<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>

CC BY-SA 3.0
2017-12-02T16:50:43.943Z
2017-12-02T16:50:43.943Z
gcorso
<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>

CC BY-SA 3.0
2017-12-02T16:37:02.960Z
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.
kiner_shah
<machine-learning><math><naive-bayes><bayes-theorem>
Why is the denominator ignored in the Bayes' rule?
gcorso
# 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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