Does generalization performance of $l^q$ regularization learning depend on $q$? A negative example

Abstract

$l^q$-regularization has been demonstrated to be an attractive technique in machine learning and statistical modeling. It attempts to improve the generalization (prediction) capability of a machine (model) through appropriately shrinking its coefficients. The shape of a $l^q$ estimator differs in varying choices of the regularization order $q$. In particular, $l^1$ leads to the LASSO estimate, while $l^{2}$ corresponds to the smooth ridge regression. This makes the order $q$ a potential tuning parameter in applications. To facilitate the use of $l^{q}$-regularization, we intend to seek for a modeling strategy where an elaborative selection on $q$ is avoidable. In this spirit, we place our investigation within a general framework of $l^{q}$-regularized kernel learning under a sample dependent hypothesis space (SDHS). For a designated class of kernel functions, we show that all $l^{q}$ estimators for $0< q < \infty$ attain similar generalization error bounds. These estimated bounds are almost optimal in the sense that up to a logarithmic factor, the upper and lower bounds are asymptotically identical. This finding tentatively reveals that, in some modeling contexts, the choice of $q$ might not have a strong impact in terms of the generalization capability. From this perspective, $q$ can be arbitrarily specified, or specified merely by other no generalization criteria like smoothness, computational complexity, sparsity, etc..Comment: 35 pages, 3 figure