A high-dimensional CLT in $\mathcal {W}_2$ distance with near optimal convergence rate

Abstract

Let $X_1,\ldots ,X_n$ be i.i.d. random vectors in $\mathbb {R}^d$ with $\Vert X_1\Vert \le \beta$. Then, we show that

$$\begin{aligned} \frac{1}{\sqrt{n}}\left( X_1 + \cdots + X_n\right) \end{aligned}$$

converges to a Gaussian in quadratic transportation (also known as “Kantorovich” or “Wasserstein”) distance at a rate of $O \left( \frac{\sqrt{d} \beta \log n}{\sqrt{n}} \right)$, improving a result of Valiant and Valiant. The main feature of our theorem is that the rate of convergence is within $\log n$ of optimal for $n, d \rightarrow \infty$.

Appendix

1.1 Proof of Proposition 1.2

Proof

Let $\ell _n = \frac{\beta }{\sqrt{n}}$, and consider the lattice $L = \ell _n \mathbb {Z}^d$. For any $x \in \mathbb {R}^d$, let $d_L(x)$ denote the minimum Euclidean distance from x to L. Note that $S_n$ takes values in L. Thus, letting $\rho$ denote the density of Z, we have v

$$\begin{aligned} \mathcal {W}_2(S_n, Z) \ge \int \rho (x) d_L(x)~{ dx}. \end{aligned}$$

To estimate the right hand side, for any $y \in L$, let $Q_n(y)$ denote the cube of side length $\ell _n$ centered at y (which is also the set of points in $\mathbb {R}^d$ closer to y than to any other point in L). We find that

$$\begin{aligned} \frac{1}{{{\mathrm{Vol}}}Q_n(y)} \int _{Q_n(y)} d_L(x) ~{ dx}&= \frac{\ell _n}{2^d} \int _{[-1,1]^d} \Vert x\Vert ~{ dx} = \frac{\ell _n}{2^d} \int _{[-1,1]^d} \sqrt{x_1^2 + \cdots + x_d^2} ~{ dx} \nonumber \\&\ge \frac{\ell _n}{2^d} \int _{[-1,1]^d} \frac{1}{\sqrt{d}}\left( |x_1| + \cdots + |x_d|\right) ~{ dx} = \frac{1}{2} \ell _n \sqrt{d}. \end{aligned}$$

(9)

Next, let M be large enough so that,

$$\begin{aligned} \int _{[-M,M]^d} \rho (x) ~{ dx} \ge \frac{1}{2}, \end{aligned}$$

and let

$$\begin{aligned} r_n = \inf _{\begin{array}{c} x, y \in [-2M, 2M]^d \\ \Vert x - y\Vert \le \sqrt{d} \ell _n \end{array}} \frac{\rho (x)}{\rho (y)}. \end{aligned}$$

(10)

Note that since $\rho$ is positive and continuous, we have $\lim _{n \rightarrow \infty } r_n = 1$.

Assume now that n is sufficiently large so that $\ell _n < M$. Combining (10) with (9), we have for each $y \in L \cap [-M,M]^d$ that

$$\begin{aligned} \int _{Q_n(y)} \rho (x) d_L(x) ~{ dx}&\ge \frac{r_n}{{{\mathrm{Vol}}}Q_n(y)} \int _{Q_n(y)} \rho (x) ~{ dx} \cdot \int _{Q_n(y)} d_L(x) ~{ dx} \\&\ge \frac{r_n \ell _n \sqrt{d}}{2} \int _{Q_n(y)} \rho (x) ~{ dx}. \end{aligned}$$

Summing over all such y yields

$$\begin{aligned} \mathcal {W}_2(S_n, Z)&\ge \int \rho (x) d_L(x) ~{ dx} \ge \int _{[-2M,2M]^d} \rho (x) d_L(x) ~{ dx} \\&\ge \sum _{y \in L \cap [-M,M]^d} \int _{Q_n(y)} \rho (x) d_L(x) ~{ dx} \\&\ge \frac{r_n \ell _n \sqrt{d}}{2} \int _{[-M,M]^d} \rho (x) ~{ dx} \ge \frac{r_n \beta \sqrt{d}}{4 \sqrt{n}}. \end{aligned}$$

Multiplying both sides by $\sqrt{n}$ and taking limits gives the result. $\square$

1.2 Proof of Proposition 1.4

Proof

We prove the result with $C = 5$. Let $A \subset \mathbb {R}^d$ be a given convex set. For a parameter $\epsilon$ to be specified later, define

$$\begin{aligned} A^\epsilon= & {} \left\{ x \in \mathbb {R}^d \mid \sup _{a \in A} \Vert x - a\Vert \le \epsilon \right\} \\ A_\epsilon= & {} \left\{ x \in \mathbb {R}^d \mid \inf _{a \in \mathbb {R}^d \setminus A} \Vert x - a\Vert \ge \epsilon \right\} . \end{aligned}$$

