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Minimax risk of matrix denoising by singular value thresholding

Research paper by David Donoho, Matan Gavish

Indexed on: 04 Nov '14Published on: 04 Nov '14Published in: Mathematics - Statistics



Abstract

An unknown $m$ by $n$ matrix $X_0$ is to be estimated from noisy measurements $Y=X_0+Z$, where the noise matrix $Z$ has i.i.d. Gaussian entries. A popular matrix denoising scheme solves the nuclear norm penalization problem $\operatorname {min}_X\|Y-X\|_F^2/2+\lambda\|X\|_*$, where $\|X\|_*$ denotes the nuclear norm (sum of singular values). This is the analog, for matrices, of $\ell_1$ penalization in the vector case. It has been empirically observed that if $X_0$ has low rank, it may be recovered quite accurately from the noisy measurement $Y$. In a proportional growth framework where the rank $r_n$, number of rows $m_n$ and number of columns $n$ all tend to $\infty$ proportionally to each other ($r_n/m_n\rightarrow \rho$, $m_n/n\rightarrow \beta$), we evaluate the asymptotic minimax MSE $\mathcal {M}(\rho,\beta)=\lim_{m_n,n\rightarrow \infty}\inf_{\lambda}\sup_{\operatorname {rank}(X)\leq r_n}\operatorname {MSE}(X_0,\hat{X}_{\lambda})$. Our formulas involve incomplete moments of the quarter- and semi-circle laws ($\beta=1$, square case) and the Mar\v{c}enko-Pastur law ($\beta<1$, nonsquare case). For finite $m$ and $n$, we show that MSE increases as the nonzero singular values of $X_0$ grow larger. As a result, the finite-$n$ worst-case MSE, a quantity which can be evaluated numerically, is achieved when the signal $X_0$ is "infinitely strong." The nuclear norm penalization problem is solved by applying soft thresholding to the singular values of $Y$. We also derive the minimax threshold, namely the value $\lambda^*(\rho)$, which is the optimal place to threshold the singular values. All these results are obtained for general (nonsquare, nonsymmetric) real matrices. Comparable results are obtained for square symmetric nonnegative-definite matrices.