3

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 $U^{T}v=\langle U,v \rangle?$

StubbornAtom
  • 17,052
Learning Math
  • 2,346
  • 1
    $U$ is a vector right? So $Uv$ is what? A real number given by the euclidean product? – Stockfish Jul 23 '20 at 08:14
  • @Stockfish Ah sorry, yes $U$ is a vector it's indeed $U^{T}v$, the scalar product. – Learning Math Jul 23 '20 at 08:15
  • 1
    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. – Sarvesh Ravichandran Iyer Jul 23 '20 at 08:18
  • 1
    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 $[-1,1]$. – Jan Bohr Jul 23 '20 at 08:28
  • @JanBohr Thanks for your comment - I need to think; but it seems this question says that the projection is uniformly distributed if I'm not mistaken? https://math.stackexchange.com/questions/1019286/if-u-is-uniformly-distributed-on-s2-then-its-first-component-is-uniforml?noredirect=1&lq=1 – Learning Math Jul 23 '20 at 09:40
  • @астонвіллатересалисбон It seems the answer is yes from this question asked before? https://math.stackexchange.com/questions/2647552/projections-of-uniformly-distributed-mathbbr3-unit-vector-have-uniform-dis – Learning Math Jul 23 '20 at 10:28
  • 1
    @LearningMath That is for $d=3$, right? – Sarvesh Ravichandran Iyer Jul 23 '20 at 11:07
  • @астонвіллатересалисбон Thanks again! Yes it is. So for $d=2, d\ge 4,$ it's no longer true...? And if it's not the uniform distribution, can we find the general PDF for $U^{T}v?$ – Learning Math Jul 23 '20 at 11:18
  • I think it can be done in general, but I can't do it, I'm stuck myself! (Of course, for $d=2$, Bohr told you it is the distribution induced by the random variable $\theta \to \cos \theta$, which is not uniform). – Sarvesh Ravichandran Iyer Jul 23 '20 at 11:24
  • 1
    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 $N(0,1)$. – kimchi lover Jul 23 '20 at 13:50
  • @kimchilover Thanks for your comment, could you please elaborate it a bit more, or write it as answer please? – Learning Math Jul 23 '20 at 14:08

1 Answers1

4

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.

kimchi lover
  • 24,277
  • 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 $U:= Z/||Z|| \sim \mathcal{U}nif(S^{d-1}).$ – Learning Math Jul 23 '20 at 16:33
  • 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 $v?$ – Learning Math Jul 23 '20 at 16:58
  • 1
    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. – kimchi lover Jul 23 '20 at 18:01
  • Thanks for editing your answer! One last question though (as mentioned in my comment above): how do we show that for any uniform distribution, there's a normal distribution $N$ with zero mean identify covariance matrix, so that the uniform distribution is the ration $N/||N||?$ You're using that argument in your answer, no? – Learning Math Jul 23 '20 at 19:41
  • 1
    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. – kimchi lover Jul 23 '20 at 20:16
  • Thank you. But what I was pointing to was slightly different, and in fact the other direction of the statement. Yes, if $Z \sim \mathcal{N}(0, I_d),$ then $\frac{Z}{||Z||}\sim Unif(S^{d-1}), ||Z|| \sim \chi_d.$ You mentioned this direction surely. But in order to finish the argument: we indeed need the other direction: $U \sim Unif(S^{d-1}) \implies \exists Z\sim \mathcal{N}(0,I_d) $ so that $U= \frac{Z}{||Z||}.$ I'm not sure if it's in your argument? Do you've a source that proves it? I can ask a separate question and link it if you prefer. Thanks again! – Learning Math Jul 24 '20 at 08:04
  • P.S. I think one way to answer the question is to simply define $Z$ by: $Z:= U. \chi_d, \chi_d$ being a chi-distribution with $d$ degrees of freedom. It still remains to prove that $Z \sim \mathcal{N}(0,I_d),$ but I believe it's proven somewhere, although I don't know where. What do you think? – Learning Math Jul 24 '20 at 09:56
  • 1
    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. – kimchi lover Jul 24 '20 at 10:46
  • I submitted a separate question here - https://math.stackexchange.com/questions/3767725/is-a-uniform-distribution-on-a-sphere-always-a-norm-scaled-normal-mathcaln0. However, I'm not after the proof of Maxwell's theorem, my point is that you started with the assumption that any uniform distribution $U$ arises as the norm scaled $N(0,I_d)$ distribution, which requires a proof. – Learning Math Jul 24 '20 at 12:36