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Numerical evaluation of the following integral of a 3D gaussian $G$ seems to result in a 1D Gaussian $g$: $$\int_{0}^{2\pi}\int_{0}^{\pi}G(R,\phi,\theta)\sin\theta\ \text{d}\theta \ \text{d}\phi= g(R)$$ where the 3D Gaussian in spherical coordinates with mean $\mu$ and covariance matrix $\Sigma$ $$G(R,\phi,\theta)=H \exp\left[-\frac{1}{2}(X-\mu)^{T}\cdot \Sigma^{-1}\cdot (X-\mu)\right]$$ $$X=\begin{bmatrix}R\cos\phi\sin\theta\\\ R\sin\phi\sin\theta\\\ R\cos\theta\end{bmatrix}$$ and the 1D Gaussian with mean $R_\mu$ and variance $\sigma^2$ $$g(R)=H'\exp\left[-\frac{(R-R_\mu)}{2\sigma^2}\right]$$ The question is: is this true? I realize this is a forum for more fundamental mathematics, but I've asked this question on applied math and physics forums without any luck (link). I'd be grateful if someone can give me a hit on how to approach this. My attempt to solve this problem involved transforming $G$ so that $\mu=(0\ 0\ R_\mu)$ (adapting $\Sigma$ in the process) and substitute $z=\cos\theta$. However I couldn't solve the integral over $z$ and $\phi$.

Edit The special case where $\Sigma=\sigma^2I$ proves that this is not a Gaussian but it is very close. For $R_\mu\gg\sigma$ we find (link) $$\int_{0}^{2\pi}\int_{0}^{\pi}G(R,\phi,\theta)\sin\theta\ \text{d}\theta \ \text{d}\phi= \frac{g(R)}{RR_\mu}(1-\exp(-\frac{2RR_{\mu}}{\sigma^2}))\approx\frac{g(R)}{RR_\mu}$$ If I could find a similar result for any $\Sigma$, it would solve my problem.

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Please try Unfortunately, it's just not appropriate for this site. – Deane Yang Sep 27 '12 at 17:32
I tried that already without result. – Wox Sep 27 '12 at 20:24
I suggest linking questions that you cross post.… – j.c. Sep 28 '12 at 8:55

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