standard gaussian random vectors are orthogonally invariant

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Theorem

Suppose $x \sim N (0, I_{d})$ is a standard gaussian random vector. For any $a \in S^{d - 1}$ , we have

Law (⟨ a, x ⟩) = N (0, 1)

In particular, this does not depend on $a$ and $x$ is orthogonally invariant

Proof

For $Q \in O (d)$ , we have (from gaussian random vectors are closed under linear transformations) that

Law (Q x) = N (0, Q^{T} I_{d} Q) = N (0, I_{d}) = Law (x)

Now, consider a special case where $a$ is the first row of $Q$ (and we can find an orthogonal resulting $Q$ via Gram-Schmidt). Then we see that the law is independent of $a$ as desired.

File	Last Modified
normalized standard gaussian random vectors have the orthogonally invariant distribution on the unit sphere	2025-09-09
Random Matrix Lecture 01	2025-09-11

standard gaussian random vectors are orthogonally invariant

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