# 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 $$x=(x_1,\ldots,x_d)$$ be uniformly distributed on the unit-sphere in $$\mathbb R^d$$.

Question. What is a good upper-bound for Wasserstein distance between $$N(0,1/d)$$ and the marginal distribution of $$x_1$$ ?

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.

• Thanks very much. Commented Aug 10, 2021 at 20:33
• I'd be interested in what you think of my general method below (which gives a slightly weaker bound). Commented Aug 12, 2021 at 15:17

## Slightly weaker bound via a more robust method

Below I present an alternative method with slightly worse upper-bound than in the accepted answer, namely $$d^{-1+o(1)}$$ instead of $$d^{-1}$$.

The advantage of the new method is that it can be applied to (almost all) one-dimensional projections of arbitrary isotropic log-concave distributions (thanks to Klartag's CLT for convex bodies, namely Theorem 1.3 of this paper).

Disclaimer. It's possible I've made a mistake some. Thanks in advance for catching.

Let $$F$$ be the marginal cdf of $$\sqrt{d}x_1$$ and let $$F^c := 1 - F$$ be its survival function. Let $$\Phi$$ be the standard gaudssian cdf.

We will make use of the following fact (see ref https://mathoverflow.net/a/315232/78539)

Fact 1 (Approximation). $$Ko(x_1,N(0,1)) := \sup_{t \in \mathbb R}|F(t) - \Phi(t)| \le C/d$$, for an absolute constant $$C>0$$ which doesn't depend on $$d$$.

We shall also need the following fact (which follows concentration of $$\mathcal O(1)$$-Lipschitz transformations of log-concave random vectors)

Fact 2 (Concentration). There exists an absolute constant $$b>0$$ such that $$\max(F(-t),\Phi(-t),F^c(t),\Phi^c(t)) \le e^{-bt^2}$$ for sufficiently large $$t>0$$.

Now, for any $$T > 0$$, one computes $$W_1(\sqrt{d}x_1,N(0,1)) = \int_{-\infty}^\infty |F(t)-\Phi(t)|dt = A_1(T) + A_2(T) + A_3(T),$$ where the first is a a classical result (e.g see Proposition 2.17 of Santambrogio's OTAM), and the $$A_k(T)$$'s are defined by $$\begin{split} A_1(T) &:= \int_{-\infty}^{-T}|F(t)-\Phi(t)|dt,\\ A_2(T) &:= \int_{-T}^T|F(t)-\Phi(t)|dt,\\ A_3(T) &:= \int_{T}^\infty|F(t)-\Phi(t)|dt=\int_{T}^\infty|F^c(t)-\Phi^c(t)|dt. \end{split}$$

Thanks to Fact 1, we know that $$A_2(T) = \mathcal O(T/d)$$. On the other hand, thanks to Fact 2, we have for sufficiently large $$T>0$$, $$A_1(T) \le \int_{-\infty}^{-T} \max(F(t),\Phi(t))dt \le \int_{-\infty}^{-T} e^{bt^2}dt = \mathcal O(e^{-bT^2}).$$

By a symmetric argument, we also have $$A_3(T) = \mathcal O(e^{-bT^2})$$. Taking $$T=\sqrt{(\log d)/b}$$ then gives $$A_2(T) = \mathcal O(\sqrt{\log d}/d)$$ and $$A_1(T),A_3(T) = \mathcal O(1/d)$$. Thus $$W_1(x_1,N(0,1/d)) = W_1(\sqrt{d} x_1,N(0,1)) = \mathcal O(\frac{\sqrt{\log d}}{d})=\mathcal O(d^{-1+o(1)}).$$

## Extension to one-dimensional projections of isotropic log-concave distributions

Let $$x$$ be a random vector in $$\mathbb R^d$$ with an arbitrary isotropic log-concave density. Recall that by isotropy, we mean

• $$\mathbb E [x] = 0$$, and
• $$\mathbb E |x^\top u|^2=1$$ for any unit-vector $$u \in \mathbb R^d$$.

Now, for any unit-vector $$u \in \mathbb R^d$$, let $$x_u := x^\top u$$ be the projection of $$x$$ in the direction of $$u$$, and let $$F_u$$ be its cdf. Finally, let $$\sigma_d$$ be the uniform distribution on the unit-sphere in $$\mathbb R^d$$. We will prove the following

Theorem. For a proportion $$1-e^{-\mathcal O(d^{0.99})}$$ of unit-vectors $$u$$ in $$\mathbb R^d$$ (w.r.t the uniform distribution on the unit-sphere in $$\mathbb R^d$$), it holds that $$W_1(x_u,N(0,1)) = \mathcal O((\log d)^{-1/2+o(1)})=o(1).$$

Thanks to Theorem 1.3 of Klartag's paper (referenced in the preamble), we know that for $$\varepsilon_d = \left(\dfrac{\log \log d}{\log d}\right)^{1/2}=o(1)$$,

Fact 1' (Approximation). There exists a measurable subset $$\mathcal U$$ of unit-vectors with $$\sigma_d(\mathcal U) \ge 1 - e^{-\mathcal O(d^{0.99})}$$ such that $$Ko(x_u,N(0,1)) \le \mbox{TV}(x_u,N(0,1)) = \mathcal O(\varepsilon_d)$$ for all $$u \in \mathcal U$$.

Just as in Fact 2, we have the following property which is a direct consequence of the concentration of $$\mathcal O(1)$$-Lipschitz transformations of log-concave random vectors,

Fact 2' (Concentration). There exists an absolute constant $$b>0$$ such that $$\max(F_u(-t),\Phi(-t),F_u^c(t),\Phi^c(t)) \le e^{-bt^2}$$ for all $$u \in \mathcal U$$ and for sufficiently large $$t>0$$.

Emulating the proof of the spherical case, but using the cutoff $$T=\sqrt{\log (1/\varepsilon_d)/b}$$ instead, we obtain $$\sup_{u \in \mathcal U}W_1(x_u,N(0,1))) = \mathcal O(\varepsilon_d\sqrt{\log(1/\varepsilon_d)}) = \mathcal O((\log d)^{-1/2+o(1)}) = o(1).$$ Noting that $$\sigma_d(\mathcal U)\ge 1 - e^{-\mathcal O(d^{0.99})}$$, the proof is complete. $$\quad\quad\quad\quad\quad\quad\Box$$