Marginal density of uniform spherical distribution

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.

Answer 2

With $X$ uniformly distributed over the unit $n$-sphere, the joint probability distribution of all $n$ elements of $X$ is a Dirac delta function, $$P(X_1,X_2,\ldots X_n)\propto\delta\left(1-\sum_{j=1}^n X_j^2\right).\qquad\qquad(1)$$ Now you integrate out elements one by one, to obtain the marginal distribution $P_k$ of $k<n$ elements. The first integration gives $$P_{n-1}(X_2,X_3,\ldots X_n)\propto\left(1-\sum_{j=2}^n X_j^2\right)^{-1/2}\theta\left(1-\sum_{j=2}^n X_j^2\right),\qquad(2)$$ with $\theta$ the unit step function. The second integration gives $$P_{n-2}(X_3,\ldots X_n)\propto\theta\left(1-\sum_{j=3}^n X_j^2\right),$$ the third integration $$P_{n-3}(X_4,\ldots X_n)\propto\left(1-\sum_{j=4}^n X_j^2\right)^{1/2}\theta\left(1-\sum_{j=4}^n X_j^2\right),$$ and so on. Each additional integration increases the power by 1/2, $$P_{k}(X_{n-k+1},\ldots X_n)\propto\left(1-\sum_{j=n-k+1}^n X_j^2\right)^{(n-k)/2-1}\theta\left(1-\sum_{j=n-k+1}^n X_j^2\right).$$ This is the answer in the OP _{(without rescaling the radius of the $n$-sphere, so $r^2/n\mapsto r^2$).}

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_{As requested in the comments, a more detailed exposition of the various steps. • **First step:** the delta function. Denote the surface measure on the unit $n$-sphere as $d\Omega$, and $\int d\Omega=A_n$ the surface area. Uniformity of a distribution on the unit $n$-sphere means uniformity with measure $d\Omega$. I maintain that the joint probability distribution of the components of the vector ${\mathbf X}=(X_1,X_2,\ldots X_n)$, uniformly distributed on the unit $n$-sphere, is given by Eq. (1) with normalization constant $2/A_n$. Let us check this by calculating the expectation value of an arbitrary function $f$ of ${\mathbf X}$. For that purpose I transform to hyperspherical coordinates $r,\phi_1,\phi_2,\ldots\phi_{n-1}$, $$\mathbb{E}[f(\mathbf{X})]=\int dX_1\int dX_2\cdots\int dX_n \,f(X_1,X_2,\ldots X_n)P(X_1,X_2,\ldots X_n)$$ $$\qquad=\int_0^\infty r^{n-1} dr \int d\Omega\, f(r,\phi_1,\phi_2,\ldots\phi_{n-1})\frac{2}{A_n}\delta(1-r^2)$$ $$\qquad=\frac{1}{A_n}\int d\Omega\, f(r=1,\phi_1,\phi_2,\ldots\phi_{n-1}).$$ In the last step I used that $\int_0^\infty r^{n-1}\delta(1-r^2)\,dr=1/2$ for $n\geq 2$. • **Second step:** integration of the delta function, to arrive at Eq. (2). From now on I will ignore the normalization constants, these can easily be recovered at the end. Let me abbreviate $\sum_{j=2}^n X_j^2=s_2$. The marginal distribution $P_1(X_2,X_3,\ldots X_n)$ is obtained by definition upon integration of $P(X_1,X_2,X_3,\ldots X_n)$ over $X_1$. I carry out this integration in cartesian coordinates, changing variables to $q=X_1^2$, $$P_1(X_2,X_3,\ldots X_n)\propto \int_{-\infty}^\infty dX_1\delta(1-s_2-X_1^2),$$ $$\qquad=\int_0^\infty\delta(1-s_2-q)\frac{dq}{\sqrt q}=(1-s_2)^{-1/2}\theta(1-s_2).$$ • **Third and following steps:** The following steps, subsequent integrations of $X_2,X_3,\ldots$ are now immediate consequences of the integral $$\int_0^a(a^2-x^2)^p\,dx=c_p a^{1+2p}.$$}

Answer 3

Let $G_1,\dots,G_n$ be iid standard normal random variables. Then the random vector \begin{equation*} (Y_1,\dots,Y_n):=\Big(\frac{G_1}{\sqrt{\sum_1^n G_j^2}},\dots, \frac{G_n}{\sqrt{\sum_1^n G_j^2}}\Big) \end{equation*} is uniformly distributed on the unit sphere $\mathbb S^{n-1}$. Let \begin{equation*} \begin{aligned} Z_i&:=Y_i=\frac{G_i}{\sqrt{\sum_1^n G_j^2}}&\text{ if }i\le k,\\ Z_i&:=G_i&\text{ if }i> k. \end{aligned} \end{equation*} We want to find the joint pdf of $(Z_1,\dots,Z_k)$, which is the same as the joint pdf of $(X_1,\dots,X_k)/\sqrt n$.

