Let $x$ be a random vector uniformly distributed on the unit sphere $\mathbb{S}^{D1}$ and let $\mathcal{V}$ be a linear subspace of dimension $m$. Then it is known that the euclidean norm of the orthogonal projection $(x)_{\mathcal{V}}$ of $x$ onto $\mathcal{V}$ satisfies \begin{align} (1\epsilon) \sqrt{m/D} \le \left\ (x)_{\mathcal{V}} \right\_2 \le (1+\epsilon) \sqrt{m/D} \end{align} with probability at least $1  2\exp\left(m\epsilon^2 / 2\right)$. What bothers me with this result is that for $m=0 \, (m=D)$ i do not get that the norm is zero (one) with probability one, as one would intuitively expect from such a result. I wonder if a tigher projection theorem is available in the literature. Many thanks.
2

$\begingroup$ Isn't the distribution know to be a beta distribution? $\endgroup$ – user83457 Sep 19 '17 at 11:49

$\begingroup$ @michael: Very interesting, i didn't know the distribution was known. Could you please point to a reference? $\endgroup$ – Manos Sep 19 '17 at 14:11

2$\begingroup$ I don't have a reference but if any m dimensional projection will have the same distribution, so if you represent your uniform as $\frac{\bar X}{\bar X} $ where the $X_i$ are iid normal the projection on the first m coordinates is $\frac {(X_1,...X_m)}{\bar X}$ whose norm square is $\frac {X_1^2 + ... + X_m^2}{X_1^2 + ... + X_m^2 + X_{m+1}^2 + ... + X_N^2}$ which is of the form $\chi_1^2/(\chi_1^2 + \chi_2^2)$, $\chi_1^2, \chi_2^2$ independent chi squareds, which I think (per Wikipedia) makes them beta, so its the norm squared, not the norm. $\endgroup$ – user83457 Sep 19 '17 at 15:09