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dohmatob
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Let $g \sim N(0, \frac 1 d I)$$g \sim N(0, (1/d)I_d)$ be independent of $x$. Then $g_1 \overset{d}{=} \|g\| x_1$$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| \le \sqrt{\mathbb E|x_1|^2} = \sqrt{1/d}$$\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.

Let $g \sim N(0, \frac 1 d I)$ be independent of $x$. Then $g_1 \overset{d}{=} \|g\| x_1$, 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 $\mathbb E|x_1| \le \sqrt{\mathbb E|x_1|^2} = \sqrt{1/d}$.

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.

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dohmatob
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Let $g \sim N(0, \frac 1 d I)$ be independent of $x$. Then $g_1 \overset{d}{=} \|g\| x_1$, 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 $$\mathbb E|x_1 - \|g\|x_1| = \mathbb E|1 - \|g\|| \mathbb E|x_1| = O(d^{-1}).$$$$\mathbb E|x_1 - \|g\|x_1| = (\mathbb E|1 - \|g\||) \mathbb E|x_1| = O(d^{-1}), $$ where we have used the fact that $\mathbb E|x_1| \le \sqrt{\mathbb E|x_1|^2} = \sqrt{1/d}$.

Let $g \sim N(0, \frac 1 d I)$ be independent of $x$. Then $g_1 \overset{d}{=} \|g\| x_1$, 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 $$\mathbb E|x_1 - \|g\|x_1| = \mathbb E|1 - \|g\|| \mathbb E|x_1| = O(d^{-1}).$$

Let $g \sim N(0, \frac 1 d I)$ be independent of $x$. Then $g_1 \overset{d}{=} \|g\| x_1$, 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 $\mathbb E|x_1| \le \sqrt{\mathbb E|x_1|^2} = \sqrt{1/d}$.

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Alf
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Let $g \sim N(0, \frac 1 d I)$ be independent of $x$. Then $g_1 \overset{d}{=} \|g\| x_1$, 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 $$\mathbb E|x_1 - \|g\|x_1| = \mathbb E|1 - \|g\|| \mathbb E|x_1| = O(d^{-1}).$$