5
$\begingroup$

Suppose $x\in R^n$ follows the uniform distribution on the surface of a unit sphere centered at 0 and $a$ is a vector in $R^n$. What can we say about the distribution of the following quantity?

$$\sum_i x_i^2 a_i $$

Mean seems to be $\frac{\sum_i a_i}{n}$, what about the variance?

This quantity is the value of quadratic form with eigenvalues $a_1,a_2,\ldots$ when evaluated at some point on the surface of unit sphere.

enter image description here notebook

Improve this question
$\endgroup$
9
  • $\begingroup$ So are you only asking about the variance of $\sum_i x_i^2 a_i$? Also, what exactly do you mean by "sample from Gaussian, normalize"? $\endgroup$
    – mhdadk
    Commented Jun 24, 2021 at 23:51
  • 1
    $\begingroup$ I mean that it's a uniform distribution on a sphere, I've edited with a more precise definition from Mardia $\endgroup$
    – Yaroslav Bulatov
    Commented Jun 24, 2021 at 23:53
  • 1
    $\begingroup$ @MattF. it's the value of $x'Ax$ where $A$ is a diagonal matrix with $a_i$ as diagonal entries $\endgroup$
    – Yaroslav Bulatov
    Commented Jun 25, 2021 at 1:19
  • 2
    $\begingroup$ The variance is $$\frac{2}{n^2(n+2)}\sum_{i<j}(a_i-a_j)^2$$ and the mean, as you said, is $\frac{1}{n}\sum a_i$. I may write up a proof tomorrow if no one gives a reference first. $\endgroup$
    – user225256
    Commented Jun 25, 2021 at 2:53
  • 2
    $\begingroup$ A multiple of $(x_1^2,x_2^2\ldots,x_p^2)$ has a Dirichlet$(1/2,1/2,\ldots,1/2)$ distribution when the sphere is centered at the origin (as strongly implied by your quotation). That will lend itself to further analysis. It yields the moments easily. $\endgroup$
    – whuber ♦
    Commented Jun 25, 2021 at 14:03

1 Answer 1

Reset to default
4
$\begingroup$

Let's start with $m(a)=E[\langle x^2,a\rangle]$ and $s(a)=E[\langle x^2,a\rangle^2]$, so that the variance will be $s(a)-m(a)^2$.

We will calculate these by expressing $x$ in $n$-dimensional spherical coordinates, so \begin{align} & & x_1&=\cos(\phi_1)\\ 0\ \le\ &\phi_1\le\pi & x_2&=\sin(\phi_1)\cos(\phi_2)\\ 0\ \le\ &\phi_2\le\pi & x_3&=\sin(\phi_1)\sin(\phi_2)\cos(\phi_3)\\ &\vdots & \vdots\\ 0\ \le\ & \phi_{n-1}\le\pi & x_{n-1}&=\sin(\phi_1)\cdots\sin(\phi_{n-2})\cos(\phi_{n-1})\\ 0\ \le\ &\phi_n\le2\pi & x_n&=\sin(\phi_1)\cdots\sin(\phi_{n-2})\sin(\phi_{n-1})\\ \end{align} and the element of $n-1$-dimensional surface area is $$dS = \sin^{n-2}(\phi_1)\sin^{n-3}(\phi_2)\cdots \sin(\phi_{n-2})\ d\phi_1\, d\phi_2 \cdots d\phi_n$$

We begin with some special cases, letting $b_i$ be the vector that agrees with $a$ on the $i^{th}$ coordinate and is zero elsewhere. First, \begin{align} m(b_1)&=\frac {\int\cdots\int a_1\cos^2(\phi_1)\, dS} {\int\cdots\int dS}\\ &=\frac{\int a_1\cos^2(\phi_1)\sin^{n-2}(\phi_1)\,d\phi_1}{\int \sin^{n-2}(\phi_1)\,d\phi_1}\\ &=a_1\left(1-\frac{\int \sin^n(\phi_1)\,d\phi_1} {\int \sin^{n-2}(\phi_1)\,d\phi_1}\right)\\ &=a_1\left(1-\frac{n-1}{n}\right)\\ &=a_1/n \end{align} where the integrals over higher $\phi_i$ cancel, and the final ratio uses integration by parts. Since $m$ must be symmetric in all the coordinates, we get \begin{align} m(b_i)&=b_i/n\\ m(a)=m\left(\sum b_i\right)&=\sum m(b_i)=\frac{1}{n}\sum a_i \end{align} Similarly, \begin{align} s(b_i)&=3a_i^2/(n^2+2n)\\ s(b_i+b_j)-s(b_i)-s(b_j)&=2a_ia_j/(n^2+2n) \end{align}

Now by the polynomial identity $$\left(\sum x_i\right)^2=\sum_i x_i^2+\sum_{i<j} \left((x_i+x_j)^2 - x_i^2-x_j^2\right)$$ we also have \begin{align} s(a) &=s\left(\sum b_i\right)\\ &=\sum_i s(b_i)+\sum_{i<j}\left(s(b_i+b_j)-s(b_i)-s(b_j)\right)\\ &=\sum_i \frac{3a_i^2}{n^2+2n}+\sum_{i<j}\frac{2a_ia_j}{n^2+2n} \end{align}

So finally, the variance is \begin{align} v&=s(a)-m(a)^2\\ &=\sum_{i}a_i^2\left(\frac{3}{n^2+2n}-\frac{1}{n^2}\right) + \sum_{i<j} a_ia_j\left(\frac{2}{n^2+2n}-\frac{2}{n^2}\right)\\ &=\frac{2}{n^2(n+2)}\left((n-1)\sum_{i}a_i^2 - 2\sum_{i<j} a_ia_j\right)\\ &=\frac{2}{n^2(n+2)}\sum_{i<j}(a_i-a_j)^2 \end{align}

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ +1. Re the expectations: because the $x_i$ are exchangeable, all the $x_i^2$ have the same expectation. Since these $n$ squares sum to unity, that common expectation must be $1/n.$ Re the covariances: the covariance matrix of $(x_i^2)$ is readily obtained from the marginal distribution derived at stats.stackexchange.com/a/520811/919. Some people might find these approaches more congenial than resorting to spherical coordinates. $\endgroup$
    – whuber ♦
    Commented Jun 25, 2021 at 18:30
  • $\begingroup$ that looks intense....but matches what I'm getting numerically, thanks! $\endgroup$
    – Yaroslav Bulatov
    Commented Jun 25, 2021 at 18:44
  • 2
    $\begingroup$ @whuber, that is slicker! I am just happy that these calculations turned out nicer than most with high-dimensional spherical coordinates: no multiple integrals, no gamma function, and no alternating appearances of $\sqrt{\pi}$. $\endgroup$
    – user225256
    Commented Jun 25, 2021 at 19:05
  • $\begingroup$ Indeed. But it is a little sobering to look at the full distribution for various vectors $(a_i)$ ;-). $\endgroup$
    – whuber ♦
    Commented Jun 25, 2021 at 19:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.