Обсуждение участника:Riabenko

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[[http://isi.cbs.nl/glossary/index.htm Глоссарий статистических терминов ISI]]
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<tex>\mathbb{E}\left[1\left\{K_j^{(b)}\leq\alpha\gamma\right\}\right] \leq \mathbb{P}\left[\left.P_j^{(b)}\leq\alpha\gamma\right|S\subseteq \tilde{S}^{(b)}\right].</tex>
<tex>\mathbb{E}\left[1\left\{K_j^{(b)}\leq\alpha\gamma\right\}\right] \leq \mathbb{P}\left[\left.P_j^{(b)}\leq\alpha\gamma\right|S\subseteq \tilde{S}^{(b)}\right].</tex>

Версия 11:45, 12 мая 2010

[Глоссарий статистических терминов ISI]


\mathbb{E}\left[1\left\{K_j^{(b)}\leq\alpha\gamma\right\}\right] \leq \mathbb{P}\left[\left.P_j^{(b)}\leq\alpha\gamma\right|S\subseteq \tilde{S}^{(b)}\right].

\mathbb{E}\left[1\left\{K_j^{(b)}\leq\alpha\gamma\right\}\right] = 1\cdot\mathbb{P}\left[K_j^{(b)}\leq\alpha\gamma\right] + 0\cdot\mathbb{P}\left[K_j^{(b)}>\alpha\gamma\right] = \mathbb{P}\left[K_j^{(b)}\leq\alpha\gamma\right] = \{\text{using the definition of } K_j^{(b)} \text{ and complete probability formula}\} =

= \mathbb{P}\left[\left.P_j^{(b)}\leq\alpha\gamma\right|S\subseteq \tilde{S}^{(b)}\right] + \mathbb{P}\left[1\leq\alpha\gamma\left|S\not\subseteq \tilde{S}^{(b)}\right.\right] = \mathbb{P}\left[\left.P_j^{(b)}\leq\alpha\gamma\right|S\subseteq \tilde{S}^{(b)}\right]

\mathbb{P}\left[\left.P_j^{(b)}\leq\alpha\gamma\right|S\subseteq \tilde{S}^{(b)}\right] = \mathbb{P}\left[\left.\tilde{P}_j^{(b)}\leq\frac{\alpha\gamma}{\left|\tilde{S}^{(b)}\right|}\right|S\subseteq \tilde{S}^{(b)}\right]

\mathbb{P}\left[\left.P_j^{(b)}\leq\alpha\gamma\right|S\subseteq \tilde{S}^{(b)}\right] = \frac {\alpha\gamma} { \left| \tilde{S}^{(b)} \right| }.

\tilde{P}_j^{(b)}

S\subseteq \tilde{S}^{(b)}

\beta_j=0

\tilde{S}^{(b)}

\tilde{P}_j^{(b)}\leq\frac{\alpha\gamma}{\left|\tilde{S}^{(b)}

Источник — «http://machinelearning.ru/wiki/index.php?title=%D0%9E%D0%B1%D1%81%D1%83%D0%B6%D0%B4%D0%B5%D0%BD%D0%B8%D0%B5_%D1%83%D1%87%D0%B0%D1%81%D1%82%D0%BD%D0%B8%D0%BA%D0%B0:Riabenko»
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