Обсуждение участника:Riabenko

Материал из MachineLearning.

(Различия между версиями)
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TMP:
<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 11:47, 12 мая 2010

Глоссарий статистических терминов ISI


TMP: \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)}

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