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Let $B^{n} _p= ${$ (x_1, \dots, x_n) : |x_1|^p + \dots |x_n|^p = 1 $} be the unit ball in $\mathbb{R}^n$ in the $\ell^p$ norm.

If $X_1,\dots,X_n$ are iid $\exp(1)$ -distributed random variables, then $(X_1/D,\dots,X_n/D)$, where $D =X_1+ \dots + X_n $ is uniformly distributed in $B^{n}_1$.

If $X_1,\dots,X_n$ are iid normally distributed with mean 0, then $(X_1/D,\dots,X_n/D)$, where $D = (X_1^2+\dots+X_n^2)^{1/2}$, is uniformly distributed in $B^{n}_2$.

Is there a choice of $X_1,\dots , X_n$ iid such that $ ( X_1 / D, \dots, X_n/D)$, where $D = (|X_1|^p + \dots + |X_n|^p)^{1/p} $ is uniformly distributed in $B^{n} _p$ for arbitrary $p$?

I would be happy with any sensible common generalization of the two statements above. I have no particular reason to believe there is such a generalization - I'm just hoping that two so similar and neat examples have similarly nice generalizations.

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    $\begingroup$ you are probably looking for: mathoverflow.net/questions/9185/… $\endgroup$
    – Suvrit
    Feb 7, 2012 at 21:16
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    $\begingroup$ so it seems that you are looking for uniform distribution on the surface of an $\ell_p$ ball (not in the ball). $\endgroup$
    – Suvrit
    Feb 7, 2012 at 22:10
  • $\begingroup$ I think you intend to normalize by $D^{1/p}$ instead of $D$, if I'm not mistaken. Also, $B_1^n$ is not the standard simplex. To generate uniformly on the $\ell_1$ ball you need to do something like multiply each coordinate $X_i$ by iid random variables $\epsilon_i$ uniform on {−1,+1}. $\endgroup$
    – cardinal
    Feb 8, 2012 at 3:22
  • $\begingroup$ Reposting a link mentioned in a previous comment so that it appears in the "Linked" questions list: How to generate random points in $\ell_p$ balls? $\endgroup$ May 6, 2023 at 9:57

3 Answers 3

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The result you want, I think, is in Stationarity, Isotropy and Sphericity in $l_p^*$. It is behind a pay-wall, but the form of the distribution is stated in the abstract.

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  • $\begingroup$ This paper fails to do the job since the OP wants independent $X_i$. $\endgroup$ Feb 8, 2012 at 17:28
  • $\begingroup$ Sure, I misread the question as asking what form of iid distribution leads to particular forms of exchangeability via de Finetti representation theorems. This is the context I'm most used to seeing the normal distribution referred to a "spherically symmetric". The paper I linked to is the natural extension of that idea to the $l_p$ setting. $\endgroup$
    – R Hahn
    Feb 8, 2012 at 17:58
  • $\begingroup$ The link to springerlink.com is broken. I'm also unable to find any snapshot saved on the Wayback Machine. Presumably it is meant to point to the following article: Berman, Simeon M., Stationarity, isotropy and sphericity in $\ell_p$, Z. Wahrscheinlichkeitstheor. Verw. Geb. 54, 21-23 (1980). Zbl 0441.60012. $\endgroup$ May 6, 2023 at 9:51
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If by uniform measure you mean $(n-1)$-dimensional Hausdorff measure on the sphere, the answer is no. As a consequence of the results of this paper by Barthe, Csörnyei, and Naor, under mild regularity assumptions the only measure on the boundary of any convex body which can be generated in this way is the "cone measure" on the $\ell_p$ sphere for $1 \le p < \infty$, which coincides with uniform measure only for $p=1,2$.

