Answers.Body: <p>The answer from @kiner_shah in the comments has solved it:</p>
  
  <blockquote>
    <p>They have eliminated it because in comparing probability for best outcome, it would just introduce additional division and the divisor p(x) is same for all candidates. For example, if I tell you to compare the largest of the following numbers: 3/2, 5/2, 1/2, you will without thinking pick 5/2 since the divisors are same and 5 is greatest of all.</p>
  </blockquote>
  
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2017-12-02T16:50:43.943Z
Answers.LastActivityDate: 2017-12-02T16:50:43.943Z
Answers.UserDisplayName: gcorso
Body: <p>The naïve Bayes' generative algorithm is often represented by the following formula:</p>
  <p><span class="math-container">$$\text{argmax}_{y} p(y|x) = \text{argmax}_y \frac{p(x|y)p(y)}{p(x)} \approx \text{argmax}_y p(x|y)p(y)$$</span></p>
  <p>Why do we have <span class="math-container">$p(x)=1$</span> which allows the approximation <span class="math-container">$\text{argmax}_y \frac{p(x|y)p(y)}{p(x)} \approx \text{argmax}_y p(x|y)p(y)$</span>?</p>
  
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2017-12-02T16:37:02.960Z
Comments.Text: They have eliminated it because in comparing probability for best outcome, it would just introduce additional division and the divisor `p(x)` is same for all candidates. For example, if I tell you to compare the largest of the following numbers: 3/2, 5/2, 1/2, you will without thinking pick 5/2 since the divisors are same and 5 is greatest of all.
Comments.UserDisplayName: kiner_shah
Tags: <machine-learning><math><naive-bayes><bayes-theorem>
Title: Why is the denominator ignored in the Bayes' rule?
UserDisplayName: gcorso
ThreadText: # Why is the denominator ignored in the Bayes' rule?
  
  (Asked by: username_0 on 2017-12-02)
  
  The naïve Bayes' generative algorithm is often represented by the following formula:
  
  $$\text{argmax}\_{y} p(y|x) = \text{argmax}\_y \frac{p(x|y)p(y)}{p(x)} \approx \text{argmax}\_y p(x|y)p(y)$$
  
  Why do we have $p(x)=1$ which allows the approximation $\text{argmax}\_y \frac{p(x|y)p(y)}{p(x)} \approx \text{argmax}\_y p(x|y)p(y)$?
  
  **Tags:** machine-learning, math, naive-bayes, bayes-theorem
  > **Comments:**
  > **username_1**: They have eliminated it because in comparing probability for best outcome, it would just introduce additional division and the divisor `p(x)` is same for all candidates. For example, if I tell you to compare the largest of the following numbers: 3/2, 5/2, 1/2, you will without thinking pick 5/2 since the divisors are same and 5 is greatest of all.  
  
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  ## 1 answers
  
  ### Answer by username_0 on 2017-12-02. Score: 4 | Accepted answer
  
  The answer from @kiner\_shah in the comments has solved it:
  
  > They have eliminated it because in comparing probability for best outcome, it would just introduce additional division and the divisor p(x) is same for all candidates. For example, if I tell you to compare the largest of the following numbers: 3/2, 5/2, 1/2, you will without thinking pick 5/2 since the divisors are same and 5 is greatest of all.
  
  
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