Probability of n hypothesis

1. Sep 13, 2007

Aldebaran2

Ciao all,

I have a conjunction of n non-independant hypothesis:

h1 and h2 and .... and hn.

I want to calculate the probability of such conjunction. To do so, I used the product rule p(h1 and h2) =p(h1).p(h2|h1)

p(h1 and h2 and .... and hn) = p(h1).p(h2|h1).p(h3|h1 and h2 )....

Do you think that this is correct?

Aldebaran

2. Sep 13, 2007

EnumaElish

I think it is.

3. Sep 14, 2007

Aldebaran2

Proof

Yes the proof was pretty simple:

p(a1 and a2 and... and an)
= p(an | a1 and a2 and... and an-1). p(a1 and a2 and... and an-1).
= p(an | a1 and a2 and... and an-1). p(an-1 | a1 and a2 and... and an-2).p(a1 and a2 and... and an-2).

:rofl: