Spherical Symmetry and Rotation Invariance

Symmetry in Every Direction

A vector distribution can look the same after its coordinates are rotated or reflected. This is stronger than saying that its coordinates have similar marginal distributions: the entire joint distribution must remain unchanged. Such symmetry is useful when no direction in the coordinate system should be privileged. It also has a precise signature in the characteristic function: the function depends on a vector argument only through its length.

We work in $\mathbb{R}^d$, with $d\geq 1$. For $x\in\mathbb{R}^d$, write $\|x\|=(x^\mathsf{T}x)^{1/2}$. An orthogonal matrix $Q$ satisfies $Q^\mathsf{T}Q=I_d$; it preserves lengths and includes both rotations and reflections. In particular, considering every orthogonal matrix is essential: invariance under rotations alone need not include reflection symmetry.

Definition: A random vector $X$ in $\mathbb{R}^d$ has a spherical distribution about the origin if $QX$ has the same distribution as $X$ for every $d\times d$ orthogonal matrix $Q$. Equivalently, $QX\overset{d}{=}X$ for every such $Q$. Its characteristic function is $\varphi_X(t)=E[\exp(i t^\mathsf{T}X)]$, for $t\in\mathbb{R}^d$.

The origin is part of this definition: a vector shifted away from the origin need not be spherical about the origin. No density or moment assumption is required. Thus spherical distributions may be discrete, continuous, or mixtures, and the definition still applies.

Characteristic Functions and Spherical Symmetry

The key geometric fact is that any two vectors of the same length can be carried to one another by an orthogonal transformation. Consequently, if the distribution of $X$ is unchanged by all such transformations, the characteristic function cannot distinguish arguments with equal length.

Theorem: If $X$ has a spherical distribution in $\mathbb{R}^d$, there is a function $\psi:[0,\infty)\to\mathbb{C}$ such that $$ \varphi_X(t)=\psi(\|t\|),\qquad t\in\mathbb{R}^d. $$ Conversely, if the characteristic function of $X$ has this form, then $X$ has a spherical distribution.

Proof. First suppose $X$ is spherical. If $\|s\|=\|t\|$, there is an orthogonal matrix $Q$ such that $s=Qt$. Therefore

\varphi_X(s) =E[\exp(i t^\mathsf{T}Q^\mathsf{T}X)] =\varphi_{Q^\mathsf{T}X}(t) =\varphi_X(t).

The final equality follows because $Q^\mathsf{T}$ is orthogonal and $Q^\mathsf{T}X\overset{d}{=}X$. Thus the characteristic function is constant on every set of vectors with a fixed length. Set $e_1=(1,0,\ldots,0)^\mathsf{T}$ and define $\psi(r)=\varphi_X(re_1)$ for $r\geq0$. For any $t$, choose an orthogonal matrix carrying $\|t\|e_1$ to $t$; the equality just proved gives $\varphi_X(t)=\psi(\|t\|)$.

For the converse, let $Q$ be any orthogonal matrix. The characteristic function of $QX$ at $t$ is

\varphi_{QX}(t) =E[\exp(i t^\mathsf{T}QX)] =\varphi_X(Q^\mathsf{T}t) =\psi(\|Q^\mathsf{T}t\|) =\psi(\|t\|) =\varphi_X(t).

The third equality uses the assumed radial form, and the fourth uses preservation of length by $Q^\mathsf{T}$. By uniqueness of characteristic functions, $QX$ and $X$ have the same distribution. This holds for every orthogonal $Q$, so $X$ is spherical. $\square$

The theorem is about the full distribution, not just its first few moments. Its converse is particularly useful: to check spherical symmetry, one may sometimes find it easier to establish that the characteristic function depends only on $\|t\|$ than to compare distributions after every transformation.

Worked Examples

Worked Example: An Isotropic Normal Vector

Let $X=(X_1,\ldots,X_d)^\mathsf{T}$, where the coordinates are independent and each has distribution $N(0,\sigma^2)$, with $\sigma^2>0$. Verify spherical symmetry and find its characteristic function.

The joint density is

f_X(x)=(2\pi\sigma^2)^{-d/2} \exp\left(-\frac{x^\mathsf{T}x}{2\sigma^2}\right).

For any orthogonal $Q$, $\|Qx\|=\|x\|$ and $|\det Q|=1$. The density of $QX$, obtained by the change of variables $x=Q^\mathsf{T}y$, is $f_X(Q^\mathsf{T}y)|\det Q^\mathsf{T}|=f_X(y)$. Hence $QX$ has the same density as $X$, and $X$ is spherical.

Independence gives the characteristic function as a product of the coordinate characteristic functions:

\varphi_X(t) =\prod_{j=1}^d \exp\left(-\frac{\sigma^2t_j^2}{2}\right) =\exp\left(-\frac{\sigma^2}{2}\sum_{j=1}^d t_j^2\right) =\exp\left(-\frac{\sigma^2\|t\|^2}{2}\right).

This depends on $t$ only through its length, as the theorem requires. For example, in dimension $3$ with $\sigma^2=4$, at any $t$ with $\|t\|=2$, the value is $\exp(-8)$, regardless of the direction of $t$.

Worked Example: A Uniform Distribution on a Disk

Let $X$ be uniform on the disk $D_R=\{x\in\mathbb{R}^2:\|x\|\leq R\}$, where $R>0$. Its density is $f_X(x)=1/(\pi R^2)$ on the disk and zero outside. Verify spherical symmetry and calculate its mean and covariance.