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  • $\begingroup$ The link to cims.nyu.edu seems to be broken, but a snapshot is saved on the Wayback Machine. The full citation might be: Barthe, Franck; Csörnyei, Marianna; Naor, Assaf, A note on simultaneous polar and Cartesian decomposition, Milman, V. D. (ed.) et al., Geometric aspects of functional analysis. Proceedings of the Israel seminar (GAFA) 2001–2002. Berlin: Springer (ISBN 3-540-00485-8/pbk). Lect. Notes Math. 1807, 1-19 (2003). Zbl 1036.52004. $\endgroup$ May 6, 2023 at 9:55
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[Similar to this answer, except for change of notation and form of random vector studied]

Let ${ p \in [1, \infty) .}$ So ${ \lVert x \rVert _{p} := (\sum _{1} ^{n} \vert x _i \vert ^{p}) ^{\frac{1}{p}} }$ is a norm on ${ \mathbb{R} ^{n} }.$

For any ${ n \geq 1 ,}$ let ${ (B _{p} ^n) ^{+} }$ denote the positive part of ${ n }$ dimensional unit ball namely $${ (B ^n _p ) ^{+} := \lbrace x \in \mathbb{R} ^{n} : \text{each } x _i> 0, \lVert x \rVert _{p} < 1 \rbrace ,}$$ and let ${ (S ^{n} _p) ^{+} }$ denote the positive part of ${ n }$ dimensional unit sphere namely $${ (S ^{n} _p) ^{+} := \lbrace x \in \mathbb{R} ^{n+1} : \text{each } x _i > 0, \lVert x \rVert _{p} = 1 \rbrace. }$$

We have a "lifting" bijection $${ \phi : (B ^{n} _p) ^{+} \to (S ^{n} _p) ^{+} ,}$$ $${ (x _1, \ldots, x _{n}) \mapsto (x _1, \ldots, x _{n}, (1-x _{1} ^{p} - \ldots - x _{n} ^{p} ) ^{\frac{1}{p}} ) }.$$
We can study how bijection ${ \phi }$ maps small boxes. (Recall box notation as here).
For any fixed ${ x \in (B ^{n} _p ) ^{+} }$ and small ${ h },$ the difference ${ \phi(x+h) - \phi(x) }$ is approximately $${ (D\phi) _{x} (h) = \underbrace{\begin{pmatrix} 1 &0 &\cdots &0 \\ 0 &1 &\cdots &0 \\ \vdots &\vdots &\ddots &\vdots \\ 0 &0 &\cdots &1 \\ \partial _{1} \phi _{n+1} &\partial _{2} \phi _{n+1} &\ldots &\partial _{n} \phi _{n+1} \end{pmatrix}} _{(n+1) \times n \text{ matrix}} h }$$ where ${ \partial _{i} \phi _{n} }$ ${ = - (x _{i}) ^{p-1} (1- x _{1} ^{p} - \ldots - x _{n} ^{p}) ^{\frac{1}{p} -1} .}$
So any small box ${ x + \text{box}(\delta _{1} e _{1}, \ldots, \delta _{n} e _{n}) }$ in ${ (B _p ^{n}) ^{+} }$ (with ${ \delta _{i} }$s positive) gets approximately mapped to ${ \phi(x) + \text{box}((D\phi) _{x} (\delta _{1} e _{1}), \ldots, (D\phi) _{x} (\delta _{n} e _{n})). }$

Now change-of-density on mapping via bijection ${ \phi }$ can be studied. Say a random vector ${ Y = (Y _1, \ldots, Y _{n}) \in (B ^{n} _p ) ^{+} }$ has density ${ f _{Y} }.$ For any ${ y \in (B ^{n} _p) ^{+} }$ the event $${ \lbrace Y \in y + \text{box}(\delta _{1} e _{1}, \ldots , \delta _{n} e _{n}) \rbrace }$$ is approximately the event $${ \lbrace \phi(Y) \in \phi(y) + \text{box}((D\phi) _{y} (\delta _{1} e _{1}), \ldots, (D\phi) _{y} (\delta _{n} e _{n})) \rbrace }$$ giving $${ f _{Y}(y) \delta _{1} \ldots \delta _{n} \approx f _{\phi(Y)} (\phi(y)) \delta _{1} \ldots \delta _{n} \sqrt{\det (D\phi) _{y} ^{T} (D\phi) _{y} } }.$$ Hence the random vector ${ Z = \phi(Y) \in (S ^{n} _p) ^{+} }$ has density $${ f _{Z} (z) = \frac{1}{\sqrt{\det( M _{z} ^{T} M _{z}) }} f _{Y} (\phi ^{-1} (z)) \text{ for } z \in (S ^n _p) ^{+} },$$ where ${ M _{z} = (D\phi) _{ \phi ^{-1} (z)} }.$ To compute this density,