The vector $(Z_1,\dots,Z_n)$ is obtained from $(G_1,\dots,G_n)$ by the transformation given by \begin{equation*} \begin{aligned} z_i&:=\frac{g_i}{\sqrt{\sum_1^n g_j^2}}&\text{ if }i\le k,\\ z_i&:=g_i&\text{ if }i> k. \end{aligned} \end{equation*} The transformation inverse to this is given by \begin{equation*} \begin{aligned} g_i&:=\sqrt{s_2}\frac{z_i}{\sqrt{1-s_1}}&\text{ if }i\le k,\\ g_i&:=z_i&\text{ if }i> k, \end{aligned} \tag{1} \end{equation*} where \begin{equation*} s_1:=\sum_1^k z_j^2,\quad s_2:=\sum_{k+1}^n z_j^2. \end{equation*} The Jacobian determinant of the inverse transformation is \begin{equation*} J=\det(cM)=c^k\det M, \end{equation*} where \begin{equation*} c:=s_2^{1/2}(1-s_1)^{-3/2},\quad M:=(1-s_1)I_k+UU^T, \end{equation*} $I_k$ is the $k\times k$ identity matrix, and $U:=[z_1,\dots,z_k]^T$.

Write $U=|U|Qe_1$, where $|U|=\sqrt{s_1}$ is the Euclidean norm of $U$, $Q$ is some orthogonal matrix, and $e_1:=(1,0,\dots,0)$. Then it is clear that the matrix $M$ is similar to $N:=(1-s_1)I_k+|U|^2e_1e_1^T=(1-s_1)I_k+s_1e_1e_1^T$, whence $\det M=\det N=(1-s_1)^{k-1}$. So, \begin{equation*} J=s_2^{k/2}(1-s_1)^{-k/2-1}. \tag{2} \end{equation*} Also, the joint pdf of $(G_1,\dots,G_n)$ is given by \begin{equation*} (2\pi)^{-n/2}\exp\Big\{-\frac12\sum_1^n g_j^2\Big\}. \end{equation*} So, in view of (1) and (2), the joint pdf of $(Z_1,\dots,Z_n)$ is given by \begin{equation*} f_n(z_1,\dots,z_n) =(2\pi)^{-n/2}\exp\Big\{-\frac12\frac{s_2}{1-s_1}\Big\}s_2^{k/2}(1-s_1)^{-k/2-1}. \end{equation*} So, the joint pdf of $(Z_1,\dots,Z_k)$ is given by \begin{align*} f_k(z_1,\dots,z_k)&=\int_{\mathbb R^{n-k}}dz_{k+1}\dots dz_n\,f_n(z_1,\dots,z_n) \\ &=(2\pi)^{-n/2}(1-s_1)^{-k/2-1} \\ &\times \int_{\mathbb R^{n-k}}dz_{k+1}\dots dz_n\,s_2^{k/2}\,\exp\Big\{-\frac12\frac{s_2}{1-s_1}\Big\} \\ &\propto(1-s_1)^{(n-k)/2-1}, \end{align*} because $s_2=\sum_{k+1}^n z_j^2$. So, we have the desired result.

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More specifically, \begin{align*} f_k(z_1,\dots,z_k) =\frac{\Gamma(n/2)}{\pi^{k/2}\Gamma((n-k)/2)}(1-s_1)^{(n-k)/2-1} \end{align*} for $s_1=\sum_1^k z_j^2\in(0,1)$.

Answer 4

$\newcommand{\R}{\mathbb{R}} \newcommand{\x}{\mathbf{x}} \newcommand{\X}{\mathbf{X}}$ Here is yet another solution, which is partly informal but I think not hard to completely formalize. Its advantage is a strong and hopefully convincing appeal to geometric intuition.

Again, we are going to assume that $(X_1,\dots,X_n)$ is uniformly distributed on the unit sphere $\mathbb S^{n-1}$, rather than on $\sqrt n\,\mathbb S^{n-1}$. Let $\X_j:=(X_1,\dots,X_j)$ and $\x_j:=(x_1,\dots,x_j)$ for $\x=(x_1,\dots,x_n)\in\R^n$ and $j=1,\dots,n-1$. Let $|\cdot|$ denotes the Euclidean norm.

The main point is that the probability density $p_{n-1}(\x_{n-1})$ of $\X_{n-1}$ at a point $\x_{n-1}\in\R^{n-1}$ with $|\x_{n-1}|<1$ is proportional to the ratio $r_{n-1}(\x_{n-1}):=vol_{n-1}(dS)/vol_{n-1}(dA)$, where $vol_{n-1}$ is of course the $(n-1)$-volume, $dA$ is an infinitesimal neighborhood of the point $\x_{n-1}$ in $\R^{n-1}$ and $dS$ is the preimage of $dA$ under the projection of the upper hemisphere $\mathbb S^{n-1}_+:=\{\x\in\mathbb S^{n-1}\colon\x\cdot e_n\ge0\}$ onto the closed unit ball in $\R^{n-1}$; this projection is given by $\mathbb S^{n-1}_+\ni(\x_{n-1},u)\mapsto\x_{n-1}$; here $e_n:=(0,\dots,0,1)$ and $\cdot$ denotes the dot product. But \begin{equation} r_{n-1}(\x_{n-1})=\frac{vol_{n-1}(dS)}{vol_{n-1}(dA)}=\frac1{\cos\phi}, \end{equation} where $\phi$ is the angle between the hyperplane $\R^{n-1}\times\{0\}$ of $\R^n$ and the tangent hyperplane to $\mathbb S^{n-1}$ at the point $(\x_{n-1},\sqrt{1-|\x_{n-1}|^2})\in\mathbb S^{n-1}_+$; that is, $\phi$ is the angle between the corresponding normal vectors $e_n$ and $(\x_{n-1},\sqrt{1-|\x_{n-1}|^2})$ of these two hyperplanes.

