Let $x=(x\_1,\ldots,x\_d)$ be uniformly distributed on the unit-sphere in $\mathbb R^d$.



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> **Question.**
> What is a good upper-bound for Wasserstein distance between $N(0,1/d)$ and the marginal distribution of $x\_1$ ?
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Wasserstein distance between $N(0,1/d)$ and the marginal distribution of $x_1$ when $x=(x_1,\ldots,x_d)$ is uniform on the unit-sphere in $R^d$
Let $g \sim N(0, (1/d)I\_d)$ be independent of $x$. Then $g\_1 \overset{\mathcal L}{=} \|g\| x\_1$, so $(x\_1, \|g\|x\_1)$ is a coupling between the marginal distribution of $x\_1$ and $N(0, 1/d)$.


The norm $\|g\|$ is sharply concentrated around $1$, with fluctuations of order $1/\sqrt{d}$, so the Wasserstein distance is bounded above by
$$\mathbb E|x\_1 - \|g\|x\_1| = (\mathbb E|1 - \|g\||) \mathbb E|x\_1| = O(d^{-1}),
$$
where we have used the fact that $\mathbb E|x\_1|^2 = (1/d)\mathbb E\|x\|^2 = 1/d$ by isotropy of the distribution of $x$, so that $\mathbb E|x\_1| \le \sqrt{\mathbb E|x\_1|^2} = 1/\sqrt{d}$ thanks to Jensen's inequality.