Lemma 1: Let ${ y = (y _{1} , \ldots, y _{n}) \in (B ^{n} _p ) ^{+} }.$ With ${ \phi }$ as above, the matrix of inner products ${ (D\phi) _{y} ^{T} (D\phi) _{y} }$ has determinant $${ \det (D\phi) _{y} ^{T} (D\phi) _{y} = 1 + \frac{\sum _{1} ^{n} y _i ^{2(p-1)}}{(1-y _{1} ^{p} - \ldots - y _{n} ^{p}) ^{2(1-\frac{1}{p})}} }.$$
Pf: The matrix of inner products ${ (D\phi) _{y} ^{T} (D\phi) _{y} }$ is ${ n }$ by ${ n }.$ Its ${ (i,i) }$ entry is ${ 1 + (\partial _{i} \phi _{n+1}) ^{2} },$ and its ${ (i,j) }$ entry where ${ i \neq j }$ is ${ (\partial _{i} \phi _{n+1}) (\partial _{j} \phi _{n+1}) }.$ So we have $${ (D\phi) _{y} ^{T} (D\phi) _{y} = \text{Id} _{n} + \begin{pmatrix} \partial _{1} \phi _{n+1} \\ \vdots \\ \partial _{n} \phi _{n+1} \end{pmatrix} \begin{pmatrix} \partial _{1} \phi _{n+1} &\cdots &\partial _{n} \phi _{n+1} \end{pmatrix}}.$$ It is of the form identity plus a rank ${ 1 }$ matrix (unless all ${ y _{i} }$ are ${ 0 }$). By lemma ${ 2 }$ below, its determinant is ${ 1 }$ plus trace of this rank ${ 1 }$ matrix, i.e. \begin{align*} \det (D\phi) _{y} ^{T} (D\phi) _{y} &= 1 + \sum _{1} ^{n} (\partial _{i} \phi _{n+1}) ^{2} \\ &= 1 + \sum _{1} ^{n} (y _i) ^{2(p-1)}(1-y _{1} ^{p} - \ldots - y _{n} ^{p}) ^{2(\frac{1}{p} - 1)} , \end{align*} as needed.

Lemma 2: Let ${ A \in \mathbb{F} ^{n \times n} }$ be a rank ${ 1 }$ matrix. Then its characteristic polynomial is given by $${ f(t) = \det(tI-A) = t ^{n-1} (t - \text{tr}(A)) }.$$ Especially, ${ \text{det}(I + A) }$ ${ = (-1) ^{n} f(-1) }$ ${ = 1 + \text{tr}(A) }.$
Pf: Since ${ A }$ has kernel of dimension ${ (n-1) }$ pick a basis ${ (e _1, \ldots, e _{n-1}) }$ of ${ \text{ker}(A) }.$ Picking a ${ v \notin \text{ker}(A) }$ gives a basis ${ (e _1, \ldots, e _{n-1}; v) }$ of ${ \mathbb{F} ^{n} }.$
Now matrix of ${ A }$ w.r.t this basis has first ${ (n-1) }$ columns zero. So characteristic polynomial of ${ A }$ is ${ f(t) = t ^{n-1} (t - \lambda ) },$ where ${ \lambda }$ is bottom right entry of above matrix. So infact ${ f(t) = t ^{n-1} (t - \text{tr}(A)) },$ as needed.

