[20分] 數學高手們請進!!!請問bayes theorem 入面個分母點樣計到?

2007-09-29 3:58 am
我之前都問過類似問題, 不過我問得唔好,而家再問多次, 條式最簡單o個陣個樣係甘樣o既:

P(A | B) = P(B | A) * P(A) / P(B)

甘如果我有幾個evidence:B1,B2,B3

又想計P(A | B1 B2 B3)

甘我條式應該甘樣啦:

P(A | B1 B2 B3) = P(B1 | A) * P(B2 | A) * P(B3 | A) * P(A) / P(B)

分子就無問題喇, 甘P(B)點樣可以計番出黎呀?
如果而家我知道B1 B2 B3係independent o既話
佢地o既joint probability會唔會等於P(B)?
定係佢地o既union probability先會等於P(B)?

唔該晒
更新1:

我睇wikipedia分母o個部分: http://en.wikipedia.org/wiki/Naive_bayes_classifier 佢寫P(F1, ... ,Fn) 甘我一路用個B字黎代表...其實都一樣: P(B1,B2 ... Bn) 甘樣其實我想問B1 B2 B3 依d 係咪可以計番個分母出黎架

更新2:

P(F1, ... ,Fn) <<< 其實佢甘樣寫 係咪等於P(F1 and F2 and F3 ... and Fn) 定係還是等於P(F1 or F2 or F3 ... or Fn) 邊個先o岩架?

回答 (1)

2007-09-29 10:21 am
✔ 最佳答案
If B1 B2 B3係independent, P(B) = P(B1andB2andB3) = P(B1)*P(B2)*P(B3)
**providing B = B1, B2 and B3 all happen.

Union probability:
P(B1orB2orB3)
= P(B1) + P(B2) + P(B3) - [P(B1andB2) + P(B1andB3) + P(B2andB3)]+
P(B1andB2andB3)
=P(B1) + P(B2) + P(B3) - [P(B1)P(B2) + P(B1)P(B3) + P(B2)P(B3)] + P(B1)P(B2)P(B3)**


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