Timeline for N-Dimensioned Normal CDF and Mahalanobis Distance
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 23, 2015 at 0:02 | comment | added | Bruno | I did it in R and got the same result. And thanks for the MATLAB tip (duh for me!). I actually e-mailed Michaël Bensimhoun about this, but haven't heard back... so I'm hoping this community will shed some light. :) | |
| Jan 22, 2015 at 23:46 | comment | added | ocramz |
you are right, the lower limit for CDF integration is $-\infty$. The documentation for mvncdf mentions that the CDF is evaluated for every row of X, and you have a column vector, but I guess Matlab silently reshapes it.. sorry, I can't come up with an explanation right now. Oh and don't forget that A\b is in general cheaper than inv(A)*b, but for small cases it's fine ;)
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| Jan 22, 2015 at 23:41 | comment | added | Bruno | Elliptical coordinates I should say. Any ideas on why the results don't match? | |
| Jan 22, 2015 at 23:32 | comment | added | Bruno | Thanks for the comment. If you read pages 2 --4 (or at least the formulas), you'll see that there's a change of variables (from Cartesian to polar) to perform the integration (that's the whole trick to get rid of the double integrals). And that formula, $F(r) = p = 1 - \exp(-r^2/2)$ should be the final formula, as I understand it. So by just plugging in the $r$ value, it should return the CDF whose upper limit of integration is the z point (lower limit is $-\infty$). Right? | |
| Jan 22, 2015 at 23:21 | review | First posts | |||
| Jan 22, 2015 at 23:23 | |||||
| Jan 22, 2015 at 23:20 | history | answered | ocramz | CC BY-SA 3.0 |