On Mon, Feb 25, 2008 at 8:18 PM, Ed Porter <[EMAIL PROTECTED]> wrote: > > As you all know the Naïve Bayes formula for the conditional probability of H > given evidence E1, E2,...EN is > > p(H|E1,E2,...EN) = p(H) * p(E1|H)/p(E1) * p(E2|H)/p(E2) *...* > p(EN|H)/p(EN)
Hi Ed, This variant is more restricted than is necessary in some cases: it needs not just conditional independence of Ei on hypothesis, that is P(Ei|Ej,H)=P(Ei|H), but also mutual independence of Ei, that is P(E1,...,En)=P(E1)*...*P(En). Using P(Ei,H)=P(Ei|H)*P(H)=P(H|Ei)*P(Ei), or P(Ei|H)/P(Ei)=P(H|Ei)/P(H) you expressed the formula in terms of P(H|Ei) and P(H) rather than P(Ei), P(Ei|H) and P(H). It's clearly equivalent, but it also introduces (n-1)'th power of P(H), which will have to figure in MAP and looks more cumbersome. -- Vladimir Nesov [EMAIL PROTECTED] ------------------------------------------- agi Archives: http://www.listbox.com/member/archive/303/=now RSS Feed: http://www.listbox.com/member/archive/rss/303/ Modify Your Subscription: http://www.listbox.com/member/?member_id=8660244&id_secret=95818715-a78a9b Powered by Listbox: http://www.listbox.com
