Tightness of the maximum likelihood semidefinite relaxation for angular synchronization

Abstract

Maximum likelihood estimation problems are, in general, intractable optimization problems. As a result, it is common to approximate the maximum likelihood estimator (MLE) using convex relaxations. In some cases, the relaxation is tight: it recovers the true MLE. Most tightness proofs only apply to situations where the MLE exactly recovers a planted solution (known to the analyst). It is then sufficient to establish that the optimality conditions hold at the planted signal. In this paper, we study an estimation problem (angular synchronization) for which the MLE is not a simple function of the planted solution, yet for which the convex relaxation is tight. To establish tightness in this context, the proof is less direct because the point at which to verify optimality conditions is not known explicitly. Angular synchronization consists in estimating a collection of n phases, given noisy measurements of the pairwise relative phases. The MLE for angular synchronization is the solution of a (hard) non-bipartite Grothendieck problem over the complex numbers. We consider a stochastic model for the data: a planted signal (that is, a ground truth set of phases) is corrupted with non-adversarial random noise. Even though the MLE does not coincide with the planted signal, we show that the classical semidefinite relaxation for it is tight, with high probability. This holds even for high levels of noise.

Appendix: Wigner matrices are discordant

This appendix is a proof for Proposition 3.3, namely, that for arbitrary $z\in \mathbb {C}^n$ such that $|z_1| = \cdots = |z_n| = 1$, complex Wigner matrices are z-discordant (Definition 3.1) with high probability.

A matrix W is z-discordant if and only if ${\text {diag}}(z)^* W {\text {diag}}(z)$ is $\mathbbm {1}$-discordant. Since ${\text {diag}}(z)^* W {\text {diag}}(z)$ has the same distribution as W (owing to complex normal random variables having uniformly random phase), we may without loss of generality assume $z = \mathbbm {1}$ in the remainder of the proof.

• 1. $\Pr \left\{ \left\| W\right\| _{\mathrm {op}} > 3 n^{1/2} \right\} \le e^{-n/2}$.

  Although tail bounds for the real version of this are well-known (see for example [15, 56]) and they mostly hold verbatim in the complex case, for the sake of completeness we include a classical argument, based on Slepian’s comparison theorem and Gaussian concentration, for a tail bound in the complex valued case.

  We will bound the largest eigenvalue of W. It is clear that a simple union bound argument will allow us to bound also the smallest, and thus bound the largest in magnitude. Let $\lambda _+ = \max _{v\in \mathbb {C}^n: \Vert v\Vert =1}v^*W v$ denote the largest eigenvalue of W. For any unit-norm $u,v\in \mathbb {C}^n$, the real valued Gaussian process $X_v = v^*Wv$ satisfies:

  $$\begin{aligned} \mathbb {E}\left( X_v - X_u\right) ^2&= \mathbb {E}\left( \sum _{i<j}W_{ij}\left( \overline{v_i}v_j -\overline{u_i}u_j\right) + W_{ji}\left( \overline{v_j}v_i -\overline{u_j}u_i\right) \right) ^2\\&= \sum _{i<j}\mathbb {E}\left[ W_{ij}\left( \overline{v_i}v_j -\overline{u_i}u_j\right) + W_{ji}\left( \overline{v_j}v_i -\overline{u_j}u_i\right) \right] ^2. \end{aligned}$$

  The variable $W_{ij}$ has uniformly random phase, hence so does $W_{ij}^2$, so that $\mathbb {E}W_{ij}^2 = 0$. As a result,

  $$\begin{aligned} \mathbb {E}\left( X_v - X_u\right) ^2&= \sum _{i<j}2\mathbb {E}\left| W_{ij}\right| ^2\left| \overline{v_i}v_j -\overline{u_i}u_j\right| ^2 \\&= 2\sum _{i<j}\left| \overline{v_i}v_j -\overline{u_i}u_j\right| ^2 \\&\le \sum _{i,j}\left| \overline{v_i}v_j -\overline{u_i}u_j\right| ^2. \end{aligned}$$

