Proof: If the marginal distribution is Gaussian, the joint distribution is clearly spherical. Below we show the converse: if the joint distribution is spherical, the marginal distribution is Gaussian.
Let the marginal distribution be $g(x)$, so $p(x,y) = g(x)g(y)$. Spherical means the gradient of the joint distribution is proportional to $(x,y)$. That is $$\nabla p(x,y) = \begin{bmatrix} g'(x) g(y) \\ g(x) g'(y)\end{bmatrix} = c(x,y) \begin{bmatrix}x \\ y \end{bmatrix}.$$
To eliminate the constant of proportionality $c(x,y)$, we divide the first term in the gradient by the second, to get $$ {g'(x) g(y) \over g(x) g'(y)} = {x \over y}.$$ Recognizing $g'(x)/g(x) = d\ln g(x)/dx$, we get $$ {d \ln g(x)/ dx \over d \ln g(y)/dy } = {x \over y}.$$
This motivates expressing $g(x) = e^{\phi(x)}$, for some unknown $\phi(x)$. Then $d \ln g(x)/dx = \phi'(x),$ and we get $${\phi'(x) \over \phi'(y)} = {x \over y}.$$
We then set $y$ to 1, and get $$ {d \phi(x) \over dx} = \phi'(1) x \implies d\phi(x) = \phi'(1) x dx \implies \phi(x) = {\phi'(1) \over 2}x^2 + c,$$ which then implies that $$ g(x) = e^{\phi(x)} \propto e^{\phi'(1) {x^2 \over 2}},$$ and the proof is complete.
$$\blacksquare$$
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