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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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    $\begingroup$ A uniform distribution along a (whole) line would have infinite integral. $\endgroup$
    – geodude
    Jun 24, 2015 at 12:12
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    $\begingroup$ I meant a uniform distribution on a sub-interval of that line. $\endgroup$ Jun 24, 2015 at 12:13
  • $\begingroup$ A uniform distribution in some cube? $\endgroup$ Jun 24, 2015 at 12:19
  • $\begingroup$ This has the property only for lines parallel to the edges of the cube. $\endgroup$ Jun 24, 2015 at 12:22
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    $\begingroup$ If 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. $\endgroup$ Jun 24, 2015 at 13:17

2 Answers 2

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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 $n\leq 3$.

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  • $\begingroup$ @ChristianRemling I'm saying that by scaling we may assume the variance-covariance matrix is the identity matrix, or for conveneince $(1/3)I$. $\endgroup$
    – Will Sawin
    Jun 24, 2015 at 18:39
  • $\begingroup$ 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. $\endgroup$ Jun 24, 2015 at 19:03
  • $\begingroup$ @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. $\endgroup$
    – Will Sawin
    Jun 24, 2015 at 19:05
  • $\begingroup$ Right. So for $n=3$ the argument shows that any such measure is supported by the sphere. $\endgroup$ Jun 24, 2015 at 19:07
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    $\begingroup$ @ChristianRemling But doesn't the uniform measure on the sphere work? $\endgroup$
    – Will Sawin
    Jun 24, 2015 at 20:36
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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.

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