I am stuck with a problem in multivariable statistics. The problem can be stated as follows:
For a spherically symmetric distribution in $\mathbb{R}^d$, it can be specified completely by the function $F(r)=\Pr(\|X\|>r)$. For example, 2d standard Gaussian distribution has $F(r)=\exp(-r^2/2)$. Now $X_1, X_2, \dots, X_d$ i.i.d. follow this distribution. These $d$ random points form a hyperplane in $\mathbb{R}^d$. What is the probability $H(p)$ that the distance from the origin to this hyperplane is larger than $p$ for $p>0$?
The answer should be written as integration about $F$. I found the solution for $d=2$ in a German paper [1], which says $$ H(p)=\frac{2}{\pi} \int_p^{\infty} \arccos\frac{p}{r} |d(F^2(r))| $$ I successfully derive the formula for $d=3$ as: \begin{align} H(p)= \int_p^{+\infty}\frac{2p }{\pi r \sqrt{r^2-p^2}}G_3^{3}(r)dr \end{align} where \begin{align} G_3(r)=\int_r^{+\infty} \sqrt{1-\frac{r^2}{y^2}} |d F(y)| \end{align} which represents $P(x_1^2+x_2^2>r^2)$.
In $d=3$, I can write $H(p)$ in a similar form as $d=2$: \begin{align} H(p)=\frac{2}{\pi} \int_p^{\infty} \textrm{arccos}\frac{p}{r} |d(G_3^{3}(r))| \end{align} Then I encounter problem to obtain $H(p)$ for $d>3$, since the derivation for $d=2,3$ relies on geometric insights. Any suggestions on this problem?
Now I guess for $d\geq 2$: \begin{align} G_d(r) &= \int_r^{+\infty} (1-\frac{r^2}{y^2})^{\frac{d}{2}-1}|dF(y)|\\ H(p) & =\frac{2}{\pi} \int_p^{\infty} \textrm{arccos}\frac{p}{r} |d(G_d^{d}(r))| \end{align} However, I found difficulty to prove this conjecture.
[1] Carnal, Henri. "Die konvexe Hülle von n rotationssymmetrisch verteilten Punkten." Zeitschrift für Wahrscheinlichkeitstheorie und verwandte Gebiete 15.2 (1970): 168-176.
