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Let $a, b, c \in \mathrm R$ be such, that $a^2+b^2+c^2=1$. Let $U=(X,Y,Z)$ be a random vector, about what we know is only that $aX+bY+cZ$ is uniformly distributed on $(-1, 1)$ line (for each a,b,c satisfying the condition $a^2+b^2+c^2=1$). Show $U$ is uniformly distributed on unit sphere $S_3$.


I totally have no idea where to start. Everything i can conclude by myself is for that $U$ distribution, if there exists positive mass outside of $S_3$, then $aX+bY+cZ$ whould not be uniformly distributed on $(-1,1)$
Jakub Wierzbicki
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  • For each of them – Jakub Wierzbicki Aug 22 '20 at 20:03
  • Sorry for ambiguity, i have edited the post. – Jakub Wierzbicki Aug 22 '20 at 20:06
  • Thanks for the edit. You could start by verifying that the uniform distribution on $S_3$ has the desired property. – kimchi lover Aug 22 '20 at 20:14
  • Okay, I've got this. Let $p=(a,b,c), ||p||=1$ and $u=(x,y,z)$. Let''s assume $u$ comes from uniform distribution on $S_3$, which implies $||u||=1$ for sure. Let $\alpha=\measuredangle (p, v)$. Then $p \cdot u = \cos(\alpha)$. $\mathrm F(t)=\mathrm P(aX+bY+cZ<t)=\mathrm P(\cos(\alpha)<t)$. With the last equality it's straightforward to compute that probability as the measure of the surface of the sphere slice, which leds to $F(t)=\mathbb{1}_{(-1, 1)}(t) \cdot \frac{1-t}{2} $ – Jakub Wierzbicki Aug 22 '20 at 23:09
  • Good. Is there some other distribution on $\mathbb R^3$ with the same 1-dimensional margins? If not, you are done. – kimchi lover Aug 22 '20 at 23:12
  • Uniform distribution on $(-1,1)\mathrm x(-1,1)\mathrm x(-1,1)$ has the same margins as that one on $S_3$. Also: I made a mistake above and $F(t)$ should be equal to $\frac{t+1}{2}$ – Jakub Wierzbicki Aug 22 '20 at 23:32
  • The issue is, can there be a different distribution with the same margins? – kimchi lover Aug 22 '20 at 23:38
  • By marginal distribution you mean distribution $\mu_{X_i}(A)=\mu(R_1,...,R_{i-1}, A,R_{i+1},..R_n)$ for a random vector $(X_1, ... X_n)$ with values in $R_1xR_2...xR_n$ and so on? – Jakub Wierzbicki Aug 22 '20 at 23:52
  • In that sense, are'nt uniform distributions on cube, and on sphere different? – Jakub Wierzbicki Aug 23 '20 at 00:01

1 Answers1

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The trick is that the distribution of a $\mathbb R^n$-valued random vector is completely specified by the distributions of all the linear forms in $X$, that is, by the distributions of $\langle \alpha, X\rangle$ for all $\alpha\in \mathbb (R^n)^*$. (To know the distribution of $\langle \alpha, X\rangle$ is to know the 1-dimensional characteristic function $\phi_\alpha(t)=E\exp(it\langle \alpha, X\rangle)$. To know all the $\phi_\alpha$ functions is enough to determine the multivariate characteristic function $\phi(\tau)=E\exp(i\langle\tau,X\rangle)$, using the formula $\phi(t\alpha)=\phi_\alpha(t)$.) This is the Cramér–Wold theorem ; it is the basic reason why CAT scans work.

In the present problem, a simple calculation shows that the uniform distribution on the surface of the unit sphere is such that for every $\alpha$, the marginal distribution of $\langle\alpha,X\rangle$ is $U[-\|\alpha\|,\|\alpha\|]$. These marginals are what you get by integrating the multivariate distribution of $X$ on hyperplanes orthogonal to $\alpha$. The C-W result shows this is the only distribution with this property.

kimchi lover
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