large-Scale Supervised Sparse Principal Component Analysis

Introduction

The sparse PCA is a variant of the classical PCA, which assumes sparsity in the feature space. It has several advantages such as easy to interpret, and works for really high-dimensional data. The main issue about sparse PCA is that it is computationally expensive. Many algorithms have been proposed to solve the sparse PCA problem, and the authors introduced a fast block coordinate ascent algorithm with much better computational complexity.

1 Drawbacks of Existing techniques

Existing techniques include ad-hoc methods(e.g. factor rotation techniques, simple thresholding), greedy algorithms, SCoTLASS, the regularized SVD method, SPCA, the generalized power method. These methods are based on non-convex optimization and they don't guarantee global optimum.

A semi-definite relaxation method called DSPCA can guarantee global convergence and has better performance than above algorithms, however, it is computationally expensive.

2 Contribution of this paper

This paper solves DSPCA in a computationally easier way, and hence it is a good solution for large scale data sets. This paper applies a block coordinate ascent algorithm with computational complexity [math]\displaystyle{ O(\hat{n^3}) }[/math], where [math]\displaystyle{ \hat{n} }[/math] is the intrinsic dimension of the data. Since [math]\displaystyle{ \hat{n} }[/math] could be very small compared to the dimension [math]\displaystyle{ n }[/math] of the data, this algorithm is computationally easy.

Primal problem

The sparse PCA problem can be formulated as [math]\displaystyle{ max_x \ x^T \Sigma x - \lambda \| x \|_0 : \| x \|_2=1 }[/math].

This is equivalent to [math]\displaystyle{ max_z \ Tr(\Sigma Z) - \lambda \sqrt{\| Z \|_0} : Z \succeq 0, Tr Z=1, Rank(Z)=1 }[/math].

Replacing the [math]\displaystyle{ \sqrt{\| Z \|_0} }[/math] with [math]\displaystyle{ \| Z \|_1 }[/math] and dropping the rank constraint gives a relaxation of the original non-convex problem:

[math]\displaystyle{ \phi = max_z Tr (\Sigma Z) - \lambda \| Z \|_1 : Z \succeq 0 }[/math], [math]\displaystyle{ Tr(Z)=1 \qquad (1) }[/math] .

Fortunately, this relaxation approximates the original non-convex problem to a convex problem.

Here is an important theorem used by this paper:

Theorem(2.1) Let [math]\displaystyle{ \Sigma=A^T A }[/math] where [math]\displaystyle{ A=(a_1,a_2,......,a_n) \in {\mathbb R}^{m \times n} }[/math], we have [math]\displaystyle{ \psi = max_{\| \xi \|_2=1} }[/math] [math]\displaystyle{ \sum_{i=1}^{n} (({a_i}^T \xi)^2 - \lambda)_+ }[/math]. An optimal non-zero pattern corresponds to the indices [math]\displaystyle{ i }[/math] with [math]\displaystyle{ \lambda \lt (({a_i}^T \xi)^2-\lambda)_+ }[/math]

Block Coordinate Ascent Algorithm

There is a row-by-row algorithm applied to the problems of the form [math]\displaystyle{ min_X \ f(X)-\beta \ log(det X): \ L \leq X \leq U, X \succ 0 }[/math].

Problem (1) can be written as [math]\displaystyle{ {\frac 1 2} {\phi}^2 = max_X \ Tr \Sigma X - \lambda \| X \|_1 - \frac 1 2 (Tr X)^2: X \succeq 0 \qquad (2) }[/math] .

In order to apply the row by row algorithm, we need to add one more term [math]\displaystyle{ \beta \ log(det X) }[/math] to (2) where [math]\displaystyle{ \beta\gt 0 }[/math] is a penalty parameter.

That is to say, we address the problem [math]\displaystyle{ \ max_X \ Tr \Sigma X - \lambda \| X \|_1 - \frac 1 2 (Tr X)^2 + \beta \ log(det X): X \succeq 0 \qquad (3) }[/math]

By matrix partitioning, we could obtain the sub-problem:

[math]\displaystyle{ \phi = max_{x,y} \ 2(y^T s- \lambda \| y \|_1) +(\sigma - \lambda)x - {\frac 1 2}(t+x)^2 + \beta \ log(x-y^T Y^{\dagger} y ):y \in R(Y) }[/math].

The sub-problem can be simplified to be

Here is the algorithm:

Convergence and complexity

1. The algorithm is guaranteed to converge to the global optimizer.

2. The complexity for the algorithm is [math]\displaystyle{ O(Kn^{3}) }[/math], where [math]\displaystyle{ K }[/math] is the number of sweeps through columns (fixed, typically [math]\displaystyle{ K=5 }[/math])