Maxwell's Theorem - Calvin Khor

In a recent 3Blue1Brown video, it was mentioned that the 2D (centered) Gaussian are the unique distributions which are radial and have independent marginals. On a google I found out that this is known as Maxwell’s theorem, but the linked Wikipedia page is a little lacking. This blogpost does a better job. The blogpost says that the solution to $ℎ (𝑥) ℎ (𝑦) = ℎ (\sqrt{𝑥^{2} + 𝑦^{2}})$ is clearly of the form $𝑒^{𝑎 𝑥^{2}}$ . For anyone else who doesn’t want to spend the 5 seconds necessary to unpack this, you should set $𝐻 (𝑥) ≔ ℎ (\sqrt{𝑥})$ (without loss of generality $𝑥 \geq 0$ ) to find the equivalent equation $𝐻 (𝑥^{2}) 𝐻 (𝑦^{2}) = 𝐻 (𝑥^{2} + 𝑦^{2}),$ which is essentially the defining characteristic of the exponential, i.e. $𝐻 (𝑥) = 𝑒^{𝑎 𝑥}$ .

However there are some pedantic points that one might bring up. The first is that the exponential is not the only solution: search for exponential in this Math.SE post. This can indeed be ignored since the other solutions will not lead to a probability distribution. The second is more interesting, and comes back to the rough state of the Wikipedia page. The proof given there is as follows:

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, both $𝑋_{1}, 𝑋_{2}$ has the same probability measure. Let it be $𝜇$ . If $𝜇$ is a Dirac delta distribution at zero, then it’s a gaussian distribution, just degenerate. Now assume that it is not.
By Lebesgue’s decomposition theorem, we decompose it to a sum of regular measure and an atomic measure: $𝜇 = 𝜇_{𝑟} + 𝜇_{𝑠}$ . We need to show that $𝜇_{𝑠} = 0$ , with a proof by contradiction.
Suppose $𝜇_{𝑠}$ contains an atomic part, then there exists some $𝑥 \in ℝ$ such that $𝜇_{𝑠} ({𝑥}) > 0$ . By independence of $𝑋_{1}, 𝑋_{2}$ , the conditional variable $𝑋_{2} | {𝑋_{1} = 𝑥)$ is distributed the same way as $𝑋_{2}$ . Suppose $𝑥 = 0$ , then since we assumed $𝜇$ is not concentrated at zero, $Pr (𝑋_{2} 𝑛 = 0) > 0$ , and so the double ray ${(𝑥_{1}, 𝑥_{2}) : 𝑥_{1} = 0, 𝑥_{2} 𝑛 = 0}$ has nonzero probability. Now by rotational symmetry of $𝜇 \times 𝜇$ , any rotation of the double ray also has the same nonzero probability, and since any two rotations are disjoint, their union has infinite probability, contradiction.
So now let $𝜇$ have probability density function $𝜌$ , and the problem reduces to solving the functional equation
$𝜌 (𝑥) 𝜌 (𝑦) = 𝜌 (𝑥 cos 𝜃 + 𝑦 sin 𝜃) 𝜌 (𝑥 sin 𝜃 - 𝑦 cos 𝜃)$

Ignoring that the proof isn’t complete, the Wikipedia article points out that that it is assumed that there is a distribution function. What the earlier blogpost actually gives a proof for is the following result:

Theorem (Maxwell) The (centered) gaussians are the only absolutely continuous distributions on $ℝ^{𝑛}$ which are radial and have independent marginals.

Indeed, the Dirac mass at $0$ is a different solution, although it is quite natural as a kind of ‘degenerate’ Gaussian. It seems that the possible existence of singular continuous distributions is not settled. Some tangentially related links for if I come back to this (or the interested reader):

Related Posts