  Note that, since $\Vert u\Vert =\Vert v\Vert =1$,

  $$\begin{aligned} \sum _{i,j}\left| \overline{v_i}v_j -\overline{u_i}u_j\right| ^2&= \sum _{ij}\left[ |v_i|^2|v_j|^2 + |u_i|^2|u_j|^2 - \overline{v_i}v_ju_i\overline{u_j} - v_i\overline{v_j}\overline{u_i}u_j\right] \\&= 2- 2\left| v^*u\right| ^2 \\&\le 2\left( 2 - 2\left| v^*u\right| \right) \\&\le 4\left( 1 - \mathfrak {R}\left[ v^*u \right] \right) \\&= 2\Vert v-u \Vert ^2. \end{aligned}$$

  This means that we can use Slepian’s comparison theorem (see for example [39, Cor. 3.12]) to get

  $$\begin{aligned} \mathbb {E}\lambda _+ \le \sqrt{2}\mathbb {E}\max _{\tilde{v}\in \mathbb {R}^{2n}:\Vert \tilde{v}\Vert =1} \tilde{v}^T g \le 2\sqrt{n}, \end{aligned}$$

  (5.1)

  where g is a standard Gaussian vector in $\mathbb {R}^{2n}$ and $\tilde{v}$ is a vector in $\mathbb {R}^{2n}$ obtained from $v\in \mathbb {C}^{n}$ by stacking its real and imaginary parts. Since $\left| \Vert W_1\Vert - \Vert W_2\Vert \right| \le \Vert W_1-W_2\Vert \le \Vert W_1-W_2\Vert _{\mathrm {F}}$, Gaussian concentration [39] gives

  $$\begin{aligned} \Pr \left\{ \lambda _+ - \mathbb {E}\lambda _+ \ge t \right\} \le e^{-t^2/2}. \end{aligned}$$

  (5.2)

  Using (5.1) and (5.2) gives

  $$\begin{aligned} \Pr \left\{ \left\| W\right\| _{\mathrm {op}} > 3 n^{1/2} \right\} \le e^{-n/2}. \end{aligned}$$

• 2. $\Pr \left\{ \Vert W\mathbbm {1}\Vert _\infty > 3\sqrt{n \log n} \right\} \le 2n^{-5/4}$.

  The random vector given by $\frac{1}{(n-1)^{1/2}}W\mathbbm {1}$ is jointly Gaussian where the marginal of each entry is a standard complex Gaussian. By a suboptimal union bound argument, the maximum absolute value among k standard complex Gaussian random variables (not necessarily independent) is larger than t with probability at most $2ke^{-t^2/4}$. Hence,

  $$\begin{aligned} \Pr \left\{ \Vert W\mathbbm {1}\Vert _\infty> 3\sqrt{n\log n} \right\}&\le \Pr \left\{ \left\| \frac{1}{(n-1)^{1/2}}W\mathbbm {1}\right\| _\infty > 3 \sqrt{\log n} \right\} \\&\le 2ne^{-\frac{9}{4} \log n} = 2n^{-5/4}. \end{aligned}$$

To support the discussion following Proposition 3.3, we further argue that

$$\begin{aligned} \Pr \left\{ |\mathbbm {1}^* W \mathbbm {1}| > n^{3/2} \right\} \le e^{-n/2}. \end{aligned}$$

It is easy to see that $\mathbbm {1}^* W \mathbbm {1}$ is a real Gaussian random variable with zero mean and variance $2\frac{n(n-1)}{2} = n(n-1)$. This implies that:

$$\begin{aligned} \Pr \left\{ |\mathbbm {1}^* W \mathbbm {1}|> n^{3/2} \right\}&\le \Pr \left\{ \frac{1}{(n(n-1))^{1/2}} |\mathbbm {1}^* W \mathbbm {1}| > n^{1/2} \right\} \\&\le \frac{1}{\sqrt{2\pi }}\frac{1}{n^{1/2}}e^{-\frac{n}{2}} \le e^{-n/2}. \end{aligned}$$