Ball [1] showed a $4d^{1/4}$ upper bound⁶ for the Gaussian surface area of any convex set in $\mathbb {R}^d$. Hence,⁷

$$\begin{aligned} \mathbf {P}\left( Z \in A^\epsilon \setminus A\right) \le 4 \epsilon d^{1/4},\quad \text {and}\quad \mathbf {P}\left( Z \in A \setminus A_\epsilon \right) \le 4 \epsilon d^{1/4}. \end{aligned}$$

We may regard T as being coupled to Z so that $\mathbf {E}\Vert T - Z\Vert ^2 = \mathcal {W}_2(T, Z)^2$. Then,

$$\begin{aligned} \mathbf {P}(T \in A)&\le \mathbf {P}(\Vert T - Z\Vert \le \epsilon , \; T \in A) + \mathbf {P}(\Vert T - Z\Vert > \epsilon ) \\&\le \mathbf {P}(Z \in A^\epsilon ) + \epsilon ^{-2} \mathcal {W}_2(T, Z)^2 \\&\le \mathbf {P}(Z \in A) + 4 \epsilon d^{1/4} + \epsilon ^{-2} \mathcal {W}_2(T, Z)^2 \end{aligned}$$

Similarly,

$$\begin{aligned} \mathbf {P}(Z \in A)&\le \mathbf {P}(Z \in A_\epsilon ) + 4 \epsilon d^{1/4} \\&\le \mathbf {P}(\Vert T - Z\Vert \le \epsilon , \; Z \in A_\epsilon ) + \mathbf {P}(\Vert T - Z\Vert > \epsilon ) + 4 \epsilon d^{1/4} \\&\le \mathbf {P}(T \in A) + \epsilon ^{-2}\mathcal {W}_2(T, Z)^2 + 4 \epsilon d^{1/4}. \end{aligned}$$

Thus,

$$\begin{aligned} |\mathbf {P}(T \in A) - \mathbf {P}(Z \in A)| \le \epsilon ^{-2}\mathcal {W}_2(T, Z)^2 + 4 \epsilon d^{1/4}, \end{aligned}$$

and taking $\epsilon = d^{-1/12} \mathcal {W}_2(T, Z)^{2/3}$ gives the result. $\square$

1.3 Proof of Lemma 3.3

Proof

Let $A'$ and $B'$ be independent copies of A and B. Then,

$$\begin{aligned} \mathbf {E}\Big ( f(A, B) + f(A', B') - f(A, B') - f(A', B) \Big )^2 \ge 0. \end{aligned}$$

Expanding yields

$$\begin{aligned} 4 \mathbf {E}f(A, B)^2 + 4 (\mathbf {E}f(A, B))^2&= 4 \mathbf {E}f(A, B)^2 + 2 \mathbf {E}f(A, B) f(A', B') \\&\quad +\,2 \mathbf {E}f(A, B') f(A', B) \\&\ge 2 \mathbf {E}f(A, B)f(A, B') + 2 \mathbf {E}f(A, B) f(A', B) \\&\quad +\,2 \mathbf {E}f(A', B') f(A, B') + 2 \mathbf {E}f(A', B') f(A', B) \\&= 2 \mathbf {E}f_B(A)^2 + 2 \mathbf {E}f_A(B)^2 \\&\quad +\,2 \mathbf {E}f_A(B)^2 + 2 \mathbf {E}f_B(A)^2 \\&= 4 \mathbf {E}f_B(A)^2 + 4 \mathbf {E}f_A(B)^2, \end{aligned}$$

as desired. $\square$

1.4 Proof of Equation (2)

Proof

We proceed by induction on the dimension d, retracing the argument of [23], section 3. The base case $d = 1$ is immediate from Theorem 3.1.

Assume now that the inequality holds in $d - 1$ dimensions. For the inductive step, we can follow the same argument used to prove Theorem 3.1 (see [23], section 3). The argument proceeds by first comparing Y to another $\mathbb {R}^d$-valued random variable ${\hat{Y}}$ sharing the first $d - 1$ coordinates of Y, but whose last coordinate is independently drawn from $\mathcal {N}(0, \sigma _d)$.

Fix a $(d - 1)$-dimensional vector ${\hat{x}}$, and let $T_{{\hat{x}}}$ denote a random variable distributed as the last coordinate of Y conditioned on the first $d - 1$ coordinates being equal to ${\hat{x}}$. Let ${\hat{\rho }}({\hat{x}}) = \int _{-\infty }^\infty \rho ({\hat{x}}, t) { dt}$. Then, the density of $T_{{\hat{x}}}$ at t is given by

$$\begin{aligned} \frac{f({\hat{x}}, t) \cdot \rho ({\hat{x}}, t)}{f_{(d)}({\hat{x}}, 0) \cdot {\hat{\rho }}({\hat{x}})}. \end{aligned}$$

