Let $O$ be a $d$-dimensional rotation matrix (i.e., it has real entries and $OO^T = O^TO = I$). Let $\mathbf{x}$ be a uniformly random bitstring of length $d$, i.e., $\mathbf{x} \sim U(\{0,1\}^d)$. In other words, $\mathbf{x}$ is a vertex of the Hamming cube, selected uniformly at random. I would like to show that there exists a $C > 0$ such that $$\mathbb{P}\left[\|O\mathbf{x}\|_1 \leq \frac{d}{4}\right] \leq 2^{-Cd}.$$ I am horribly stuck, any ideas on how to approach this problem would be very much appreciated. Below are some of my own attempts. This question is cross-posted at math stack exchange here.

Observation 1: If $O = I$, then the statement holds.

If $O = I$, then $\|O\mathbf{x}\|_1 = \|\mathbf{x}\|_1$ is simply the number of ones in the bitstring. Among the $2^d$ choices for $\mathbf{x}$, the number of choices that satisfies $\|\mathbf{x}\|_1 \leq d/4$ is

$$1 + \binom{d}{1} + \binom{d}{2} + \cdots + \binom{d}{\lfloor d/4\rfloor} \leq 2^{dH(\lfloor d/4\rfloor/d)} \leq 2^{dH(1/4)},$$ hence the probability is upper bounded by $2^{-d(1-H(1/4))}$. Here, $H(\cdot)$ is the binary entropy function, i.e., $H(p) = -p\log_2(p) - (1-p)\log_2(1-p)$.

Observation 2: Numerical experiments support this result. Below is a plot of the probability versus the dimension, where $O$ is selected at random:

Plot of the probability versus the dimension

The blue line is the probability. The orange line is the bound derived in the case where $O = I$.

For comparison, here is the same numerical experiment, but with $O = I$:

enter image description here

Thus, it appears that the introduction of $O$ decreases the probability.

Both plots are obtained by sampling $100000$ $\mathbf{x}$'s at random. The code is here:

import numpy as np
import matplotlib.pyplot as plt
import random
from scipy.stats import ortho_group

H = lambda p : -p * np.log2(p) - (1-p) * np.log2(1-p)
C = 1 - H(1/4)

N = 100000
ds,Ps = [],[]
for d in range(2,40):
    O = ortho_group.rvs(dim = d)
    # O = np.eye(d)
    P = 0
    for _ in range(N):
        x = random.choices(range(2), k = d)
        if np.linalg.norm(O @ x, ord = 1) <= d/4: P += 1/N

fig = plt.figure()
ax = fig.gca()
ax.plot(ds, Ps)
ax.plot(ds, [2**(-C*d) for d in ds])
  • 1
    $\begingroup$ I don't know about your question, but it should not be too difficult to show that, on average over $O$, your question has a positive answer. Your question then boils down to concentration inequalities for the $\ell_1$ norm of a random uniform point on the unit sphere if $\mathbf{R}^d$, which are certainly well-known. This at least explains your numerical evidence. $\endgroup$ Apr 16, 2021 at 16:05
  • $\begingroup$ Thanks for the reply! Indeed, on average over $O$ it should work out. :) But in this case I'm specifically looking for an argument that works for all $O$ (or find an $O$ for which the statement doesn't hold, of course :p) $\endgroup$
    – arriopolis
    Apr 16, 2021 at 16:49
  • $\begingroup$ Is the $\frac 1 4$ in your question important, or are you happy with another constant, say $\frac 1 {10}$? $\endgroup$ Apr 16, 2021 at 16:53
  • $\begingroup$ I would also be fine with changing d/4 to d/10 or d/10000. If this is still too hard, then even though d/log(d) would be somewhat weaker, it would still be interesting. What interests me most is the method behind the proof, not the precise constants. :) $\endgroup$
    – arriopolis
    Apr 16, 2021 at 16:56
  • 2
    $\begingroup$ Maybe one can use the Sauer-Shelah lemma and argue that if the probability is too large, then there is a subcube $\{0,1\}^{cn}$ "inside" the $\ell_1$ unit ball, which is not possible because e.g. of mean width estimates? $\endgroup$ Apr 16, 2021 at 17:26

3 Answers 3


Here is an attempt to the problem for a worst-case $O$, with worse constants. So fix $O$, letting $o_i$ denote its $i$th row, and take $X$ random in $\{0,1\}^d$.

  1. We claim that $E |\langle o_i, X\rangle| \ge cst$. To see this, write $$\langle o_i, X\rangle = \langle o_i, \frac{{\bf 1}}{2}\rangle + \langle o_i, (X - \frac{{\bf 1}}{2})\rangle$$ and assume WLOG the first term on the RHS is non-negative. The second term on the RHS is a weighted sum of Rademacher random variables, and so with probability at least $\frac{1}{20}$ it is above its standard deviation, which is $\Omega(1)$ (see for example this paper of Oleszkiewicz)

  2. Adding over all $i$'s, the result holds in expectation: $E \|OX\|_1 \ge cst\cdot d$. But since the function $x \mapsto \|Ox\|_1$ is $\sqrt{d}$-Lipschitz wrt $\ell_2$ (and convex), we should be able to use concentration to say that the probability that we get below this mean minus $\frac{cst \cdot d}{2}$ is at most $e^{-cst' \cdot d}$ (see for example Corollary 4.23 of van Handel's notes). This gives the result.

