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Suppose there are three points in 3D space, each with coordinates $A_i=(X_i,Y_i,Z_i)\leadsto \mathcal{N}(\mu_i,\tau^2\mathbb{I}_3)$. We compute the distance between the three points, e.g. $D_{ij} = \|A_i-A_j\|$. Then $D_{ij}/(\sqrt{2}\tau)$ follows a noncentral chi distribution with 3 degrees of freedom, and noncentrality parameter $\|\mu_i-\mu_j\|/(\sqrt{2}\tau)$.

What is the distribution for, say, $(D_{12},D_{13})$ ?

Details and hints: Define $$ G=\left(\begin{matrix} X_1 - X_2 & X_1 - X_3\\ Y_1 - Y_2 & Y_1 - Y_3\\ Z_1 - Z_2 & Z_1 - Z_3\\ \end{matrix}\right)$$ Then $W=G^TG$ follows a non-central Wishart distribution with 3 degrees of freedom, positive-definite correlation matrix $$\Psi=\left(\begin{matrix} 2\tau^2 & \tau^2 \\ \tau^2 & 2\tau^2 \end{matrix}\right)$$ and noncentrality matrix $\Theta=A^{-1}(\sum_{i=1}^3 \mu_i \mu_i^T)A^{-1}$ where $A$ is the square root of $\Sigma$.

With these definitions, I am asking what is the PDF for $\text{diag}(W)$, the squared distances? A change of variables would then allow to get the PDF of the distances themseleves.

I found a paper that does what I want, but only in the case where $\forall i, \mu_i = \mu$. Another paper gives the Laplace transform of the wanted PDF, but I cannot calculate either the reverse transform or the expectation I need. Related distributions include the bivariate Ricean (or Rician), bivariate Rayleigh, or bivariate noncentral chi (or chi-square) distributions.

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  • $\begingroup$ Do you assume the $A_i$ are independent? Also, what is $\Sigma$? I think there is a trick that can be used: if you can find the PDF of $G$, and it is a function of $G^{\top}G$, then you can compute the PDF of $W=G^{\top}G$. $\endgroup$ – shabbychef Jul 22 '13 at 21:49

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