Exercise 4.18 - Naive Bayes with mixed features

Answers

For question (a):

\begin{aligned} p (y = 1 | x_{1} = 0, x_{2} = 0) & = \frac{p (y = 1) \cdot p (x_{1} | y = 1) \cdot p (x_{2} = 0 | y = 1)}{p (x_{1} = 0, x_{2} = 0)} \\ = \frac{0.5 \cdot 0.5 \cdot \frac{\exp (- 0.5)}{\sqrt{2 π}}}{p (x_{1} = 0, x_{2} = 0)} . \end{aligned}

Similarly,

p (y = 2 | x_{1} = 0, x_{2} = 0) = \frac{0.25 * 0.5 * \frac{1}{\sqrt{2 π}}}{p (x_{1} = 0, x_{2} = 0)},

p (y = 3 | x_{1} = 0, x_{2} = 0) = \frac{0.25 * 0.5 * \frac{\exp (- 0.5)}{\sqrt{2 π}}}{p (x_{1} = 0, x_{2} = 0)} .

A normalization yields the final result by eliminating $p (x_{1} = 0, x_{2} = 0)$ .

For question (b), we have:

p (y = 1 | x_{1} = 0) = 0.5,

p (y = 2 | x_{1} = 0) = 0.25,

p (y = 3 | x_{1} = 0) = 0.25,

since $x_{1}$ yields no more information for the classification label.

For question (c), we have:

p (y = 1 | x_{2} = 0) \propto 0.5 * \frac{\exp (- 0.5)}{\sqrt{2 π}},

p (y = 2 | x_{2} = 0) \propto 0.25 * \frac{1}{\sqrt{2 π}},

p (y = 3 | x_{2} = 0) \propto 0.25 * \frac{\exp (- 0.5)}{\sqrt{2 π}} .

One can observe that unlike $p (y | x_{1} = 0)$ , $p (y | x_{1} = 0)$ is different from the prior on labels.

solour_lfq

2021-03-24 13:42