So we have:

Theorem 1 [Change of density from ${ (B _p ^n) ^{+} }$ to ${ (S _p ^{n}) ^{+} }$]
Let ${ Y \in (B ^{n} _p) ^{+} }$ be a random vector with density ${ f _Y }.$ Then the vector ${ Z = \phi(Y) \in (S ^{n} _p) ^{+} }$ has density \begin{align*} f _Z (z) &= f _{Z _1, \ldots, Z _{n+1}} (z _1, \ldots, z _{n+1}) \\ &= \frac{1}{\sqrt{\det M _{z} ^{T} M _{z} }} f _{Y} (\phi ^{-1} (z)) \\ &= \left( 1 + \frac{\sum _{1} ^{n} z _i ^{2(p-1)}}{(1- z _{1} ^{p} - \ldots - z _{n} ^{p}) ^{2(1-\frac{1}{p})}} \right) ^{- \frac{1}{2}} f _{Y _1, \ldots, Y _{n}} (z _1, \ldots, z _n). \end{align*} for ${ z = (z _1, \ldots, z _{n+1}) \in (S ^{n} _p) ^{+} }.$ Equivalently, $${ f _{Z} (z) = \left( 1 + \sum _{i=1} ^{n} \left(\frac{z _i}{z _{n+1}}\right) ^{2(p-1)} \right) ^{-\frac{1}{2}} f _{Y} (z _1, \ldots, z _n) \text{, for } z \in (S _p ^n) ^{+} . }$$



Now let ${ X _1, X _2, \ldots \geq 0 }$ be independent positive random variables with ``nice" densities ${ f _1, f _2, \ldots }$ respectively. We will soon pick appropriate ${ f _i }$s. Like here but slightly different, consider the map $${ G : (0, \infty) ^{n+1} \to (B ^n _p) ^{+} \times (0, \infty) ,}$$ $${ (x _1, \ldots, x _{n+1}) \mapsto (y _1, \ldots, y _{n+1}) = \left(\frac{x _1}{(\sum _{1} ^{n} x _{i} ^{p} + x _{n+1} ^{p}) ^{\frac{1}{p}} }, \ldots, \frac{x _n}{(\sum _{1} ^{n} x _{i} ^{p} + x _{n+1} ^{p}) ^{\frac{1}{p}} } , (\sum _{1} ^{n} x _{i} ^{p} + x _{n+1} ^{p}) ^{\frac{1}{p}} \right) }$$ which can be inverted as $${ G ^{-1} : (y _1, \ldots, y _{n+1}) \mapsto (x _1, \ldots, x _{n+1}) = (y _1 y _{n+1}, \ldots, y _{n} y _{n+1}, [(y _{n+1}) ^{p} - (y _1 y _{n+1}) ^p - \ldots - (y _n y _{n+1}) ^{p}] ^{\frac{1}{p}} ) }.$$ The last coordinate is ${ x _{n+1} = (y _{n+1}) (1- y _{1} ^{p} - \ldots - y _{n} ^{p}) ^{\frac{1}{p}} }.$

The random vector $${ (Y _1, \ldots, Y _{n+1}) = G(X _1, \ldots, X _{n+1}) \in (B ^n _p) ^{+} \times (0, \infty) }$$ has density $${ f _{Y _1, \ldots, Y _{n+1}} (y _1, \ldots, y _{n+1}) = f _{X _1, \ldots, X _{n+1}} (G ^{-1}(y)) \vert \det D(G ^{-1}) _{y} \vert }$$ that is $${ f _{Y _1, \ldots, Y _{n+1}} (y _1, \ldots, y _{n+1}) }$$ $${ = f _{1} (y _1 y _{n+1}) \ldots f _{n} (y _{n} y _{n+1}) f _{n+1} ( (y _{n+1}) (1- y _{1} ^{p} - \ldots - y _{n} ^{p}) ^{\frac{1}{p}}) \vert \det D(G ^{-1}) _{y} \vert .}$$
Computing ${ \det D(G ^{-1}) _{y} }$ like here we get change of density $${ f _{Y _1, \ldots, Y _{n+1}} (y _1, \ldots, y _{n+1}) }$$ $${ = f _{1} (y _1 y _{n+1}) \ldots f _{n} (y _{n} y _{n+1}) f _{n+1} ( (y _{n+1}) (1- y _{1} ^{p} - \ldots - y _{n} ^{p}) ^{\frac{1}{p}}) (y _{n+1} ^{n}) (1- y _{1} ^{p} - \ldots - y _{n} ^{p}) ^{\frac{1}{p} -1} }.$$

