I guess that the result is derived from integrating over the sphere, or by some symmetry argument. I would be happy to understand both.
2026-03-31 20:05:43.1774987543
Why $\Bbb E_x [x_i^2] = 1/d$ when $x \sim Unif(\Bbb S^{d-1})$?
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Note that, by linearity of expectation and the definition of $S^{d-1}$, $$ \sum \mathbb{E}(x_i^2) = \mathbb{E}(\sum x_i^2) = \mathbb{E}(|x|) = 1. $$ Since there are $d$ many summands on the left hand side and symmetry demands $\mathbb{E}(x_i^2) = \mathbb{E}(x_j^2)$ for all $1 \leq j \leq d$, you have $\mathbb{E}(x_i^2) = \frac{1}{d}$. That's the symmetry idea, and probably the path of least resistance in proving this.