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In the context of Variational Bayesian Inference I am facing the following problem:

Let $\alpha$ follow a "von Mises" distribution with mean $\mu$ and concentration $\kappa$. Does there exist a formula for the calculation of the expectation values $E \left[ \sin \alpha \right]$ and $E \left[ \cos \alpha \right]$?

Many thanks in advance

Matthias
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  • Good question. It's obvious that there's a simple formula for the median value of $\tan\alpha$. Whether that's also an expected value, I'm not sure, but if it is, then it's clear what the expected value is. With sine and cosine, on the other hand, it's obvious that they have expected values, and that $E(\cos\alpha,\sin\alpha)]$ is the median value of $\tan\alpha$. But the hard part is the expected value of $\sqrt{(E(\cos\alpha))^2+(E(\sin\alpha))^2}$. For all I know at the moment, that's not all that hard, but it's where most of the work is in solving this problem. – Michael Hardy May 30 '13 at 22:57

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@Matthias: Certainly. $E[\cos\alpha] = r\cos\mu$, where $r$ is the mean resultant length. For the von Mises distribution $r = A(\kappa) = \frac{I_1(\kappa)}{I_0(\kappa)}$, where $I_p(.)$ is the modified Bessel function of the first kind of order $p$. Similarly, $E[\sin\alpha] = r\sin\mu$.

Abas
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