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Assume $\mathbf{U} = (U_{1}, U_{2}, U_{3})' \sim unif(S^{2})$. How would I show that $U_{1} \sim unif(-1,1)$?

I don't know if I'm confusing myself, because I can't see this as being true for higher dimensions.

pidgeot
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    I don't see how the claim could be true: the surface area of the subset of $\boldsymbol U$ such that $0 \le U_1 < 1/2$ is clearly larger than the subset corresponding to $1/2 \le U_1 < 1$. – heropup Nov 12 '14 at 23:13
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    There is a well known and classical theorem that says that projection from a sphere to a cylinder is area preserving. This fact plus the rotational symmetry of the situation implies what you want. – Nate Nov 12 '14 at 23:16
  • @Nate The theorem is true for spheres of all dimensions, not just $2,$ right? – Learning Math Jul 22 '20 at 12:05
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    @LearningMath Nope, it is special for dimension 2. For a 1 dimensional sphere, projecting the uniform measure onto one of the axes concentrates a bit toward the poles. As the dimension increases, a larger and larger percentage of the mass becomes concentrated on the equator and in fact for large enough dimension, 99% of the surface of a unit sphere is within .001 of the equator. – Nate Jul 22 '20 at 15:49
  • @Nate Sorry, I think I made a mistake in my asking my question to you. Yes, I'm aware of the concentration of measure near the equator, but my question was: is the orthogonal projection of the uniform distribution on any dimensional sphere is uniform on $(-1,1),$ right? If not, what is it? – Learning Math Jul 22 '20 at 20:30
  • @LearningMath I am confused as to the difference between what you are asking now and what I was addressing in my previous comment. The answer is no, the projection of the uniform distribution on $S^n$ to $(-1,1)$ is not uniform unless $n=2$. – Nate Jul 23 '20 at 14:43

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Relating to the comment of Nate, there's also the "well known" formula for the area of a surface of revolution, namely if $f:[a,b]\rightarrow \mathbb R$ is smooth and positive, we can rotate its graph $\Gamma_f = \{(x,y)\ |\ x \in [a,b],\ y = f(x)\}$ around the $x$-axis and obtain a surface $\Sigma$. The area of that surface is $$ \mathrm{Area}(\Sigma) = 2\pi \int_a^bf(x)\sqrt{1+f'(x)^2}dx. $$ It's likely that your familiar with that from calculus or classical differential geometry. Alternatively, you can consult Wikipedia http://en.wikipedia.org/wiki/Surface_of_revolution

In particular, the function $g:[-1,1]\rightarrow\mathbb R,\ x \mapsto \sqrt{1-x^2}$ produces the sphere $S^2$ as a surface of revolution. We have to be careful about the end points $-1$ and $1$ of the interval $[-1,1]$, but since we're in a probabilistic setting and consider uniform distribution on $S^2$, we can ignore these points by invoking a measure zero argument.

Now, let's denote uniform probability measure on $S^2$ by $P$, i.e. for $A \subseteq S^2$ Lebesgue measurable, we have $$ P(A) = \mathrm{Area}(A) / \mathrm{Area}(S^2) = \mathrm{Area}(A)/4\pi $$ (you can get $\mathrm{Area}(S^2)$ from the formula above :-). Moreover, let's denote Lebesgue measure on $\mathbb R$ by $\lambda$ and uniform probability measure on $[-1,1]$ by $Q$, i.e. for $B \subseteq [-1,1]$ Lebesgue measurable, we have $$ Q(B) = \lambda(B) / \lambda([-1,1]) = \lambda(B) / 2. $$ Finally, let $$ T:S^2\rightarrow [-1,1],\ (x,y,z) \mapsto x $$ be the projection onto the first coordinate. We want to show that the image of $P$ under $T$ is $Q$, i.e. $\forall B \subseteq [-1,1]$ Lebesgue measurable: $P(T^{-1}(B)) = Q(B)$. By construction of Lebesgue measure, this is equvalent to $$ \forall [a,b] \subseteq [-1,1]:\ P(T^{-1}([a,b])) = Q([a,b]).$$

Now, $T^{-1}([a,b])$ is the slice of $S^2$ "around" the interval $[a,b]$ on the $x$-axis. We can calculate its area with the formula from above, namely $$ \mathrm{Area}(T^{-1}([a,b])) = 2\pi\int_a^bg(x)\sqrt{1+g'(x)}dx. $$ Here, the integrand is $$g(x)\sqrt{1+g'(x)^2} = \sqrt{1-x^2}\sqrt{1+\frac{x^2}{1-x^2}} = \sqrt{1-x^2}\sqrt{\frac1{1-x^2}} = 1. $$ From this we get $$ P(T^{-1}([a,b])) = \frac{\mathrm{Area}(T^{-1}([a,b]))}{4\pi} = \frac{2\pi(b-a)}{4\pi} = \frac{b-a}2 = Q([a,b]), $$ which is exactly what we want.

Michael Hardy
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jflipp
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