Marginal Density of Uniform Spherical Distribution in Probability

Question

Suppose that $X$ is distributed uniformly in the scaled $n$-sphere $\sqrt{n} \mathbf{S}^{n-1} \subset \mathbf{R}^n$. Then apparently the distribution of $(X_1, \dots, X_k)$, the first $k < n$ coordinates of $X$ has density $p(x_1, \dots, x_k)$ with respect to Lebesgue measure in $\mathbf{R}^k$, moreover if $r^2 = x_1^2 + \cdots + x_k^2$, then it is proportional to
$$
\left(1 – \frac{r^2}{n}\right)^{(n-k)/2 – 1}, \quad \text{if}~0 \leq r^2 \leq n,
$$
and otherwise is 0. I tried to compute this using the fact that $(X_1, \dots, X_k) \stackrel{\rm d}{=} \sqrt{n} (g_1, \dots, g_k)/\sqrt{g_1^2 + \cdots + g_n^2}$, when $g_i$ are iid standard normal variables, but it was somewhat unclear to me how to use even this representation to compute the density. Can anyone sketch the details for me?

Answer 1

$\newcommand{\R}{\mathbb{R}} \newcommand{\x}{\mathbf{x}} \newcommand{\X}{\mathbf{X}}$ This is to present a formalization of the answer by Carlo Beenakker, without explicit use of the delta function.

We are going to assume that $\X=(X_1,\dots,X_n)$ is uniformly distributed on the unit sphere $\mathbb S^{n-1}$, rather than on $\sqrt n\,\mathbb S^{n-1}$.

For each real $t\in(0,1)$, define the measures $\mu_t$ and $\nu_t$ over $\R^n$ by the conditions \begin{equation*} \int f\,d\mu_t=\int_{\R^n}d\x f(\x)1_{1-t<|\x|^2\le1} \end{equation*} and \begin{equation*} \int f\,d\nu_t=\frac{\int f\,d\mu_t}{\int d\mu_t} \end{equation*} for all (say) nonnegative continuous functions $f\colon\R^n\to\R$, where $|\cdot|$ denotes the Euclidean norm. Then $\nu_t$ is a probability measure converging (as $t\downarrow0$) to the Haar measure $h$ on the unit sphere in $\R^n$, in the sense that (say) \begin{equation*} \int f\,d\nu_t\to\int f\,dh \end{equation*} for all nonnegative continuous functions $f\colon\R^n\to\R$.

Take now any function $f\colon\R^n\to\R$ such that \begin{equation*} f(\x)=g(\x_{n-1}) \end{equation*} for some nonnegative continuous function $g\colon\R^{n-1}\to\R$ and all $\x\in\R^n$, where $\x_j:=(x_1,\dots,x_j)$ for $\x=(x_1,\dots,x_n)\in\R^n$ and $j=1,\dots,n-1$. Then
\begin{align*} \int f\,d\mu_t&=\int_{\R^{n-1}}d\x_{n-1}\,g(\x_{n-1})\int_\R du\, 1_{1-t<|\x_{n-1}|^2+u^2\le1} \\ &=\int_{\R^{n-1}}d\x_{n-1}\,g(\x_{n-1}) (1+o(1))t\,(1-|\x_{n-1}|^2)^{-1/2}\,1_{|\x_{n-1}|<1}. \end{align*} Also, \begin{equation*} \int d\mu_t=\int_{\R^n}d\x\, 1_{1-t<|\x|^2\le1}\propto(1+o(1))t, \end{equation*} where $\propto$ means an equality up to a constant factor, depending only on $n,k$. So, \begin{equation*} \int f\,dh=\lim_{t\downarrow0} \int f\,d\nu_t\propto\int_{\R^{n-1}}d\x_{n-1}\,g(\x_{n-1}) (1-|\x_{n-1}|^2)^{-1/2}\,1_{|\x_{n-1}|<1}. \end{equation*} Thus, the joint pdf of $\X_{n-1}=(X_1,\dots,X_{n-1})$ is given by \begin{equation*} p_{n-1}(\x_{n-1})\propto(1-|\x_{n-1}|^2)^{-1/2}\,1_{|\x_{n-1}|<1}. \end{equation*} Now successively integrating $p_{n-1}(\x_{n-1})$ ($n-1-k$ times) in $x_{n-1},\dots,x_{k+1}$ and each time using the formula \begin{equation*} \int_0^{b^{1/2}}(b-u^2)^p du=c_p b^{p+1/2} \end{equation*} for real $b>0$ and $p>-1$ with $c_p:=\int_0^1(1-u^2)^pdu\in(0,\infty)$ (so that $1/2$ is added to the exponent $p$ after such an integration), we see that the joint pdf of $\X_k=(X_1,\dots,X_k)$ is given by \begin{equation*} p_k(\x_k)\propto(1-|\x_k|^2)^{(n-k)/2-1}\,1_{|\x_k|<1}, \end{equation*} as desired.