Hence the vector $${ (Y _1, \ldots, Y _n) = \left(\frac{X _1}{(\sum _{1} ^{n} X _{i} ^{p} + X _{n+1} ^{p}) ^{\frac{1}{p}} }, \ldots, \frac{X _n}{(\sum _{1} ^{n} X _{i} ^{p} + X _{n+1} ^{p}) ^{\frac{1}{p}} } \right) \in (B ^{n} _p) ^{+} }$$ has density \begin{align*} f _{Y _1, \ldots, Y _n} (y _1, \ldots, y _n) &= \int _{0} ^{\infty} f _{Y _1, \ldots, Y _{n+1}} (y _1, \ldots, y _{n+1}) \, dy _{n+1} \\ &=(1- y _{1} ^{p} - \ldots - y _{n} ^{p}) ^{\frac{1}{p} -1} \int _{0} ^{\infty} f _{1} (y _1 u) \ldots f _{n} (y _{n} u ) f _{n+1} ( u (1- y _{1} ^{p} - \ldots - y _{n} ^{p}) ^{\frac{1}{p}}) u ^{n} \, du \end{align*}.

So the further transformed vector $${ Z =\phi(Y _1, \ldots, Y _n) \in (S ^{n} _p) ^{+}, }$$ that is $${ Z = \left(\frac{X _1}{(\sum _{1} ^{n} X _{i} ^{p} + X _{n+1} ^{p}) ^{\frac{1}{p}} }, \ldots, \frac{X _n}{(\sum _{1} ^{n} X _{i} ^{p} + X _{n+1} ^{p}) ^{\frac{1}{p}} } , \frac{X _{n+1}}{(\sum _{1} ^{n} X _{i} ^{p} + X _{n+1} ^{p}) ^{\frac{1}{p}} } \right) \in (S ^n _p) ^{+} ,}$$ has density \begin{align*} f _{Z} (z ) &= f _{Z _1, \ldots, Z _{n+1}} (z _1, \ldots, z _{n+1}) \\ &= \left( 1 + \frac{\sum _{1} ^{n} z _i ^{2(p-1)}}{(1- z _{1} ^{p} - \ldots - z _{n} ^{p}) ^{2(1-\frac{1}{p})}} \right) ^{- \frac{1}{2}} f _{Y _1, \ldots, Y _{n}} (z _1, \ldots, z _n) \\ &= \left( 1 + \frac{\sum _{1} ^{n} z _i ^{2(p-1)}}{(1- z _{1} ^{p} - \ldots - z _{n} ^{p}) ^{2(1-\frac{1}{p})}} \right) ^{- \frac{1}{2}} (1-z _{1} ^{p} - \ldots - z _{n} ^{p}) ^{\frac{1}{p}-1} \\ &\, \int _{0} ^{\infty} f _{1} (z _1 u) \ldots f _{n} (z _{n} u ) f _{n+1} ( u (1- z _{1} ^{p} - \ldots - z _{n} ^{p}) ^{\frac{1}{p}}) u ^{n} \, du \\ &= \underbrace{\left(\sum _{i=1} ^{n+1} z _i ^{2(p-1)} \right) ^{-\frac{1}{2}}} _{=: \, A(z)} \underbrace{\int _{0} ^{\infty} f _{1} (z _1 u) \ldots f _{n} (z _{n} u ) f _{n+1} ( u (1- z _{1} ^{p} - \ldots - z _{n} ^{p}) ^{\frac{1}{p}}) u ^{n} \, du} _{=: \, B(z)} \end{align*} for ${ z = (z _1, \ldots, z _{n+1}) \in (S ^{n} _p) ^{+} }.$

It is useful to see in which cases above density ${ f _{Z} (z) }$ is constant over ${ z \in (S ^{n} _p) ^{+} }.$

If densities ${ f _{1} (t), \ldots, f _{n+1} (t) }$ are all proportional to ${ e ^{-{t ^p}} }$ (for ${ t \geq 0 }$), the integral term ${ B(z) }$ is constant over ${ z \in (S ^{n} _p) ^{+} }.$ If further ${ p \in \lbrace 1,2 \rbrace },$ the sum term ${ A(z) }$ is constant over ${ z \in (S ^{n} _p) ^{+} }$. This gives:

