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bayes_theorem [2013/03/30 20:00]
jongbor [Decompose a conditional probability into a product]
bayes_theorem [2014/04/19 15:41] (current)
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 $$ P(A,B) = P(A|B)P(B) = P(B|A)P(A) $$ $$ P(A,B) = P(A|B)P(B) = P(B|A)P(A) $$
  
-This can be rearranged as:+This can be rearranged as the **product rule**:
  
 $$ P(A,B) = P(A)P(B|A) $$ $$ P(A,B) = P(A)P(B|A) $$
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 $$ P(A,B|C) = \frac{P(A,​B,​C)}{P(C)} = \frac{P(A,​C)}{P(C)}\frac{P(A,​B,​C)}{P(A,​C)} = P(A|C)P(B|A,​C) $$ $$ P(A,B|C) = \frac{P(A,​B,​C)}{P(C)} = \frac{P(A,​C)}{P(C)}\frac{P(A,​B,​C)}{P(A,​C)} = P(A|C)P(B|A,​C) $$
  
-Luckily, this is the same as $P(A,B) = P(A)P(B|A)$,​ but adding $C$ in the conditional part of the probability.+Luckily, this is the same as the product rule $P(A,B) = P(A)P(B|A)$,​ but adding $C$ in the conditional part of the probability.
  
 ==== Generalization to $n$ variables ==== ==== Generalization to $n$ variables ====