Unify latex align (#63)

docs/lectures/naive_bayes.md

@@ -179,12 +179,12 @@ With that, we have **derived Bayes Rule**.
  Let $A$ be the event that a tweet is positive and $B$ be the event that a tweet contains the word "amazing".
 
  $$
- \begin{aligned}
+ \begin{align}
  P(A) &= 0.4 \\
  P(B) &= 0.13 \\
  P(B|A) &= 0.25 \\
  P(A|B) &= \frac{P(B|A)P(A)}{P(B)} = \frac{0.25 \times 0.4}{0.13} = 0.7692
- \end{aligned}
+ \end{align}
  $$
 
  The probability that the tweet "amazing to be here" is positive is 0.7692.
@@ -280,23 +280,23 @@ We end up with the following table:
  Let's calculate the probability $P(\text{happy}|\text{pos})$ of the word "happy" given that the tweet is **positive**.
 
  $$
- \begin{aligned}
+ \begin{align}
  P(\text{happy}|\text{pos}) &= \frac{freq(\text{happy},\text{pos}) + 1}{N_{pos} + |V|} \\
  &= \frac{2 + 1}{11 + 9} \\
  &= \frac{3}{20} \\
  &= 0.15
- \end{aligned}
+ \end{align}
  $$
 
  Let's calculate the probability $P(\text{happy}|\text{neg})$ of the word "happy" given that the tweet is **negative**.
 
  $$
- \begin{aligned}
+ \begin{align}
  P(\text{happy}|\text{neg}) &= \frac{freq(\text{happy},\text{neg}) + 1}{N_{neg} + |V|} \\
  &= \frac{0 + 1}{11 + 19} \\
  &= \frac{1}{20} \\
  &= 0.05
- \end{aligned}
+ \end{align}
  $$
 
 ## Ratio of Probabilities
@@ -361,26 +361,26 @@ where
  We have the following ratios of probabilities:
 
  $$
- \begin{aligned}
+ \begin{align}
  \frac{P(\text{I}|\text{pos})}{P(\text{I}|\text{neg})} &= \frac{0.2}{0.2} = 1.0 \\
  \frac{P(\text{am}|\text{pos})}{P(\text{am}|\text{neg})} &= \frac{0.15}{0.15} = 1.0 \\
  \frac{P(\text{happy}|\text{pos})}{P(\text{happy}|\text{neg})} &= \frac{0.15}{0.05} = 3.0 \\
  \frac{P(\text{because}|\text{pos})}{P(\text{because}|\text{neg})} &= \frac{0.1}{0.1} = 1.0 \\
  \frac{P(\text{I}|\text{pos})}{P(\text{I}|\text{neg})} &= \frac{0.2}{0.2} = 1.0 \\
  \frac{P(\text{love}|\text{pos})}{P(\text{love}|\text{neg})} &= \frac{0.1}{0.05} = 2.0 \\
- \end{aligned}
+ \end{align}
  $$
 
  Note that the words "ice" and "cream" are not in the vocabulary, so we ignore them.
 
  Given these ratios, we can calculate the likelihood of the tweet being positive as follows:
 
  $$
- \begin{aligned}
+ \begin{align}
  P(\text{pos}|\text{tweet}) &= \prod_{i=1}^{m} \frac{P(w_i|pos)}{P(w_i|neg)} \\
  &= 1.0 \times 1.0 \times 3.0 \times 1.0 \times 1.0 \times 2.0 \\
  &= 6.0
- \end{aligned}
+ \end{align}
  $$
 
 ## Prior
@@ -568,26 +568,26 @@ Words that **do not appear** in the vocabulary are **ignored**. They are conside
  We have the following **ratios of probabilities**:
 
  $$
- \begin{aligned}
+ \begin{align}
  \log \frac{P(\text{I}|\text{pos})}{P(\text{I}|\text{neg})} &= \log \frac{0.2}{0.2} = \log 1.0 = 0.0 \\
  \log \frac{P(\text{am}|\text{pos})}{P(\text{am}|\text{neg})} &= \log \frac{0.15}{0.15} = \log 1.0 = 0.0 \\
  \log \frac{P(\text{happy}|\text{pos})}{P(\text{happy}|\text{neg})} &= \log \frac{0.15}{0.05} = \log 3.0 = 1.0986 \\
  \log \frac{P(\text{because}|\text{pos})}{P(\text{because}|\text{neg})} &= \log \frac{0.1}{0.1} = \log 1.0 = 0.0 \\
  \log \frac{P(\text{I}|\text{pos})}{P(\text{I}|\text{neg})} &= \log \frac{0.2}{0.2} = \log 1.0 = 0.0 \\
  \log \frac{P(\text{love}|\text{pos})}{P(\text{love}|\text{neg})} &= \log \frac{0.1}{0.05} = \log 2.0 = 0.6931 \\
- \end{aligned}
+ \end{align}
  $$
 
  Note that the words "ice" and "cream" are not in the vocabulary, so we ignore them.
 
  Given these ratios, and considering the **log prior**, we can calculate the **log likelihood** of the tweet being positive as follows:
 
  $$
- \begin{aligned}
+ \begin{align}
  \log \frac{P(pos)}{P(neg)} + \sum_{i=1}^{m} \log \frac{P(w_i|pos)}{P(w_i|neg)} &= 0.0 + 0.0 + 0.0 + 1.0986 + 0.0 + 0.0 + 0.6931\\
  &= 1.0986 + 0.6931 \\
  &= 1.7917
- \end{aligned}
+ \end{align}
  $$
 
  Since $1.7917 > 0$, the tweet is classified as **positive**.