii correspond to one standard deviation in the direction of the semi-principal axes?

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    $\begingroup$ The correct calculation is described in the question at stats.stackexchange.com/questions/89512. Other closely related questions include stats.stackexchange.com/questions/24380 and stats.stackexchange.com/questions/62092. Your question needs to be refined a little: to what does "one standard deviation" refer? After all, these are bivariate data. $\endgroup$
    – whuber
    Commented Apr 17, 2014 at 18:21
  • $\begingroup$ @whuber I've updated the question. What I meant is that when we are talking about the 95% confidence ellipse the square root of the eigenvalues is scaled by the appropriate chi squared value. So, when these values are not scaled (just square roots of the eigenvalues), do the lengths of the radii correspond to one standard deviation? $\endgroup$
    – vikkor
    Commented Apr 18, 2014 at 8:07

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No. To get the radii, you take the square root of the eigenvalues (i.e. the singular values). See http://en.wikipedia.org/wiki/Confidence_region

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    $\begingroup$ I've wrongly described what I did. In fact, I'm taking the square root of the eigenvalues. I've updated the question. $\endgroup$
    – vikkor
    Commented Apr 18, 2014 at 8:00

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