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Consider the density $f(x,y)=\large\frac{1}{2\pi}\frac{1}{\sqrt{1-x^2-y^2}}$ on the unit disk centered at the origin. There is a particular characterization of this distribution: it is the unique circularly symmetric distribution whose projections onto any line through the origin are uniformly distributed.

Showing the projections of $f(x,y)$ are uniform is simple calculus. However I can't seem to think of an elegant proof for uniform projections implying $f$ must have the above density. If we let $R$ denote any rotation of coordinates $x,y\rightarrow x',y'$, then by assumption, $f(x,y)=R\circ f(x,y):=f(x',y')$. It would then suffice to show that $f(x,0)=\large\frac{1}{2\pi}\frac{1}{\sqrt{1-x^2}}$. Writing out the projected density as $u(x)$:


and we must have that $u(x)=1/2$ (being uniform on $[-1,1]$). Differentiating in $x$ with Leibnitz's rule seems to get nowhere. I've also tried considering the characteristic function of $f$ but got nowhere.

If it helps, $f(x,y)$ arises from projecting the uniform distribution on the sphere to the $x,y$ plane. So in essence this is Archimedes rule for the sphere inscribed in a cylinder: the surface areas are equal. My gut feeling is that this is not entirely dissimilar from showing that the multivariate Gaussian distribution is the only rotationally invariant distribution with independent components.

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Your $u(x)$ is wrong. If it were given by that integral, it could never be constant, since that integral is monotonic in $x$ if $f$ is non-negative. What you want is $f(x,y)=g(\sqrt{x^2+y^2})=g(r)$. Unfortunately that doesn't seem to make things any easier. – joriki Jan 3 '13 at 23:56
@joriki: Whoops! Thanks for pointing that out. I've adjusted it above. – Alex R. Jan 3 '13 at 23:58
When you say "...showing that the multivariate Gaussian distribution is the only rotationally invariant distribution" presumably you want an extra condition on top of that (I think it's "...whose components are independent"?). – Qiaochu Yuan Jan 4 '13 at 0:31
@Qiaochu Yuan: corrected, thanks. – Alex R. Jan 4 '13 at 0:33
up vote 4 down vote accepted

This is not an elementary proof, but you can use the projection-slice theorem here, and the fact that essentially you are taking the Radon transform of your distribution. If $f(x,y)$ is the distribution, and $Rf_\vec\alpha(s)$ is its integral over the line $\vec{x}\cdot\vec{\alpha} = s$ (here $\vec\alpha$ is a unit vector), then $$ \mathcal{F}(Rf_\vec\alpha)(\sigma) = \mathcal{F}f(\sigma \vec\alpha),$$ where the lhs is the one-dimensional Fourier transform of $Rf_\vec\alpha(s)$, and the rhs is the two-dimensional Fourier transform of $f(x,y)$.

Now, you start with $$ Rf_\vec\alpha(s) = \frac{1}{2}\mathbb{I}_{-1\leq s\leq 1},$$ then $$ \mathcal{F}(Rf_\vec\alpha)(\sigma) = \frac{\sin \sigma}{\sigma}, $$ so $$ \mathcal{F}f(\vec k) = \frac{\sin |\vec k|}{|\vec k|}.$$ Now take the inverse Fourier transform ($\vec n_\theta$ is the unit vector $(\cos\theta,\sin\theta$)): $$ \begin{align} f(\vec x) &= \int \frac{d\vec k}{(2\pi)^2} e^{-i(\vec x\vec k)} \frac{\sin|\vec k|}{|\vec k|} \\&= \iint \frac{dr d\theta}{(2\pi)^2}e^{-i r (\vec x\vec n_\theta)\sin r} \\&=\frac{1}{4\pi^2}\int_0^{2\pi}d\theta \int_0^\infty \sin r e^{-i r (\vec x\vec n_\theta)}dr \\&= \frac{1}{4\pi^2} \int_0^{2\pi}d\theta \frac{1}{1-(\vec x\vec n_\theta)^2} \\&= \frac{1}{4\pi^2}\int_0^{2\pi} \frac{d\theta}{1-|\vec x|^2\cos^2\theta} \\&= \frac{1}{2\pi} \frac{1}{\sqrt{1-|\vec x|^2}} \end{align}$$ The only issue here is that because $f$ is discontinuous, one of the integrals here only converges when $(\vec x\vec n_\theta)$ has a negative imaginary part: $$ \int_0^\infty \sin r e^{-i a r}dr = \frac{1}{1-a^2}, \qquad (\Im a < 0). $$ To evaluate it, I took $a=(\vec x\vec n_\theta)-i\epsilon$, and then took the limit $\epsilon\to0^+$.

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Very cool! Thanks! – Alex R. Jan 4 '13 at 2:22

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