# Baye's Theorem w/ Multiple Events

1. May 17, 2010

### dnbwise

So, Bayes' Theorem states

P(H|D) = P(D|H) X P(H) / [P(D|H) X P(H) + P(D|~H) x P(~H)], where ~ = negation

If my hypothesis consists of two events - S and O - would this be

P(S,O|D) = P(D|S,O) X P(S,O) / [P(D|S,O) X P(S,O) + P(D|~S,~O) x P(~S,~O)]

or this

P(S,O|D) = P(D|S,O) X P(S,O) / [P(D|S,O) X P(S,O) + P(D|~S,O) x P(~S,O) + P(D|S,~O) x P(S,~O) + P(D|~S,~O) x P(~S,~O)] ?

2. May 18, 2010

### ych22

the second formulation looks right.

3. May 20, 2010

### SW VandeCarr

Show how P(D)=P(D|H)P(H)+P(D|1-H)P(1-H)

Last edited: May 20, 2010
4. May 20, 2010

### ych22

HnH' = 1

D = HnD+ H'nD

P(D)= P (HnD) + (H'nD)= P(D | H) P(H) + P(D | H') P(H')

5. May 20, 2010

### SW VandeCarr

Can you derive this from Bayes' Theorem?:

P(D)=P(D|H)P(H)/P(H|D) is equivalent to P(D)=P(D|H)P(H)+P(D|1-H)P(1-H)?

EDIT: I think it's more clear to use probability notation P(1-H) rather than logical notation P(~H) in what is essentially a conditional probability problem.

Last edited: May 20, 2010
6. May 25, 2010

### luma

I think you meant

HuH' = 1

Given that D = HnD + H'nD can we assume that P(D) = P(HnD) + P(H'nD) ?

HuH' = 1
or H' = 1 - H
HnD = P(D|H)P(H)
H'nD = P(D|H - 1)(1 - P(H))

P(D) = P(D|H)P(H) + P(D|H - 1)(1 - P(H))

but bayes theorem says P(D)=P(D|H)P(H)/P(H|D)

so i think it's Ok

7. May 25, 2010

### SW VandeCarr

Note:

D=P(D|H)P(H)+P(D|1-H)P(1-H) is in terms of joint probabilities (terms cancel);

D=P(D|H)P(H)/P(H|D) is in terms of a conditional probability.

You can express D either way and you can write a conditional probability in terms of joint probabilities as the OP has done. However, you cannot write D in terms of Bayes Theorem without a conditional probability.

8. May 26, 2010

### SW VandeCarr

Sorry, that's P(D) where D appears in the previous post.