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@@ -2,25 +2,16 @@
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$$
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\begin{aligned} P(H(\boldsymbol{x}) \neq f(\boldsymbol{x})) &=\sum_{k=0}^{\lfloor T / 2\rfloor} \left( \begin{array}{c}{T} \\ {k}\end{array}\right)(1-\epsilon)^{k} \epsilon^{T-k} \\ & \leqslant \exp \left(-\frac{1}{2} T(1-2 \epsilon)^{2}\right) \end{aligned}
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$$
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-
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+[推导]:由基分类器相互独立，设X为T个基分类器分类正确的次数，因此$\mathrm{X} \sim \mathrm{B}(\mathrm{T}, 1-\mathrm{\epsilon})$
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$$
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-\begin{aligned} P(H(x) \neq f(x))=& P(x \leq\lfloor T / 2\rfloor) \\ & \leqslant P(x \leq T / 2) \end{aligned}
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+\begin{aligned} P(H(x) \neq f(x))=& P(X \leq\lfloor T / 2\rfloor) \\ & \leqslant P(X \leq T / 2)
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+\\ & =P\left[X-(1-\varepsilon) T \leqslant \frac{T}{2}-(1-\varepsilon) T\right]
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+\\ & =P\left[X-
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+(1-\varepsilon) T \leqslant -\frac{T}{2}\left(1-2\varepsilon\right)]\right]
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+ \end{aligned}
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$$
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-
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-Hoeffding's inequality
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-$$
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-\mathrm{P}(H(n) \leq(p-\varepsilon) n) \leq \exp \left(-2 \varepsilon^{2} n\right)
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-$$
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-$\newline$
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-
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-$$
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-\begin{aligned} & P\left(x \leq \frac{T}{2}\right) \\=& P\left[x-(1-\varepsilon) T \leqslant \frac{T}{2}-(1-\varepsilon) T\right]
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-\\=&
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-P\left[x-
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-(1-\varepsilon) T \leqslant -\frac{1}{2}T\left(1-2\varepsilon\right)]\right.
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-\end{aligned}
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-$$
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-根据Hoeffding's inequality
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+根据Hoeffding不等式$P(X-(1-\epsilon)T\leqslant -kT) \leq \exp (-2k^2T)$
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+令$k=\frac {(1-2\epsilon)}{2}$得
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$$
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\begin{aligned} P(H(\boldsymbol{x}) \neq f(\boldsymbol{x})) &=\sum_{k=0}^{\lfloor T / 2\rfloor} \left( \begin{array}{c}{T} \\ {k}\end{array}\right)(1-\epsilon)^{k} \epsilon^{T-k} \\ & \leqslant \exp \left(-\frac{1}{2} T(1-2 \epsilon)^{2}\right) \end{aligned}
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$$
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)s