An orthogonal transformation maps $D_R$ onto itself and preserves area, since its Jacobian has absolute determinant $1$. The transformed density is therefore again constant $1/(\pi R^2)$ on $D_R$ and zero outside. This proves invariance under every orthogonal transformation.

The disk is unchanged by $x\mapsto -x$, so symmetry and boundedness give $E[X]=0$. To calculate the second moments, write $x=(r\cos\theta,r\sin\theta)^\mathsf{T}$, with area element $r\,dr\,d\theta$. Then

E[X_1^2] =\frac{1}{\pi R^2}\int_0^R\int_0^{2\pi}r^2\cos^2\theta\,r\,d\theta\,dr =\frac{1}{\pi R^2}\cdot\frac{R^4}{4}\cdot\pi =\frac{R^2}{4}.

Interchanging sine and cosine gives $E[X_2^2]=R^2/4$. Also, $\int_0^{2\pi}\cos\theta\sin\theta\,d\theta=0$, so $E[X_1X_2]=0$. Since the mean is zero, the covariance matrix is

\operatorname{Cov}(X)= \begin{pmatrix} R^2/4&0\\ 0&R^2/4 \end{pmatrix} =\frac{R^2}{4}I_2.

The characteristic function must have the form $\psi(\|t\|)$, even though calculating that function explicitly is unnecessary to verify the symmetry.

Worked Example: A Mixture of Isotropic Normals

In $\mathbb{R}^2$, let $X$ have distribution $N_2(0,I_2)$ with probability $1/4$ and $N_2(0,4I_2)$ with probability $3/4$. Show that the mixture is spherical and calculate its characteristic function.

Each component is invariant under every orthogonal transformation, by the isotropic normal calculation. For any measurable set $A$ and orthogonal $Q$, the mixture probabilities satisfy

P(QX\in A) =\frac14P(QX\in A\mid\text{first component}) +\frac34P(QX\in A\mid\text{second component}) =P(X\in A).

Thus the mixture is spherical. Its characteristic function is the same weighted average of the component characteristic functions:

\varphi_X(t) =\frac14\exp\left(-\frac{\|t\|^2}{2}\right) +\frac34\exp\left(-\frac{4\|t\|^2}{2}\right) =\frac14e^{-\|t\|^2/2}+\frac34e^{-2\|t\|^2}.

This is a function of $\|t\|$ alone. The mixture need not be a single normal distribution for spherical symmetry to hold.

What Symmetry Implies About Covariance

Spherical symmetry also constrains moments when they exist. The covariance implication below requires a finite second moment; spherical symmetry itself does not.

Result: If $X$ is spherical and $E[\|X\|^2]<\infty$, then $E[X]=0$ and $\operatorname{Cov}(X)=cI_d$ for some $c\geq0$.

Proof. Since $-I_d$ is orthogonal, $-X\overset{d}{=}X$. Finite second moment ensures that $E[X]$ exists. Taking expectations yields $E[X]=-E[X]$, so $E[X]=0$. Let $\Sigma=\operatorname{Cov}(X)$, which is finite under the stated assumption. Invariance gives $QX\overset{d}{=}X$, and the covariance transformation rule gives $Q\Sigma Q^\mathsf{T}=\Sigma$ for every orthogonal $Q$.

Take $Q$ to be diagonal with one diagonal entry $-1$ and all others $1$. The equality forces every covariance entry linking that coordinate to a different coordinate to be zero. Applying this for each coordinate shows that $\Sigma$ is diagonal. Next, take $Q$ to be a permutation matrix that swaps any two coordinates. The same equality forces the corresponding diagonal entries to be equal. Thus $\Sigma=cI_d$, with $c\geq0$ because a covariance matrix is positive semidefinite. $\square$

Common Mistake: Equal Covariances Do Not Guarantee Symmetry

A covariance matrix proportional to the identity is a consequence of spherical symmetry when second moments exist, but it is not sufficient to establish that symmetry. For a concrete counterexample in $\mathbb{R}^2$, let $X$ assign probability $1/4$ to each of $(\sqrt{2},0)$, $(-\sqrt{2},0)$, $(0,\sqrt{2})$, and $(0,-\sqrt{2})$. Its mean is zero, and direct averaging gives $E[X_1^2]=E[X_2^2]=1$ and $E[X_1X_2]=0$. Hence $\operatorname{Cov}(X)=I_2$.

Nevertheless, a rotation by $45$ degrees sends $(\sqrt{2},0)$ to $(1,1)$, which is not in the original support. The rotated distribution therefore differs from the original one. The incorrect reasoning is “the covariance is the same in every direction, so the distribution is spherical.” Covariance records only second-order information; spherical symmetry requires invariance of the entire distribution, or equivalently a characteristic function depending only on vector length.

Key takeaway: Spherical symmetry means invariance under every orthogonal transformation. It forces a radial characteristic function, and a radial characteristic function in turn guarantees spherical symmetry. Equal covariance in every direction is useful evidence, but by itself does not establish the full distributional symmetry.

Check Your Understanding

Try these questions using the definitions and results above.

Why must a spherical vector’s characteristic function take the same value at any two arguments with equal length?
What role does uniqueness of characteristic functions play in proving the converse direction of the theorem?
For a uniform disk of radius $R$, what is the covariance matrix, and which symmetry makes the off-diagonal entry zero?
Why does the mixture in the worked example remain spherical even though it combines two different normal distributions?
Give a reason that covariance $I_2$ does not, by itself, imply spherical symmetry.