Editing Maxwell's theorem (section)

== Proof ==
We only need to prove the theorem for the 2-dimensional case, since we can then generalize it to n-dimensions by applying the theorem sequentially to each pair of coordinates.

Since rotating by 90 degrees preserves the joint distribution, <math>X_1</math> and <math>X_2</math> have the same probability measure: let it be <math>\mu</math>. If <math>\mu</math> is a Dirac delta distribution at zero, then it is in particular a degenerate gaussian distribution. Let us now assume that it is not a Dirac delta distribution at zero.

By the [[Lebesgue's decomposition theorem]], we decompose <math>\mu</math> to a sum of regular measure and an atomic measure: <math>\mu = \mu_r + \mu_s</math>. We need to show that <math>\mu_s = 0</math>; we proceed by contradiction. Suppose <math>\mu_s</math> contains an atomic part, then there exists some <math>x\in \R</math> such that <math>\mu_s(\{x\}) > 0</math>. By independence of <math>X_1, X_2</math>, the conditional variable <math>X_2 | \{X_1 = x\}</math> is distributed the same way as <math>X_2</math>. Suppose <math>x=0</math>, then since we assumed <math>\mu</math> is not concentrated at zero, <math>Pr(X_2 \neq 0) > 0</math>, and so the double ray <math>\{(x_1, x_2): x_1 = 0, x_2 \neq 0\}</math> has nonzero probability. Now, by rotational symmetry of <math>\mu \times \mu</math>, any rotation of the double ray also has the same nonzero probability, and since any two rotations are disjoint, their union has infinite probability; thus arriving at a contradiction.

Let <math>\mu </math> have probability density function <math>\rho</math>; the problem reduces to solving the functional equation

<math display="block">\rho(x)\rho(y) = \rho(x \cos \theta + y \sin\theta)\rho(x \sin \theta - y \cos\theta).</math>