Normalized vector of i.i.d. copies of $X$ uniformly distributed on the sphere means $X$ is normally distributed Let $X$ be a random variable with $\mathbb{E} X^2 = 1$. Let $X_i$ be i.i.d. copies of $X$ such that
$$
\frac{1}{\sqrt{\sum X_i^2}} \left(X_1, ..., X_N\right)
$$
is uniformly distributed on $\mathbb{S}^{N-1}$. Prove that $X = \mathcal{N}(0,1)$ in distribution. 
 A: This isn't super rigorous, but it's an old argument that is historically important (for instance, it was used in deriving the Maxwell-Boltzman distribution, I think.)
We know that the distribution of the vector $\vec X$ must be spherically symmetric, so $$ f_{\vec X}(x_1,\ldots,x_N) = g(x_1^2+\ldots +x_N^2).$$
We also know that the components of the vector are independent and that their distributions are symmetric, so that $$f_{\vec X}(x_1,\ldots,x_N) = h(x_1^2)h(x_2^2)\ldots h(x_N^2)$$ so we have a functional equation $$ g(x_1^2+\ldots +x_N^2) = h(x_1^2)\ldots h(x_N^2).$$ 
Letting $x_2=0,x_3=0\ldots x_N=0,$ we get $g(x_1^2) = h(x_1^2)h(0)^{N-1}$ so we can write $$ h(0)^{N-1}h(x_1^2+\ldots +x_N^2) =  h(x_1^2)\ldots h(x_N^2)$$ or, letting $\tilde h(x) = h(x)/h(0),$ $$ \tilde h(x_1^2+\ldots+x_N^2) = \tilde h(x_1^2)\ldots \tilde h(x_N^2).$$
It is known known that the only nice solution to the functional equation $f(x_1+\ldots +x_N) = f(x_1)\ldots f(x_N)  $ is $e^{ax}$ so we have $$ h(x^2) = h(0)e^{-bx^2}$$ and clearly $b$ must be positive for the distribution to be valid.
So finally, we have $$ f_{\vec X}(x_1,\ldots x_n) = C e^{-b(x_1^2+\ldots x_N^2)}$$
So we have $X_i$ is an $N(0,\sigma^2)$ for some variance $\sigma^2.$ Since you have $E(X^2) = 1$ it must be $N(0,1).$
