Does there exist, for any natural $n$, a probability distribution in $\mathbb{R}^n$ whose projection on any line is a uniform distribution?
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2A uniform distribution along a (whole) line would have infinite integral. – geodude Jun 24 '15 at 12:12
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1I meant a uniform distribution on a sub-interval of that line. – Erfan Salavati Jun 24 '15 at 12:13
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A uniform distribution in some cube? – kjetil b halvorsen Jun 24 '15 at 12:19
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This has the property only for lines parallel to the edges of the cube. – Erfan Salavati Jun 24 '15 at 12:22
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4If your initial distribution is a uniform distribution along a line $L_0$, then the projection on any line which is not orthogonal to $L_0$ is uniform on a subinterval of that line. – Anthony Quas Jun 24 '15 at 13:17
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@Salavati: do you know the answer for the two-dimensional case? A positive answer in three dimensions would imply a positive answer in two dimensions by projecting the measure onto the plane. So I think that it will help to consider the two-dimensional case first. – Tobias Fritz Jun 24 '15 at 13:25
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The answer for the two dimension is positive. If one choose a uniform point on the surface of a three sphere and then project it on the x-y plane, the resulting distribution has the property. – Erfan Salavati Jun 24 '15 at 13:29
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I have answered my own question for 3 dimensions in the previous comment! The uniform distribution on the surface of 3-sphere. – Erfan Salavati Jun 24 '15 at 14:34
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See http://mathoverflow.net/questions/6964/ for $n=2$ – David E Speyer Jun 24 '15 at 17:19
2 Answers
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No if $n\geq 4$. By translating we may assume that the mean is $0$, and by scaling we may assume that the variance-covariance matrix is the scalar matrix $(1/3)I$. Then all projections have mean $0$ and variance $1/3$, hence are uniform in the interval $[-1,1]$.
So if we take a random projection, the value will also be uniform in the interval $[-1,1]$.
Let $\mathbf x$ be a random variable along such a distribution. Clearly $|\mathbf x|\leq 1$ with probability $1$. The probability that the absolute value of the projection of $\mathbf x$ in a random direction is at least $1-\epsilon$ is an increasing function of $|\mathbf x|$, so it is at most the probability that the projection of a unit vector in a random direction has length at least $1-\epsilon$. This is the same as the proportional of the unit $n-1$-sphere that is in a spherical cap of depth $\epsilon$, which is $\approx \epsilon^{(n-1)/2}$.
On the other hand, because the distribution is uniform, the probability that the absolute value of the projection of $\mathbf x$ in a random direction is at least $1-\epsilon$ must be $\epsilon$. So we have $\epsilon \leq \epsilon^{(n-1)/2}$, and thus $n\leq 3$.
Will Sawin
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@ChristianRemling I'm saying that by scaling we may assume the variance-covariance matrix is the identity matrix, or for conveneince $(1/3)I$. – Will Sawin Jun 24 '15 at 18:39
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Thanks, nice argument! I think it actually also works for $n=3$ because the spherical cap $z\ge 1-\epsilon$ has prob $\sim\epsilon/2$ wrt surface measure, if I got it right. – Christian Remling Jun 24 '15 at 19:03
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@ChristianRemling Technically you have to consider both spherical caps, giving another factor of $2$. I ignore the constant term if the $n \geq 4$ case because it's irrelevant. – Will Sawin Jun 24 '15 at 19:05
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Right. So for $n=3$ the argument shows that any such measure is supported by the sphere. – Christian Remling Jun 24 '15 at 19:07
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Using Will's observation on mean and covariance, one can also assume the measure to be symmetric under rotations without loss of generality. The reason is that having projection uniform on $[-1,1]$ is a property which is invariant under taking mixtures of measures. Hence if some measure has this property, then so does its convolution with a random rotation (chosen from the Haar measure on $O(n)$). – Tobias Fritz Jun 24 '15 at 20:14
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Hence we work wlog with a measure of the form $f(r), d\Omega, dr$, where $d\Omega$ is the volume element of the sphere (and $f$ may be distributional), i.e. an integral over the uniform measures on spheres for varying radius of the sphere. For these measures, it should be a straightforward calculation to determine whether there exists some $f$ for which $f(r) , d\Omega, dr$ satisfies the OP's requirement. – Tobias Fritz Jun 24 '15 at 20:35
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1@ChristianRemling But doesn't the uniform measure on the sphere work? – Will Sawin Jun 24 '15 at 20:36
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Addendum to my comments: now I understand that Will achieves something equivalent by considering a random projection. Clever! – Tobias Fritz Jun 24 '15 at 20:46
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@WillSawin: It does, as is actually obvious from the formula $dz, d\varphi$ for the surface area measure. Somehow it seemed intuitively "clear" to me that there is more area near the equator than near the poles. – Christian Remling Jun 25 '15 at 16:00
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For $n=2$ there is a solution with a density w.r.t. Lebesgue measure : $c\ dx\ dy/\sqrt{R^2-x^2-y^2}$ on a circular disc of radius $R$. So you have three possibilities :
1°) Anthony Quas's uniform distribution on subintervals of straight lines ($n\geq 2$);
2°) uniform distribution on spheres in 3-dimensional subspaces of $n$-dimensional space ($n\geq 3$);
3°) the one above on discs in planes ($n\geq 2$).
Plus their images (push forward) by affine maps, but probably no other, if I have to guess.
Jean Duchon
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