Thus, \begin{equation} p_{n-1}(\x_{n-1})\propto r_{n-1}(\x_{n-1})\propto\frac1{\cos\phi}=(1-|\x_{n-1}|^2)^{-1/2}, \end{equation} where $\propto$ means an equality up to a constant factor, depending only on $n,k$.

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.

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Here is a picture, for $n=3$, showing the upper hemisphere $\mathbb S^{n-1}_+$; a small neighborhood of a point $\x_{n-1}$ in the projection of $\mathbb S^{n-1}_+$ to the horizontal plane $\R^{n-1}\times\{0\}$ of $\R^n$; the preimage of that neighborhood under that projection; and the normal vectors of the horizontal plane and the tangent plane to the sphere -- with $\phi$ being the angle between these two normal vectors.

Answer 5

Although great answers have already been provided, the one provided below is perhaps be the shortest possible answer to the question.

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Let $x_1,\ldots,x_k \in \mathbb R$ such that $r_k^2:=\sum_{i=1}^k x_i^2 < 1$, and note that the marginal density of $(X_1,\ldots,X_k)$ at $(x_1,\ldots,x_k)$ equals $p_n(r_k^2)$, where

$$ p_n(t) \propto \int_0^\infty\ldots \int_0^\infty dx_{k+1}\ldots dx_n\delta(\sum_{i=1}^n x_i^2-t) \,\forall t \in \mathbb R. $$

The Laplace transform w.r.t $t$ is given by $$ \begin{split} \hat{p}_n(s) = \int_0^\infty e^{-ts}p_n(t)dt &\propto \int_0^\infty e^{-s\sum_{i=1}^n x_i^2}dx_{k+1}\ldots dx_n \\ &= e^{-s\sum_{i=1}^k x_i^2}\int_0^\infty e^{-s\sum_{i=k+1}^n x_i^2}dx_{k+1}\ldots dx_n\\ &= e^{-sr_k^2}\left(\int_0^\infty e^{-sz^2}dz\right)^{(n-k)} \propto e^{-sr_k^2}s^{-(n-k)/2}. \end{split} $$ Evaluating (e.g via mathematica, etc.) the inverse Laplace transform of the last term at $t=r_k^2$, we deduce that $p_n(r_k^2) \propto (1-t)^{(n-k)/2-1}\delta(t-r_k^2)\bigg\rvert_{t=r_k^2} = (1-r_k^2)^{(n-k)/2-1}$.

Answer 6

I enjoyed thinking about these answers and this is my attempt to put them into (nonrigorous) geometrical terms. Writing the joint density compositionally as

$$p(\mathbf{x}_k \mid |\mathbf{x}| = 1)p(\mathbf{x}_{n-k} \mid \mathbf{x}_k, |\mathbf{x}| = 1) = p(\mathbf{x} \mid |\mathbf{x}| = 1) \propto 1,$$

we want to solve for the first term on the left. But since our density is proportional to a constant, this is just

$$p(\mathbf{x}_k \mid |\mathbf{x}| = 1) \propto \frac{1}{p(\mathbf{x}_{n-k} \mid \mathbf{x}_k, |\mathbf{x}| = 1)}.$$

Accordingly, instead of performing our calculation by integrating out $X_{k+1} \dots X_{n}$, we can think about it in terms of the conditional density for sampling $\mathbf{X}_{n-k}$, given $\mathbf{X}_k$ and the norm constraint $|\mathbf{X}| = 1$, denoted in the denominator above.

I propose a two step procedure. First, draw a point uniformly from within the $n-k-1$ dimensional ball with radius $$r_{n-k} = \sqrt{1 - |\mathbf{x}_k|^2}.$$ Each such vector has density proportional to $$\left(1 - |\mathbf{x}_k|^2\right)^{-(n-k-1)/2}.$$

This corresponds to the second and higher integrations in the previous answers, whereas here we directly use the formula for the volume of a ball.

Next, the $(n-k)$th coordinate must satisfy $|\mathbf{x}| = 1$, which is achieved by any point on the circle with radius $r_{n-k}$, a set with measure proportional to $r_{n-k}$. Proving this is the first integration in the previous answers.

Putting these two steps together and taking the reciprocal gives

$$p(\mathbf{x}_k \mid |\mathbf{x}| = 1) \propto \left(1 - |\mathbf{x}_k|^2\right)^{(n-k)/2 - 1}.$$