翻訳と辞書 Words near each other ・ Representative of the UNHCR and WFP, London ・ Representative of the United States to the European Office of the United Nations ・ Representative office ・ Representative Party of Alberta ・ Representative Party of Ontario ・ Representative payee ・ Representative peer ・ Representative sequences ・ Representative side ・ Representative town meeting ・ Representativeness heuristic ・ Representatives elected in the Japanese general election, 2012 ・ Representatives of Indonesia to international pageants ・ Representatives of the President of Ukraine ・ Representativz ・ Representer theorem ・ Representin ・ Representin' (album) ・ Representing Border ・ Representing NYC ・ Representing the Mambo ・ Representivity (politics) ・ Represo Formation ・ Repressed ・ Repressed memory ・ Repressilator ・ Repression ・ Repression in the Soviet Union ・ Repression of communists in the Kingdom of Romania ・ Repression of Family Members of Traitors of the Motherland Dictionary Lists mini英和辞書 mini和英辞書 Webster 1913 Latin-English FOLDOC Wikipedia English ウィキペディア

翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Representer theorem ： ウィキペディア英語版 Representer theorem In statistical learning theory, a representer theorem is any of several related results stating that a minimizer $f^$ of a regularized empirical risk function defined over a reproducing kernel Hilbert space can be represented as a finite linear combination of kernel products evaluated on the input points in the training set data. ==Formal Statement== The following Representer Theorem and its proof are due to Schölkopf, Herbrich, and Smola: Theorem: Let $\backslash mathcal$ be a nonempty set and $k$ a positive-definite real-valued kernel on $\backslash mathcal\; \backslash times\; \backslash mathcal$ with corresponding reproducing kernel Hilbert space $H\_k$. Given a training sample $(x\_1,\; y\_1),\; \backslash dotsc,\; (x\_n,\; y\_n)\; \backslash in\; \backslash mathcal\; \backslash times\; \backslash R$, a strictly monotonically increasing real-valued function $g\; \backslash colon\; [0,\; \backslash infty)\; \backslash to\; \backslash R$, and an arbitrary empirical risk function $E\; \backslash colon\; (\backslash mathcal\; \backslash times\; \backslash R^2)^n\; \backslash to\; \backslash R\; \backslash cup\; \backslash lbrace\; \backslash infty\; \backslash rbrace$, then for any $f^\; \backslash in\; H\_k$ satisfying :$$ f^ = \operatorname_ \left\lbrace E\left( (x_1, y_1, f(x_1)), ..., (x_n, y_n, f(x_n)) \right) + g\left( \lVert f \rVert \right) \right \rbrace, \quad ( *) $f^$ admits a representation of the form: :$$ f^(\cdot) = \sum_^n \alpha_i k(\cdot, x_i), where $\backslash alpha\_i\; \backslash in\; \backslash R$ for all $1\; \backslash le\; i\; \backslash le\; n$. Proof: Define a mapping :$$ \begin \varphi \colon \mathcal &\to \R^ (so that $\backslash varphi(x)\; =\; k(\backslash cdot,\; x)$ is itself a map $\backslash mathcal\; \backslash to\; \backslash R$). Since $k$ is reproducing kernel, then :$$ \varphi(x)(x') = k(x', x) = \langle \varphi(x'), \varphi(x) \rangle, where $\backslash langle\; \backslash cdot,\; \backslash cdot\; \backslash rangle$ is the inner product on $H\_k$. Given any $x\_1,\; ...,\; x\_n$, one can use orthogonal projection to decompose any $f\; \backslash in\; H\_k$ into a sum of two function, one lying in $\backslash operatorname\; \backslash left\; \backslash lbrace\; \backslash varphi(x\_1),\; ...,\; \backslash varphi(x\_n)\; \backslash right\; \backslash rbrace$, and the other lying in the orthogonal complement: :$$ f = \sum_^n \alpha_i \varphi(x_i) + v, where $\backslash langle\; v,\; \backslash varphi(x\_i)\; \backslash rangle\; =\; 0$ for all $i$. The above orthogonal decomposition and the reproducing property together show that applying $f$ to any training point $x\_j$ produces :$$ f(x_j) = \left \langle \sum_^n \alpha_i \varphi(x_i) + v, \varphi(x_j) \right \rangle = \sum_^n \alpha_i \langle \varphi(x_i), \varphi(x_j) \rangle, which we observe is independent of $v$. Consequently, the value of the empirical risk $E$ in ( *) is likewise independent of $v$. For the second term (the regularization term), since $v$ is orthogonal to $\backslash sum\_^n\; \backslash alpha\_i\; \backslash varphi(x\_i)$ and $g$ is strictly monotonic, we have :$$ \begin g\left( \lVert f \rVert \right) &= g \left( \lVert \sum_^n \alpha_i \varphi(x_i) + v \rVert \right) \\ &= g \left( \sqrt \right) \\ &\ge g \left( \lVert \sum_^n \alpha_i \varphi(x_i) \rVert \right). \end Therefore setting $v\; =\; 0$ does not affect the first term of ( *), while it strictly decreasing the second term. Consequently, any minimizer $f^$ in ( *) must have $v\; =\; 0$, i.e., it must be of the form :$$ f^(\cdot) = \sum_^n \alpha_i \varphi(x_i) = \sum_^n \alpha_i k(\cdot, x_i), which is the desired result. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Representer theorem」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.