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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Chernoff bound ： ウィキペディア英語版 Chernoff bound In probability theory, the Chernoff bound, named after Herman Chernoff but due to Herman Rubin, gives exponentially decreasing bounds on tail distributions of sums of independent random variables. It is a sharper bound than the known first or second moment based tail bounds such as Markov's inequality or Chebyshev inequality, which only yield power-law bounds on tail decay. However, the Chernoff bound requires that the variates be independent – a condition that neither the Markov nor the Chebyshev inequalities require. It is related to the (historically prior) Bernstein inequalities, and to Hoeffding's inequality. ==Example== Let be independent Bernoulli random variables, each having probability ''p'' > 1/2 of being equal to 1. Then the probability of simultaneous occurrence of more than ''n''/2 of the events has an exact value , where :$S=\backslash sum\_\; \backslash rfloor\; +\; 1\}^n\; \backslash binomp^i\; (1\; -\; p)^\; .$ The Chernoff bound shows that has the following lower bound: :$S\; \backslash ge\; 1\; -\; e^n\; \backslash left(p\; -\; \backslash frac\; \backslash right)^2\}\; .$ Indeed, noticing that , we get by the multiplicative form of Chernoff bound (see below or Corollary 13.3 in Sinclair's class notes), :$\backslash begin$ \Pr\left (\sum_^n X_k \le\left\lfloor \tfrac\right\rfloor \right ) &=\Pr\left (\sum_^n X_k\le\left(1-\left(1-\tfrac\right)\right)\mu\right ) \\ &\leq e^\left(1-\frac\right)^2} \\ &=e^\left(p-\frac\right)^2} \end This result admits various generalizations as outlined below. One can encounter many flavours of Chernoff bounds: the original ''additive form'' (which gives a bound on the absolute error) or the more practical ''multiplicative form'' (which bounds the error relative to the mean). 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「Chernoff bound」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.