Theorem 2 [Density of normalised vector on ${ (S _p ^n) ^{+} }$ for ${ p = 1,2 }$]
Let ${ p \in \lbrace 1, 2 \rbrace }.$ If ${ X _1, \ldots, X _{n+1} }$ are independent positive random variables, with densities proportional to ${ e ^{-t ^p} }$ (for ${ t \geq 0 }$), the random vector $${ Z = \left(\frac{X _1}{(\sum _{1} ^{n} X _{i} ^{p} + X _{n+1} ^{p}) ^{\frac{1}{p}} }, \ldots, \frac{X _n}{(\sum _{1} ^{n} X _{i} ^{p} + X _{n+1} ^{p}) ^{\frac{1}{p}} } , \frac{X _{n+1}}{(\sum _{1} ^{n} X _{i} ^{p} + X _{n+1} ^{p}) ^{\frac{1}{p}} } \right) \in (S ^n _p) ^{+} }$$ is uniform over ${ (S _p ^{n} ) ^{+} = \lbrace x \in \mathbb{R} ^{n+1} : \text{each } x _i > 0, \lVert x \rVert _{p} = 1 \rbrace }.$

If ${ (Z _1, \ldots, Z _{n+1}) }$ is uniform over ${ (S _p ^n) ^{+}, }$ the random vector ${ ((-1) ^{S _1} Z _1, \ldots, (-1) ^{S _{n+1}} Z _{n+1}) }$ with ${ S _i }$ independent ${ \text{Unif}\lbrace 0,1 \rbrace }$ variables is uniform over the sphere ${ S _{p} ^{n} = \lbrace x \in \mathbb{R} ^{n+1} : \lVert x \rVert _{p} = 1 \rbrace }.$ In the context of above theorem $${ ((-1) ^{S _1} Z _1, \ldots, (-1) ^{S _{n+1}} Z _{n+1}) }$$ $${ = \left(\frac{(-1) ^{S _1} X _1}{(\sum _{1} ^{n} \vert (-1) ^{S _i} X _{i} \vert ^{p} + \vert (-1) ^{S _{n+1}} X _{n+1} \vert ^{p}) ^{\frac{1}{p}} }, \ldots, \frac{(-1) ^{S _n} X _n}{(\sum _{1} ^{n} \vert (-1) ^{S _i} X _{i} \vert ^{p} + \vert (-1) ^{S _{n+1}} X _{n+1} \vert ^{p}) ^{\frac{1}{p}} } , \frac{(-1) ^{S _{n+1}} X _{n+1}}{(\sum _{1} ^{n} \vert (-1) ^{S _i} X _{i} \vert ^{p} + \vert (-1) ^{S _{n+1}} X _{n+1} \vert ^{p}) ^{\frac{1}{p}} } \right). }$$ This gives:

Theorem 3 [Density of normalised vector on ${ S _p ^n }$ for ${ p = 1,2 }$]
Let ${ p \in \lbrace 1, 2 \rbrace }.$ If ${ X _1, \ldots, X _{n+1} }$ are independent random variables, with densities proportional to ${ e ^{-\vert t \vert ^p} }$ (for ${ t \in \mathbb{R} }$), the random vector $${ Z = \left(\frac{X _1}{(\sum _{1} ^{n} \vert X _{i} \vert ^{p} + \vert X _{n+1} \vert ^{p}) ^{\frac{1}{p}} }, \ldots, \frac{X _n}{(\sum _{1} ^{n} \vert X _{i} \vert ^{p} + \vert X _{n+1} \vert ^{p}) ^{\frac{1}{p}} } , \frac{X _{n+1}}{(\sum _{1} ^{n} \vert X _{i} \vert ^{p} + \vert X _{n+1} \vert ^{p}) ^{\frac{1}{p}} } \right) \in S ^n _p }$$ is uniform over ${ S _p ^{n} = \lbrace x \in \mathbb{R} ^{n+1} : \lVert x \rVert _{p} = 1 \rbrace }.$

So random variables with density proportional to ${ e ^{-t ^2}, }$ i.e. Gaussian random variables, are geometrically significant in terms of sampling uniformly from Euclidean spheres.

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