Power Spectrum Unbiasing for Dilation-Invariant Multi-reference Alignment

Abstract

This article discusses a generalization of the 1-dimensional multi-reference alignment problem. The goal is to recover a hidden signal from many noisy observations, where each noisy observation includes a random translation and random dilation of the hidden signal, as well as high additive noise. We propose a method that recovers the power spectrum of the hidden signal by applying a data-driven, nonlinear unbiasing procedure, and thus the hidden signal is obtained up to an unknown phase. An unbiased estimator of the power spectrum is defined, whose error depends on the sample size and noise levels, and we precisely quantify the convergence rate of the proposed estimator. The unbiasing procedure relies on knowledge of the dilation distribution, and we implement an optimization procedure to learn the dilation variance when this parameter is unknown. Our theoretical work is supported by extensive numerical experiments on a wide range of signals.

Appendices

Appendix A Proof of Lemma 1

Proof

From Proposition 1 and (7)

$$\begin{aligned} Pf - \widetilde{Pf}&= (I-L_{{B_0}})^{-1}{B_1}L_{{B_2}}\left[ 3(g_\eta -\widetilde{g}_\eta )+\omega ( g_\eta '(\omega )-\widetilde{g}_\eta '(\omega ))\right] \, , \end{aligned}$$

for constants ${B_0}, {B_1}, {B_2}$ defined in (5). Letting $\Vert \cdot \Vert$ denote the spectral norm, we thus obtain:

$$\begin{aligned}&\Vert Pf-\widetilde{Pf} \Vert _2^2 \le {B_1}^2 \Vert (I-L_{{B_0}})^{-1} \Vert ^2 \Vert L_{{B_2}} \Vert ^2 \times \Vert 3(g_\eta -\widetilde{g}_\eta )+\omega ( g_\eta '(\omega )-\widetilde{g}_\eta '(\omega )) \Vert _2^2 \\&\le 2{B_1}^2 \Vert (I-L_{{B_0}})^{-1} \Vert ^2 \Vert L_{{B_2}} \Vert ^2 \times \left( 9\Vert g_\eta -\widetilde{g}_\eta \Vert _2^2 + \Vert \omega ( g_\eta '(\omega )-\widetilde{g}_\eta '(\omega )) \Vert _2^2 \right) . \end{aligned}$$

We first observe that $\Vert L_C^i \Vert =C^\frac{5i}{2}$ since

$$\begin{aligned} \Vert L_C^i g \Vert _2^2&= \int (C^{3i} g(C^i\omega ))^2\ d\omega \\&= \int C^{6i} g(\tilde{\omega })^2\ \frac{d\tilde{\omega }}{C^i} \quad \text {for}\quad \tilde{\omega } = C^i \omega \\&= C^{5i} \Vert g \Vert _2^2\, . \end{aligned}$$

Thus

$$\begin{aligned} \Vert (I-L_{{B_0}})^{-1} \Vert&= \Vert [ \Vert \big ]{ \sum _{i=0}^{\infty } L_{{B_0}}^i } \le \sum _{i=0}^{\infty } {B_0}^{\frac{5i}{2}} = \frac{1}{1-{B_0}^{\frac{5}{2}}} = O(\eta ^{-1}) \\ \Vert L_{{B_2}} \Vert&= {B_2}^{\frac{5}{2}} = O(1) \\ {B_1}&= O(\eta ) \end{aligned}$$

so that

$$\begin{aligned} 2{B_1}^2 \Vert (I-L_{{B_0}})^{-1} \Vert ^2 \Vert L_{{B_2}} \Vert ^2&= O(1)O(\eta ^{2})O(\eta ^{-2}) = O(1) \end{aligned}$$

and we obtain

$$\begin{aligned} \Vert Pf-\widetilde{Pf} \Vert _2^2&\lesssim \Vert g_\eta -\widetilde{g}_\eta \Vert _2^2 + \Vert \omega ( g_\eta '(\omega )-\widetilde{g}_\eta '(\omega )) \Vert _2^2 \, . \end{aligned}$$

$\square$

Appendix B Proof of Lemma 2

Proof

Note by assumption there exist constants $C>0$, $\omega _0\ge 1$ such that $|\widehat{h} (\omega )|\le C|\omega |^{-\alpha }$ for $|\omega |\ge \omega _0$. Also note that $\widehat{\phi }_L(\omega )=e^{-L^2\omega ^2/2}$, so that $1-\widehat{\phi }_L(\omega ) = \frac{L^2\omega ^2}{2}+O(L^3)$ for small L. We have:

$$\begin{aligned} \Vert h - h *\phi _L \Vert _2^2&= (2\pi )^{-1} \Vert \widehat{h}(1 - \widehat{\phi }_L )\Vert _2^2 \\&= \frac{1}{2\pi } \int _{|\omega |<\omega _0} |\widehat{h}(\omega )|^2|1-\widehat{\phi }_L(\omega )|^2 \ d\omega \\&\quad + \frac{1}{2\pi } \int _{|\omega |\ge \omega _0} C^2|\omega |^{-2\alpha }|1-\widehat{\phi }_L(\omega )|^2 \ d\omega \\&:= (I) + (II) \, . \end{aligned}$$

