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    "# 2.5 课后习题"
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    "1、\t考虑对于一个一维的两类问题，采用下列判定规则：\n",
    "如果 $x>\\theta$，则判为 $\\omega_{1}$，否则判为 $\\omega_{2}$。\n",
    "\n",
    "(a)\t证明此规则下的误差概率为\n",
    "$$P(\\text { error })=P\\left(\\omega_{1}\\right) \\int_{-\\infty}^{\\theta} p\\left(x | \\omega_{1}\\right) d x+P\\left(\\omega_{2}\\right) \\int_{\\theta}^{\\infty} p\\left(x | \\omega_{2}\\right) d x$$\n",
    "(b)\t通过微分计算，证明最小化 $P(error)$ 的一个必要条件是 $\\theta$ 满足\n",
    "$$p\\left(\\theta | \\omega_{1}\\right) P\\left(\\omega_{1}\\right)=p\\left(\\theta | \\omega_{2}\\right) P\\left(\\omega_{2}\\right) $$\n",
    "(c)\t此式可以唯一确定$\\theta$ 吗？\n",
    "\n",
    "(d)\t给出一个例子，说明满足此式的一个 $\\theta$ 事实上有可能使误差概率最大化。\n"
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   "source": [
    "2、\t假设我们将确定的判别函数 $\\alpha(\\mathbf{x})$ 用一个随机规则替换，也即当观察$x$时所采取的行为 $\\alpha_{i}$是随机的，其概率为$P\\left(\\alpha_{i} | \\mathbf{x}\\right)$。\n",
    "\n",
    "(a)\t证明所得的风险为\n",
    "$$R=\\int\\left[\\sum_{i=1}^{a} R\\left(\\alpha_{i} | \\mathbf{x}\\right) P\\left(\\alpha_{i} | \\mathbf{x}\\right)\\right] p(\\mathbf{x}) d \\mathbf{x}$$\n",
    "(b)\t证明与最小条件风险 $R\\left(\\alpha_{i} | \\mathbf{x}\\right)$相对应的行为$\\alpha_{i}$就是选择$P\\left(\\alpha_{i} | \\mathbf{x}\\right)=1$。由此证明随机扰动最优判定规则将得不到任何好处。\n",
    "\n",
    "(c)\t我们可以从随机扰动一个“次优”的判定规则中得到好处么？请解释原因。\n"
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    "3、\t设 $\\omega_{\\max }(\\mathbf{x})$ 为类别状态，此时对所有的$i$, $i=1, \\dots, c$，有$P\\left(\\omega_{\\max } | \\mathbf{x}\\right) \\geq P\\left(\\omega_{i} | \\mathbf{x}\\right)$。\n",
    "\n",
    "(a)\t证明 $P\\left(\\omega_{\\max } | \\mathbf{x}\\right) \\geq 1 / c$。\n",
    "\n",
    "(b)\t证明对于最小误差率判定规则，平均误差概率为\n",
    "$$P(\\text { error })=1-\\int P\\left(\\omega_{\\max } | \\mathbf{x}\\right) p(\\mathbf{x}) d \\mathbf{x}$$\n",
    "(c)\t利用这两个结论证明 $P(\\text { error }) \\leq(c-1) / c$.\n",
    "\n",
    "(d)\t描述一种情况，在此情况下有 $P(\\text { error })=(c-1) / c$.\n"
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