由第一次检测：P⊕∣cancerPcancer0.0078P⊕∣¬cancerP¬cancer0.0298将第一次检测结果作为新的先验概率，归一化得到：Pcancer∣⊕0.2074P¬cancer∣⊕0.7926第二次检测的后验概率：Pcancer∣⊕⊕P⊕∣cancer∗Pcancer∣⊕.098∗0.20740.2033P¬cancer∣⊕⊕P⊕∣¬can。 - 详解

机器学习第五次作业

6.1

由第一次检测：
$P(\oplus \mid cancer)P(cancer)=0.0078$
$P(\oplus \mid \neg cancer)P(\neg cancer)=0.0298$、

将第一次检测结果作为新的先验概率，归一化得到：
$P(cancer\mid \oplus )=0.0078/0.0376=0.2074$
$P(\neg cancer\mid \oplus )=0.0298/0.0376=0.7926$

第二次检测的后验概率：
$P(cancer\mid \oplus ,\oplus )=P(\oplus \mid cancer)\ast P(cancer\mid \oplus )=.098\ast 0.2074=0.2033$
$P(\neg cancer\mid \oplus ,\oplus )=P(\oplus \mid \neg cancer)\ast P(\neg cancer\mid \oplus )=0.03\ast 0.7926=0.0238$
归一化得：
$P(cancer\mid \oplus ,\oplus )=0.2033/0.2271=0.8952$
$P(\neg cancer\mid \oplus ,\oplus )=0.0238/0.2271=0.1048$

6.2

根据贝叶斯公式：
$$\begin{gathered} P(cancer\mid \oplus )=\frac{P(\oplus \mid cancer)\ast P(cancer)}{P(\oplus )} \\ =\frac{P(\oplus \mid cancer)\ast P(cancer)}{P(\oplus \mid cancer)P(cancer)+P(\oplus \mid \neg cancer)P(\neg cancer)} \\ =\frac{1}{\frac{P(\oplus \mid cancer)P(cancer)+P(\oplus \mid \neg cancer)P(\neg cancer)}{P(\oplus \mid cancer)\ast P(cancer)}} \\ =\frac{1}{1+\frac{P(\oplus \mid \neg cancer)P(\neg cancer)}{P(\oplus \mid cancer)\ast P(cancer)}} \end{gathered}$$
而将$P(\oplus \mid cancer)P(cancer)$和$P(\oplus \mid \neg cancer)P(\neg cancer)$归一化，得到的正是$\frac{P(\oplus \mid cancer)P(cancer)}{P(\oplus )}$和$\frac{P(\oplus \mid \neg cancer)P(\neg cancer)}{P(\oplus )}$，而前者正是$P(cancer\mid \oplus )$，因此这样是正确的