Note:

$$\begin{aligned} (I)&\le \int _{|\omega |<\omega _0} |\widehat{h}(\omega )|^2|1-\widehat{\phi }_L(\omega )|^2 \ d\omega \\&\le 2\int _0^{\omega _0} |\widehat{h}(\omega )|^2\left( \frac{L^2\omega ^2}{2}+O(L^3)\right) ^2\ d\omega \\&\le 2\left( \frac{L^4\omega _0^4}{4}+O(L^5)\right) \int _0^{\omega _0}|\widehat{h}(\omega )|^2 \ d\omega \\&\le \frac{\omega _0^4}{2}\Vert h\Vert _2^2L^4+O(L^5) \, . \end{aligned}$$

To control the second term, note

$$\begin{aligned} (II)&\le 2C^2 \int _{1}^{\infty } \omega ^{-2\alpha }\left( 1-e^{-\frac{L^2\omega ^2}{2}}\right) ^2\ d\omega \\&=2C^2 \int _{L}^{\infty } \left( \frac{L}{\tilde{\omega }}\right) ^{2\alpha }\left( 1-e^{-\frac{\tilde{\omega }^2}{2}}\right) ^2\ \frac{d\tilde{\omega }}{L} \\&=2C^2L^{2\alpha -1} \int _L^{\infty } \omega ^{-2\alpha }\left( 1-e^{-\frac{\omega ^2}{2}}\right) ^2\ d\omega \, . \end{aligned}$$

Explicit evaluation of the upper bound with a computer algebra system gives:

$$\begin{aligned} \alpha =1:&\qquad C_1L + O(L^4) \\ \alpha =2:&\qquad C_2L^3 + O(L^4) \\ \alpha =3:&\qquad C_3L^4 + O(L^5) \end{aligned}$$

Also since

$$\begin{aligned}&\frac{d}{d\alpha } \int _{1}^{\infty } \omega ^{-2\alpha }\left( 1-e^{-\frac{L^2\omega ^2}{2}}\right) ^2\ d\omega \\&\quad = \int _{1}^{\infty } -2\ln (\omega )\omega ^{-2\alpha }\left( 1-e^{-\frac{L^2\omega ^2}{2}}\right) ^2\ d\omega <0\, , \end{aligned}$$

the upper bound is decreasing in $\alpha$, and we can conclude $(II) \lesssim L^{4\wedge (2\alpha -1)}$ and the lemma is proved. $\square$

Appendix C Proof of Lemma 3

Proof

First observe:

$$\begin{aligned} \Vert x(h - h *\phi _L) \Vert ^2_2&= (2\pi )^{-1} \Vert [ \Vert \Big ]{\frac{d}{d\omega }\left( \widehat{h} - \widehat{h} \widehat{\phi }_L\right) }^2_2 \\&= (2\pi )^{-1} \Vert \widehat{h}' - \widehat{h}' \widehat{\phi }_L-\widehat{h} \widehat{\phi }_L' \Vert ^2_2 \\ {}&\lesssim \Vert \widehat{h}' - \widehat{h}' \widehat{\phi }_L\Vert _2^2 +\Vert \widehat{h} \widehat{\phi }_L' \Vert ^2_2 \, . \end{aligned}$$

To bound the first term, we apply Lemma 2 to the function xh to obtain,

$$\begin{aligned} \Vert \widehat{h}' - \widehat{h}' \widehat{\phi }_L\Vert _2^2&= 2\pi \Vert xh - (xh)*\phi _L\Vert _2^2 \\&\lesssim \Vert xh\Vert _2^2L^4 + L^{4\wedge (2\alpha -1)} \, . \end{aligned}$$

To bound the second term, note $\widehat{\phi }_L'(\omega ) = -L^2\omega e^{-L^2\omega ^2/2}$, and that $\Vert \omega ^2 e^{-L^2\omega ^2}\Vert _{\infty } = (eL)^{-1}$. Thus

$$\begin{aligned} \Vert \widehat{h} \widehat{\phi }_L' \Vert ^2_2&= L^4 \int |\widehat{h}(\omega )|^2 \omega ^2e^{-L^2\omega ^2}\ d\omega \le L^3 \Vert h \Vert _2^2 \, . \end{aligned}$$

Note we could get a higher power for L by

$$\begin{aligned} \Vert \widehat{h} \widehat{\phi }_L' \Vert ^2_2&\le L^4 \int \omega ^2|\widehat{h}(\omega )|^2 \ d\omega = L^4 \Vert \omega \widehat{h} \Vert _2^2 \lesssim L^4 \Vert h' \Vert _2^2\, , \end{aligned}$$

which proves the lemma. $\square$