Exercise 2.61

Answers

(a): $\begin{array}{l} P (D | ⋂_{i = 1}^{n} T_{i}) & = \frac{P (D) P (⋂_{i = 1}^{n} T_{i} | D)}{P (⋂_{i = 1}^{n} T_{i})} \\ = \frac{p \prod_{i = 1}^{n} a}{p \prod_{i = 1}^{n} a + q \prod_{i = 1}^{n} b} \\ = \frac{p a^{n}}{p a^{n} + q b^{n}} \end{array}$
(b): $\begin{array}{l} P (D | ⋂_{i = 1}^{n} T_{i}) & = \frac{P (D) P (⋂_{i = 1}^{n} T_{i} | D)}{P (⋂_{i = 1}^{n} T_{i})} \\ = \frac{p (P (G) P (⋂_{i = 1}^{n} T_{i} | D, G) + P (G^{c}) P (⋂_{i = 1}^{n} T_{i} | D, G^{c}))}{P (⋂_{i = 1}^{n} T_{i})} \\ = \frac{p (\frac{1}{2} + \frac{1}{2} a_{0}^{n})}{P (G) P (⋂_{i = 1}^{n} T_{i} | G) + P (G^{c}) P (⋂_{i = 1}^{n} T_{i} | G^{c})} \\ = \frac{p (\frac{1}{2} + \frac{1}{2} a_{0}^{n})}{P (G) (P (D | G) P (⋂_{i = 1}^{n} T_{i} | D, G) + P (D^{c} | G) P (⋂_{i = 1}^{n} T_{i} | D^{c}, G)) + P (G^{c}) (P (D | G^{c}) P (⋂_{i = 1}^{n} T_{i} | D, G^{c}) + P (D^{c} | G^{c}) P (⋂_{i = 1}^{n} T_{i} | D^{c}, G^{c}))} \\ = \frac{p (\frac{1}{2} + \frac{1}{2} a_{0}^{n})}{\frac{1}{2} + \frac{1}{2} (p a_{0}^{n} + (1 - p) b_{0}^{n})} \\ = \frac{p (1 + a_{0}^{n})}{1 + p a_{0}^{n} + (1 - p) b_{0}^{n}} \end{array}$

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2021-12-05 00:00