  • $\begingroup$ Can you expand a bit the second point? I do not see directly how the corollary in Handel's notes gives the result. $\endgroup$ Apr 16, 2021 at 19:43
  • 2
    $\begingroup$ BTW, another way to see that $E\|OX\|_1 \geq cd$ is by symmetrization and Khintchine's inequality. $\endgroup$ Apr 16, 2021 at 19:44
  • 1
    $\begingroup$ @MikaeldelaSalle: I may be goofing off, but it seems by letting $f(x) := \|Ox\|_1$, van Handel's corollary should give that $f(X)$ is $d$-subgaussian (it is stated in terms of gradients but we can replace by subgradients, and the subgradients of $f$ should have $\ell_2$ norm at most $\sqrt{d}$). But $d$-subgaussian RVs have concentration $Pr(f(X) \le E f(X) - \epsilon d) \le e^{-\frac{(\epsilon d)^2}{2 d}}$, and we should get the result. $\endgroup$
    – Marco
    Apr 16, 2021 at 19:57
  • $\begingroup$ Thanks, I got it. $\endgroup$ Apr 16, 2021 at 20:11
  • $\begingroup$ Thanks a lot for the response! :) This certainly looks like an interesting approach. I will have a more careful look at the details after the weekend, and then award the bounty accordingly. $\endgroup$
    – arriopolis
    Apr 17, 2021 at 11:24

We prove the weaker bound $$ \mathbf{P} \left[ \|O \mathbb{x}\|_1 \leq \frac{cd}{\sqrt{\log d}} \right] \leq 2^{-Cd} $$ for some constants $C, c$.

Define the Gaussian mean width of a compact subset $A \subset \mathbf{R}^d$ as $$ w(A) = \mathbf{E} \sup_{x \in A} \langle G,x \rangle $$ where $G$ is a standard Gaussian vector in $\mathbf{R}^d$. We use the following properties

  1. If $\pi$ is an orthogonal projection, then $w(\pi(A))\leq w(A)$.
  2. If $A = \{0,1\}^k \subset \{0,1\}^n$, then $w(A)=k/\sqrt{2\pi}$.
  3. If $A=B_1^d$ (the unit $\ell_1$ ball), then $w(A) \leq \sqrt{2\log d}$.
  4. $w$ is rotation invariant

Let $A$ be the set of $x \in \{0,1\}^d$ such that $\|Ox\|_1 \leq cd/\sqrt{\log d}$. We have $w(A) \leq w(cd/\sqrt{\log d} \cdot B_1^d) \leq c\sqrt{2}d$.

If $\mathrm{card}(A) \geq 2^{dH(1/4)}$, then the Sauer--Shelah lemma implies that there is a coordinate projection $\pi$ of rank $k=d/4$ such that $\pi(A)$ identifies with $\{0,1\}^k$. Therefore, $w(A) \geq w(\pi(A))=d/4\sqrt{2\pi}$. If we choose $c=1/8\pi$, combining both estimates gives the bound $\mathrm{card}(A) < 2^{dH(1/4)}$, as needed.

  • $\begingroup$ Thanks a lot for the response! :) This certainly looks like an interesting approach. I will have a more careful look at the details after the weekend, and then award the bounty accordingly. $\endgroup$
    – arriopolis
    Apr 17, 2021 at 11:24
  • $\begingroup$ Since Marco's answer provided a better scaling, I decided to award the bounty to them. Still, thanks for the solution you proposed! :) $\endgroup$
    – arriopolis
    Apr 20, 2021 at 0:16

Adding more detail to Mikael's point, the result seems to hold on average over $O$ because of the following:

  1. Using a Chernoff bound, we can see that the probability that for any constant $\epsilon$, with probability at least $1- e^{-C d}$ the random vector $x$ has at least $\frac{(1-\epsilon)d}{2}$ 1's.

  2. Consider a fixed vector $x \in \{0,1\}^d$. For a uniformly distributed $O$, $O x$ is uniformly distributed on the sphere of size $\sqrt{\|x\|_1}$. This has distribution essentially the same as the vector $\sqrt{\|x\|_1} (G_1,\ldots,G_d)$, where the $G_i$'s are independent Gaussians $N(0,\frac{1}{d})$ (in fact $\sqrt{\|x\|_1}$ has the same distribution as $\sqrt{\|x\|_1} (G_1,\ldots,G_d)/\|G\|_2$, or alternatively one can bypass using Gaussians and work with concentration on the sphere, as Mikael suggested).

Then in expectation (over $O$) $$E \|Ox\|_1 \approx \sqrt{\|x\|_1} \cdot \|G\|_1 = \sqrt{\|x\|_1} \cdot \sqrt{d} \sqrt{\frac{2}{\pi}},$$ where the last inequality uses the fact that the expected value of a folded Gaussian of standard deviation $\sigma$ is $\sigma \sqrt{\frac{2}{\pi}}$.

Moreover, the function $\|\cdot\|_1$ is $\sqrt{d}$ Lipschitz wrt $\ell_2$ we can use concentration of such functions over Gaussian space to get that with probability at least $1 - e^{- Cd}$ we have $$\|Ox\|_1 \ge (1-\epsilon) \sqrt{\|x\|_1} \cdot \sqrt{d} \sqrt{\frac{2}{\pi}},$$ see for example Example 4.2 in van Handel excellent notes or inequality (1.6) of Ledoux-Talagrand "Probability in Banach Spaces".

  1. Taking a union bound over the two steps above seems to give the desired result.
  • $\begingroup$ Thanks a lot for your idea! :) Do I understand correctly that this only works whenever $O$ is chosen randomly too? Ideally, I would like to find an argument in which $O$ is fixed (or find a single $O$ for which the claim doesn't hold, naturally :p). $\endgroup$
    – arriopolis
    Apr 16, 2021 at 16:45
  • 1
    $\begingroup$ Sorry, I should have read the question more carefully :). I'll edit the answer to make it clear that it does not address the actual question. $\endgroup$
    – Marco
    Apr 16, 2021 at 16:52

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