Exercise 10.5 - Bayes nets for a rainy day

Answers

We first write down the entire probability:

p (V, G, R, S) = p (V) \cdot p (G) \cdot p (R | V, G) \cdot p (S | G) .

For question (a), we have:

\begin{aligned} p (S = 0 | V = 1) & = \frac{p (S = 0, V = 1)}{p (V = 1)} \\ = \frac{1}{p (V = 1)} \sum_{R, G \in {0, 1}} p (V = 1, G, R, S = 0) \\ = 0.2 𝛼𝛾 + 0.8 𝛼𝛾 + 0.1 (1 - α) β + 0.9 (1 - α) β \\ = 𝛼𝛾 + (1 - α) β . \end{aligned}

For question (b), the answer is exactly the same. Since given $G$ , $S$ is independent from $V$ .

For question (c), we independently estimate $δ$ , $α$ , $γ$ and $β$ :

\begin{aligned} δ & = 1, \\ α & = \frac{1}{3}, \\ γ & = 1, \\ β & = 0 . \end{aligned}

from counting.

solour_lfq

2021-03-24 13:42