As a follow up to this previous question on the expectation of the von Mises-Fisher distribution, what is the variance of a von-Mises Fisher distribution as a function of the mean direction $\mu$ and concentration $\kappa$?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ The relevant moment is the circular variance. $\endgroup$– whuber ♦Commented Feb 2, 2022 at 16:10
-
$\begingroup$ @whuber are you sure that your link is the correct link? I'm struggling to see the connection between this question and that question. If it is the correct link, could you explain further? $\endgroup$– Rylan SchaefferCommented Feb 5, 2022 at 19:38
-
3$\begingroup$ Sorry: somehow I pasted the wrong link and then failed to check it--two errors on my part. The intended target is the Wikipedia article on circular statistics, en.wikipedia.org/wiki/Directional_statistics#Dispersion. $\endgroup$– whuber ♦Commented Feb 5, 2022 at 21:16
Add a comment
|