What Is the Probability of a Conjunction of N Non-Independent Hypotheses?

[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

 

[EnumaElish]

I think it is.

 

[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).