Noting that $\frac{\rho ({\hat{x}}, t)}{{\hat{\rho }}({\hat{x}})}$ is the density of $\mathcal {N}(0, \sigma _d)$ at t, the one-dimensional case of Theorem 3.1 implies

$$\begin{aligned} \mathcal {W}_2(T_{{\hat{x}}}, \mathcal {N}(0, \sigma _d))^2 \le 2 \sigma _d^2 \int _{-\infty }^\infty \frac{f({\hat{x}}, t)}{f_{(d)}({\hat{x}}, t)} \log \left( \frac{f({\hat{x}}, t)}{f_{(d)}({\hat{x}}, t)} \right) \frac{\rho ({\hat{x}}, t)}{{\hat{\rho }}({\hat{x}})} ~{ dt}. \end{aligned}$$

(11)

Since $T_{{\hat{x}}}$ and $\mathcal {N}(0, \sigma _d)$ have the same distributions as Y and ${\hat{Y}}$ conditioned on ${\hat{x}}$, we may integrate (11) over ${\hat{x}}$ to obtain

$$\begin{aligned} \mathcal {W}_2(Y, {\hat{Y}})^2&\le 2 \int _{\mathbb {R}^{d - 1}} \mathcal {W}_2(T_{{\hat{x}}}, \mathcal {N}(0, \sigma _d))^2 \cdot f_{(d)}({\hat{x}}, 0) {\hat{\rho }}({\hat{x}}) ~d{\hat{x}} \\&\le 2 \sigma _d^2 \int _{\mathbb {R}^{d - 1}} \int _{-\infty }^\infty f({\hat{x}}, t) \log \left( \frac{f({\hat{x}}, t)}{f_{(d)}({\hat{x}}, t)} \right) \rho ({\hat{x}}, t) ~{ dt} ~d{\hat{x}}. \\&= 2 \sigma _d^2 \cdot \mathbf {E}\left( f(Z) \log \frac{f(Z)}{f_{(d)}(Z)} \right) \\&= 2 \sigma _d^2 \cdot \bigg ( \mathbf {E}\left( f(Z) \log f(Z) \right) - \mathbf {E}\left( f_{(d)}(Z) \log f_{(d)}(Z) \right) \bigg ) \end{aligned}$$

Next, define $Y_{(d)}$ and $Z_{(d)}$ to be the projections onto the first $d - 1$ coordinates of Y and Z, respectively. Note that the coupling of Y to ${\hat{Y}}$ changes only d-th coordinate. Furthermore, the d-th coordinates of ${\hat{Y}}$ and Z are both distributed as $\mathcal {N}(0, \sigma _d)$ independent of the first $d - 1$ coordinates. Thus, a coupling of $Y_{(d)}$ to $Z_{(d)}$ induces a coupling of ${\hat{Y}}$ to Z in which the last coordinate does not change. Consequently,

$$\begin{aligned} \mathcal {W}_2(Y, Z)^2\le & {} 2 \sigma _d^2 \cdot \bigg ( \mathbf {E}\left( f(Z) \log f(Z) \right) - \mathbf {E}\left( f_{(d)}(Z) \log f_{(d)}(Z) \right) \bigg )\nonumber \\&+ \mathcal {W}_2(Y_{(d)}, Z_{(d)})^2. \end{aligned}$$

(12)

Now, recall that the density of $Y_{(d)}$ at a point ${\hat{x}} \in \mathbb {R}^{d - 1}$ is $f_{(d)}({\hat{x}}, 0) \cdot {\hat{\rho }}({\hat{x}})$, and so applying the inductive hypothesis to $\mathcal {W}_2(Y_{(d)}, Z_{(d)})^2$ yields

$$\begin{aligned} \mathcal {W}_2(Y_{(d)}, Z_{(d)})^2\le & {} 2 \sum _{k = 1}^{d - 1} \sigma _k^2 \cdot \mathbf {E}\left( f_{[k]}(Z_{(d)}) \log f_{[k]}(Z_{(d)}) - f_{[k - 1]}(Z_{(d)}) \log f_{[k - 1]}(Z_{(d)}) \right) \\= & {} 2 \sum _{k = 1}^{d - 1} \sigma _k^2 \cdot \mathbf {E}\left( f_{[k]}(Z) \log f_{[k]}(Z) - f_{[k - 1]}(Z) \log f_{[k - 1]}(Z) \right) . \end{aligned}$$

Substituting into (12), we obtain

$$\begin{aligned} \mathcal {W}_2(Y, Z)^2 \le 2 \sum _{k = 1}^{d - 1} \sigma _k^2 \cdot \mathbf {E}\left( f_{[k]}(Z) \log f_{[k]}(Z) - f_{[k - 1]}(Z) \log f_{[k - 1]}(Z) \right) , \end{aligned}$$

completing the induction. $\square$

1.5 Proof of Lemma 4.2

Proof

Let $C_k = (2 \pi )^{-\frac{k}{2}}$. We have

$\square$