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Let f be the pdf of a $n$ dimensional $N(0,C)$ distribution i.e up to a multiplicative constant, $f(x) = \exp(-\frac{1}{2} x'C^{-1}x)$.

Which vector fields $F$ are so that ${\rm div} (F)= f$ ?

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    $\begingroup$ first transform the coordinates so that $C=1$, and then choose $F_i={\rm erf}\,(x_i)\exp(-\sum_{j\neq i}x_j^2)$. $\endgroup$ – Carlo Beenakker Nov 27 '18 at 7:28
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    $\begingroup$ @Carlo Beenakker "div" is a metric dependent operator. By choosing a convenient orthonormal basis the best one can hope is to make $C$ diagonal. Then a suitable modification of your argument above will do the trick. $\endgroup$ – Liviu Nicolaescu Nov 27 '18 at 9:03
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    $\begingroup$ After the suggested tricks, take $F=\nabla\phi$ and you will get a Poisson equation that has a known solution. $\endgroup$ – Jon Nov 27 '18 at 16:00

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