When is $\nabla^2 f (x, y, z)= $ probability measure? When is $\nabla^2 f(x, y, z) $= probability density function ?
That is $\nabla^2 f(x, y, z)= \mu (x, y, z)$ $\int \mu (x, y, z) dxdydy = 1 $
What conditions must $f(x,y,z) $ satisfy?
It is known to physicist as 'Poisson's equation' where $ f(x, y, z)$  is the potential. But it will be of practical interest to state conditions necessary or sufficient which the potential must satisfy.
 A: This isn't a sufficient condition, but a necessary one: the first partial derivatives all need to be Lipschitz continuous, so that the second partial derivatives may be bounded. 
(Of course we require more than just boundedness, but also that the partial derivatives vanish at infinity, and even more that they have a finite integral over all of $ℝ^3$.)
I recommend rephrasing the question to ask for necessary and sufficient conditions when the integral of μ over $ℝ^3$ is finite -- any such measure can automatically be reduced to the case of a probability measure via an appropriate norming. 
Moreover, it is clearly a necessary and sufficient condition (for the weakened/generalized version of Poisson's equation) that the integrals of each of the "diagonal" second partial derivatives:
$$\int_{\mathbb{R}^3} \frac{\partial^2 f}{\partial x_i^2} < \infty,\ \forall\ i=1,\dots,n $$
are all finite. This equation is only necessary but not sufficient for the specific case where the integral over all of Euclidean space must be 1 (but again we can always reduce the finite value case to this case, so we suffer no real loss of generality by considering the finite value case).
In short, here are some checks that will allow you to automatically preclude an $f$ or even any normalization thereof, from satisfying Poisson's equation:


*

*If $\displaystyle\frac{\partial f}{\partial x_i}$ is not Lipschitz for any $i$, then Poisson's equation cannot hold.

*If $\displaystyle\frac{\partial^2 f}{\partial x_i^2}$ does not vanish at infinity for any $i$, then Poisson's equation cannot hold.

*If $\displaystyle\int_{\mathbb{R}^3}\frac{\partial^2 f}{\partial x_i^2} = \infty$ for any $i$, then Poisson's equation cannot hold.
I doubt that much better can be said than this, but I too would be interested to find out if there is a better equivalent condition!
