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Let $U \in S^{d-1} \subset \mathbb{R}^d$ follow a uniform distribution on a sphere. Let $v \in \mathbb{R}^d.$ Then is the orthogonal projection $U^{T}v=\langle U,v \rangle$ uniformly distributed, and if yes/no, how do I go about proving it? This is motivated by this relevant question for uniform distributions on 2D spheres. If it's wrong, could you give a counterexample?

If it's not uniformly distributed, how can we find the PDF of $U^{T}v=\langle U,v \rangle?$

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    $\begingroup$ $U$ is a vector right? So $Uv$ is what? A real number given by the euclidean product? $\endgroup$ – Stockfish Jul 23 at 8:14
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    $\begingroup$ Simpler question : taking $v = e_1 = (1,0,...,0)$, we have $U^Tv = U_1$. Now, the question is : if $U$ is uniformly distributed on the $d-1$ sphere, is $U_1$ uniformly distributed? Think about this one. $\endgroup$ – Teresa Lisbon Jul 23 at 8:18
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    $\begingroup$ Hint: Suppose $d=2$ and $v$ has unit-length. Then $U^Tv= \cos\theta$, where $\theta$ is the angle between $U$ and $v$. Clearly $\theta$ is uniformly distributed in $[-\pi,\pi]$, but you can check that $\cos \theta$ is not uniformly distributed on $[-1,1]$. $\endgroup$ – Jan Bohr Jul 23 at 8:28
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    $\begingroup$ @LearningMath That is for $d=3$, right? $\endgroup$ – Teresa Lisbon Jul 23 at 11:07
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    $\begingroup$ The square of your $\langle U,v\rangle$ has a "beta distribution" if $\|v\|=1$, as can be seen from the representation of $U=Z/\|Z\|$ where the coordinates of $Z$ are iid $N(0,1)$. $\endgroup$ – kimchi lover Jul 23 at 13:50
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Let $v$ be a unit vector. Since the orthogonal group acts transitively on the unit sphere, there exists a rotation matrix $R$ such that $Rv=e$, where $e=(1,0,\ldots,0)'$.

Now let $Z=(Z_1,\ldots,Z_d)'$ be a vector of independent identically distributed $N(0,1)$ random variables. The distribution of $Z$ is clearly rotationally invariant, and the normalized vector $U=Z/\|Z\|$ is uniformly distributed over the unit sphere, as is $R'U$. (See, eg, the answer by Henricus V. to this MSE question, and comments, and this one, about this trick. It is a converse to Maxwell's theorem.) The quantity $\langle U, v\rangle=\langle U,Re\rangle=\langle R' U,e\rangle,$ so the distribution of $\langle U, v\rangle$ is the same as that of $\langle U, e\rangle$. So we may as well assume $v=e$ in working out the distribution of $T=\langle U,v\rangle$.

With this notation, $T=\langle U,v\rangle=Z_1/\sqrt{\sum_{i=1}^d Z_i^2}$, and $T^2=Z_1^2/\sum_{i=1}^d T_i^2$. The quantities $Z_i^2$ are iid Gamma distributed: $Z_i^2\sim\Gamma(\frac 1 2,\frac 1 2)$ and $\sum_{i=2}^d Z_i^2\sim\Gamma(\frac{d-1}2,\frac 12)$ and hence $T^2$ has the $\operatorname{Beta}(\frac 1 2,\frac {d-1}2)$ distribution. If we write $W=T^2$, the density function of $W$ is proportional to $w^{-\frac{1}2} (1-w)^{\frac{d-1}2-1}$ for $0<w<1$, and the density of $T$ is proportional to $(1-t^2)^{(d-3)/2}$ for $-1<t<1$. Only if $d=3$ does the distribution of $T$ become uniform over its range.

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  • $\begingroup$ Thank you for your answer! To finish your line of arguments, I think we need to show that any uniform distribution $U$ on the sphere can be written as a ratio $U:= Z/||Z||, Z=(Z_1 \dots Z_d), Z_i \sim_{iid} \mathcal{N}(0,1).$ Is it obvious, sorry if it is. I know that the other side is true: if $Z \sim \mathcal{N}(0,I_d),$ then $U:= Z/||Z|| \sim \mathcal{U}nif(S^{d-1}).$ $\endgroup$ – Learning Math Jul 23 at 16:33
  • $\begingroup$ Also for a general unit vector $v, U^{T}v$ seems hard to relate with $Ue_1, e_1:=(1, 0, \dots 0).$ I was thinking of using rotation matrix to bring $v$ to $e_1, $ but couldn't make it work instantly. So I guess one way to go about it will be $ U^{T}v = v_1U_1 + \dots v_dU_d, U_i$ following square root of a $Beta(1/2, \frac{d-1}{2})$ distribution as you showed. But these Beta distributions are not independent, being component of a sphere-valued distribution. So how do we calculate the distribution of $U^{T}v$ for a general unit vector $v?$ $\endgroup$ – Learning Math Jul 23 at 16:58
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    $\begingroup$ Your instinct is right: a rotation matrix does the trick. I have edited my answer. The details and consequence are easier than your comment implies. $\endgroup$ – kimchi lover Jul 23 at 18:01
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    $\begingroup$ I have edited again, with pointers to $Z/\|Z\|$ being uniformly distributed on the sphere. This is related to the spherical symmetry of both the sphere and of the multivariate Gaussian distribution. $\endgroup$ – kimchi lover Jul 23 at 20:16
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    $\begingroup$ I think you are asking for a proof of Maxwell's theorem; my argument needs only its (easier) converse. I think you should submit a separate question. I'm having a hard time distinguishing in this comment thread exactly what you are saying. $\endgroup$ – kimchi lover Jul 24 at 10:46

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