I would like to compute the following moment for the uniform distribution defined on the hypersphere (sometimes referred to as the moment of inertia):

$\int_{\mathbb{S}^{d-1}} \theta \theta^\top \sigma(\mathrm{d}\theta),$

where $\mathbb{S}^{d-1} = \{\theta \in \mathbb{R}^d ; \|\theta\|=1\}$ is the d-dimensional hypersphere and $\sigma(\mathrm{d} \theta)$ denotes the uniform distribution on $\mathbb{S}^{d-1}$.

Numerically, I have confirmed that the answer is $\frac1{d}I$, where $I$ is the $d\times d$ identity matrix, but I failed to prove it formally.

I wanted to use the fact that if $x \sim N(0,I)$ then $\frac{x}{\|x\|}$ is uniformly distributed on the sphere, but I couldn't go very far. I would be very happy if someone can shed some light on this.


1 Answer 1


First look at the off-diagonal entries of your matrix. Writing $\theta = (\theta_1, \theta_2, \ldots, \theta_d)$ you have the integral is

$$ \int_{\mathbb{S}^d} \theta_i \theta_j \sigma(\mathrm{d}\theta) $$

where $i \neq j$. Notice that $\theta_i$ is an "odd" function under reflection of the $i$th coordinate, and similarly $\theta_j$. This tells you that the above integral vanishes (as the sphere is reflectionally symmetric).

Next look at the diagonal entries. By symmetry each of the diagonal entries should be the same. If you look at the trace of this matrix, you have the integral

$$ \int_{\mathbb{S}^d} \|\theta\|^2 \sigma(\mathrm{d}\theta) $$

so using that $\|\theta\|^2 = 1$ on the sphere you have that this total integral evaluates to $1$.

Given that each of the entries are the same, each entry along the diagonal must have magnitude $1/d$ in return.

  • $\begingroup$ Thank you for the answer! However, I'm a little bit confused about the off-diagonal part: could you write it more explicitly? $\endgroup$
    – thmusic
    Commented Feb 4, 2020 at 16:37
  • 1
    $\begingroup$ @thmusic: maybe this will help: you can do a rotation of the sphere $\theta_1 \to \theta_2$ and $\theta_2 \to - \theta_1$. Your set-up is symmetric so the moment of inertia shouldn't change. But the explicit integral went from integrating $\theta_1\theta_2$ to integrating $- \theta_2\theta_1$. If the two are equal they must be zero. $\endgroup$ Commented Feb 4, 2020 at 16:44
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    $\begingroup$ Anyway, it is the same reason that the integral $\int_{-1}^1 x f(x) ~\mathrm{d}x = 0$ for any even function $f(x)$. $\endgroup$ Commented Feb 4, 2020 at 16:48

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