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• If <math>P\left(B\right)=\sum_{i}P\left(B\cap A_{i}\right)=\sum_{i}P\left(B|A_{i}\right)P\left(A_{i}\right)</math> , then <math>P\left(A_{i}|B\right)=\frac{P\left(A_{i}\cap B\right)}{P\left(B\right)}=\frac{P\left(B|A_{i}\right)A_{i}}{\sum_{j}P\left(B|A_{j}\right)A_{j}}</math> | • If <math>P\left(B\right)=\sum_{i}P\left(B\cap A_{i}\right)=\sum_{i}P\left(B|A_{i}\right)P\left(A_{i}\right)</math> , then <math>P\left(A_{i}|B\right)=\frac{P\left(A_{i}\cap B\right)}{P\left(B\right)}=\frac{P\left(B|A_{i}\right)A_{i}}{\sum_{j}P\left(B|A_{j}\right)A_{j}}</math> | ||
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Revision as of 12:48, 16 November 2010
1.3 Bayes' theorem
• = Bayes' rule
• $ P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{P\left(B|A\right)P\left(A\right)}{P\left(B\right)} $
• If $ P\left(B\right)=\sum_{i}P\left(B\cap A_{i}\right)=\sum_{i}P\left(B|A_{i}\right)P\left(A_{i}\right) $ , then $ P\left(A_{i}|B\right)=\frac{P\left(A_{i}\cap B\right)}{P\left(B\right)}=\frac{P\left(B|A_{i}\right)A_{i}}{\sum_{j}P\left(B|A_{j}\right)A_{j}} $
