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Let $X,Y,Z$ have a standard trivariate normal distribution centered at the origin, with zero means, unit variances, and correlation coefficients $\rho_1,\rho_2,\rho_3$. Show that $$P(X>0,Y>0,Z>0)=\frac{1}{8}+\frac{1}{4\pi}(\sin^{-1}(\rho_1)+\sin^{-1}(\rho_2)+\sin^{-1}(\rho_3))$$

My idea is transform $X,Y,Z$ into i.i.d. normal. First, we have $V=\begin{bmatrix}1 & \rho_1& \rho_2\\ \rho_1 & 1 & \rho_3\\ \rho_2 & \rho_3 & 1\end{bmatrix}$. $V=P^TDP$ where columns of $P$ are eigenvectors and $D$ is a diagonal matrix. Then we have $V^{1/2}V^{1/2}=P^TD^{1/2}D^{1/2}P$. So, $V^{-1/2}=PD^{-1/2}$ since $P^TP=I$. Hence, $$\begin{bmatrix}U\\V\\W\end{bmatrix}=V^{-1/2}\begin{bmatrix}X\\Y\\Z\end{bmatrix}$$ I got stuck at this step, and I don't know how to find $V^{-1/2}$. Can someone give me a hint or suggestion? Thanks

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    See (http://mathworld.wolfram.com/TrivariateNormalDistribution.html) – 2017-02-12
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    This problem boils down to computing $$\iiint_{\mathbb{R}_+^3}\exp\left(-x^T A x\right)\,d\mu$$ where $A$ is a real $3\times 3$ symmetric and positive definite matrix, hence to computing the angles between the canonical base of $\mathbb{R}^3$ and the orthonormal base of eigenvectors for $A$. – 2017-02-12
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    @JackD'Aurizio I am having trouble to find the canonical base of $R^3$ and the orthonormal base of eigenvectors – 2017-02-12
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    @Simple: the canonical base of $\mathbb{R}^3$ is $(1,0,0),(0,1,0),(0,0,1)$, that is not hard to find. And the eigenvectors of $A$ are not really needed to compute the final integral, just their dot products with the previous elements. – 2017-02-12
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    @JackD'Aurizio I see, thanks. – 2017-02-12

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