Can you help me to understand this deduction for proving Naive Bayes is a Linear Classifier?

Question

In this tutorial on Naive Bayes Classififer in section 1.1, the author proved naive bayes is a linear classifier.

Consider binary classification where $y=0$ or $1$. Our classification rule with argmax can equivalently be expressed with log odds ratio $$f(x)=log\ \dfrac{p(y=1|x)}{p(y=0|x)}\\=log\ p(y=1|x)-log\ p(y=0|x)\\=(log\ \theta_1-log\ \theta_0)^\top x+(log\ p(y=1)-log\ p(y=0))$$ The decision rule is to classify $x$ with $y=1$ if $f(x)>0$, and $y=0$ otherwise. Note for given parameters, this is a linear function in $x$. That is to say, the Naive Bayes classifier induces a linear decision boundary in feature space $X$ . The boundary takes the form of a hyperplane, defined by $f(x) = 0$.

Can you tell me how is the third line of equation deducted from the second line? It doesn't seem clear to me.

Answer 1

In section 1, the log likelihood is given: $\log p(x|y) = x^T \log \theta_y + const$ as well as Bayes rule: $p(y|x) = \frac{p(x|y)p(y)}{p(x)}$, then it should be something like:

$$\log p(y=1|x) - \log p(y=0|x) \\ = \log \frac{p(x|y=1)p(y=1)}{p(x)} - \log \frac{p(x|y=0)p(y=0)}{p(x)} \\ = \log p(x|y=1)p(y=1) - \log p(x|y=0)p(y=0) \\ = (x^T \log \theta_1 + \log p(y=1)) - (x^T \log \theta_0 + \log p(y=0)) \\ = (\log \theta_1 - \log \theta_0)^T x + (\log p(y=1) - \log p(y=0))$$