Um, I guess I may as well go right to the answer. (Pearson) correlations -- aka correlation coefficients -- can be interpreted as cosines of angles (between random variables). For three arbitrary random variables with a joint distribution, they can be thought of as the cosines of the angles between three arbitrary vectors in some Euclidean space. So 3-space gives all possibilities in full generality. The necessary and sufficient requirement that (x,y,z) be such a triple of cosines is that the matrix ( 1 x z ) ( ( x 1 y ) ( ( z y 1 ) of cosines be positive semidefinite. This is equivalent to requiring that every upper right square matrix have nonnegative determinant. We already know that 1 >= 0, and that 1 >= x^2 (since (x,y,z) must lie in [-1,1]^3) so that leaves only the fact that (*) 2xyz + 1 - (x^2 + y^2 + z^2) >= 0 whose locus fits RWG's samosa description to a T. (I named this shape the "tetrahedral pillow" a long time ago -- you could look it up -- but "samosa" has much greater elegance.) If you have a Mac you could visualize this in Grapher by setting the frame limits to be -1 <= x,y,z <= 1 and then graphing the implicit equation 2xyz + 1 - (x^2 + y^2 + z^2) = 0. --Dan On 2013-01-31, at 8:22 AM, Dan Asimov wrote:
This reminds me very much of a question I don't recall mentioning here before:
Consider real random variables U, V, W having a joint distribution on R^3.
Assume that the means E(U) = E(V) = E(W) = 0 and the variances E(U^2) = E(V^2) = E(W^2) = 1.
Let rho denotes the Pearson correlation coefficient.
Note that rho(U,V) = E(UV), rho(V,W) = E(VW), rho(W,X) = E(WX)
PUZZLE: What is the set of possible triples of correlations
T = {(x,y,z) in [-1,1]^3},
where x = rho(U,V), y = rho(V,W), z = rho(W,U). ???
Find a closed form polynomial inequality that describes the set T as a subset of [-1,1]^3.
--Dan
RWG wrote: << The equation of the samosa! ParametricPlot3D[{Cos[t], Cos[u], Cos[u + t]}, {t, 0, 2 π}, {u, 0, π}]
(Rounded regular tetrahedron, four vertices, no edges, but contains the line segments joining the vertices.) Implicit equation: acos(z)=acos(x)+acos(y) Volume pi^2/2 Area (empirical) 5 pi, but I can't get Mma to do the integral.
This can't be new. Does this surface have a legitimate name? Can anyone prove the area?
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