I stumbled upon this problem.
Suppose you have an unknown probability density function on the two dimensional real plane, $f(x,y)$ Suppose you have three distinct points $P_1, P_2, P_3$ on the plane and that, for each triplet $(r_1, r_2, r_3)$ you know the integral of $f$ on the regions bounded by circles centered in $P_1,P_2,P_3$ and radii $r_1,r_2,r_3$. These integrals are given by the functions $g_1(r_1), g_2(r_2), g_3(r_3)$. How can I derive $f(x,y)$ from $g_1,g_2$ and $g_3$ ? What are the necessary and sufficient conditions for $g1,g2,g3$ to be coherent between them and with a generic probability density function?
Thanks
You cannot derive $f(x,y)$ from $g_1$, $g_2$, $g_3$; they contain less information than $f$.
Without loss of generality, let $P_1=(-1,0)$, $P_2=(1,0)$ and $P_3=(0,a)$, and consider the four points $P_{\rho\sigma}=(\rho x,\sigma y)$ with $\rho,\sigma\in\{-1,1\}$ and with arbitrary $x,y\in\mathbb R$. With respect to each of $P_1$, $P_2$ and $P_3$, these four points form two pairs of equidistant points. We can move probability from $P_{++}$ to $P_{+-}$ and make up for it my moving the same probability from $P_{--}$ to $P_{-+}$. This changes $f$, but none of $g_1$, $g_2$ and $g_3$. (In case you didn't mean to allow point probability masses, you can do the same thing with a bump function in the vicinity of the points $P_{\sigma\rho}$.)