Does anyone know how to get the right term? It seems like the right term is actually $P(C, x_1, x_2, \ldots, x_n)$ and not $P(C \mid x_1, x_2, \ldots, x_n)$?
How is the right term derived in this conditional probability statement?
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probability
statistics
naive-bayes
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1Yes it is, but both in the numerator and the denominator, so the $P(x_1,...,x_n)$ is divided in both. Have a look at Bayes' theorem. – 2017-02-01
1 Answers
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As Blaza said, a common devisor cancels, is all.
$\require{cancel}\begin{align}\dfrac{\mathsf P\big(C=c^1\mid \bigcap\limits_{i=1}^n (X_i=x_i)\big)}{\mathsf P\big(C=c^2\mid \bigcap\limits_{i=1}^n (X_i-x_i)\big)} ~ & = ~ \dfrac{\mathsf P\big((C=c^1)\cap\bigcap\limits_{i=1}^n(X_i=x_i)\big)/\cancel{\mathsf P\big(\bigcap\limits_{i=1}^n(X_i=x_i)\big)}}{\mathsf P\big((C=c^2)\cap\bigcap\limits_{i=1}^n(X_i=x_i)\big)/\cancel{\mathsf P\big(\bigcap\limits_{i=1}^n(X_i=x_i)\big)}} \\[1ex] &=~ \dfrac{\mathsf P(C=c^1)~\mathsf P\big(\bigcap\limits_{i=1}^n(X_i=x_i)\mid C=c^1\big)}{\mathsf P(C=c^2)~\mathsf P\big(\bigcap\limits_{i=1}^n(X_i=x_i)\mid C=c^1\big)}\\[2ex] &=~\dfrac{\mathsf P(C=c^1)}{\mathsf P(C=c^2)}\prod_{i=1}^n\dfrac{\mathsf P(X_i=x_i\mid C=c^1)}{\mathsf P(X_i=x_i\mid C=c^2)}\end{align}$
The last step relies on the conditional mutual independence of $(X_\star)$ when given a value for $C$.
Anything else?