= Solution
For an <orthogonal matrix> $A$, invariance in distribution gives
$$
\varphi_X(A^Tu)
=\varphi_{AX}(u)
=\varphi_X(u).
$$
The orthogonal group is transitive on each Euclidean sphere, so $\varphi_X(u)$ depends only on $\lVert u\rVert^2$. Define
$$
f(r)=\varphi_X(\sqrt r,e_1),
\qquad r\geq0.
$$
The <continuity of a characteristic function> makes $f$ continuous. Taking $A=-I$ also gives $X\stackrel d=-X$, and hence
$$
\varphi_X(u)=\varphi_X(-u)=\overline{\varphi_X(u)},
$$
so $f$ is real-valued. Thus
$$
\boxed{\varphi_X(u)=f(u_1^2+\cdots+u_d^2)},
$$
as for every <orthogonally invariant random vector>.
Back to article page
