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Least square regression with indefinite kernels and coefficient regularization. (English) Zbl 1225.65015
Let $(y_i,x_i)_{i=1}^m$ be i.i.d. observations with $y_i\in\Bbb R$, $x_i\in X$, $X$ being some compact metric space. The authors consider estimates $f_z$ for the regression function $f_\rho(x)=E(y_i|x_i)$, where $f_z=f_{\alpha^z}$, $f_\alpha(x)=\sum_{i=1}^m \alpha_i K(x,x_i)$, $$\alpha^z=\arg\min_{\alpha\in R^m} {1\over m}\,\sum_{i=1}^m (y_i - f_\alpha(x_i))^2+\lambda m\sum_{i=1}^m\alpha_i^2, $$ $K:X\times X\to\Bbb R$ is a continuous bounded function (kernel), $\lambda$ is a regularization parameter. Consistency of $f_z$ is demonstrated under the assumptions that $\lambda=\lambda(m)\to 0$, $\lambda^{3/2}\sqrt{m}\to\infty$ and the true regression function belongs to the closure of $\{f_\alpha\}$ in some suitable reproducing kernel Hilbert space. To analyze the rates of convergence the authors make assumptions of the form $ E\|L^{-r}f_\rho(x_i)\|^2<\infty$ for some $r>0$, where $L f(x)=E \tilde K(x,x_i)f(x_i)$, $\tilde K(x,t)=E_u K(x,u)K(t,u) $. E.g. if $r>1$ then choosing $\lambda=m^{1/5}$ they get $\|f_z-f_\rho\|_{L^2}=O(m^{-1/5})$. Results of simulations are presented for $X=[0,1]$ and the Gaussian kernel $K$.

65C60Computational problems in statistics
62J02General nonlinear regression
46E22Hilbert spaces with reproducing kernels
Full Text: DOI
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