# Multiple events and using bayes theorem

1. Oct 8, 2011

### skyflashings

I am not sure if I am doing this correctly, so please check my attempt:

I have four discrete random variables: G, H, J, L

Say, I want to find P(H|J, G, L).

Then can I write as P(H|J, G, L) = P(J, G, L|H)*P(H) = P(J|G, L, H)*P(G|L, H)*P(L|H)*P(H)

Are those equivalent? I can't find an example exactly like this one; I just want to make sure that it works this way. Thanks!

2. Oct 9, 2011

### Stephen Tashi

No. P(J,G,L | H) P(H) is equal to P(H,J,G,L), which is not the same as P(H|J,G,L)

$$P(H| J \cap G \cap L) = \frac{ P(H \cap J \cap G \cap L)}{P( J \cap G \cap L)}$$

There are various ways to rewrite the right hand side.

For example, the numerator can be written as:

$$P( H \cap J \cap G \cap L) = P(H | J \cap G \cap L) P(J \cap G \cap L)$$
$$= P(H| J \cap G \cap L) P(J | G \cap L) P(G \cap L)$$
$$= P(H| J \cap G \cap L) P(J | G \cap L) P(G | L) P(L)$$

which is similar to what you had in mind. You could also write it as you did, with P(H) as the last factor.

Last edited: Oct 9, 2011
3. Oct 9, 2011

### skyflashings

Ok, I see how it unfolds now.

One of the reasons I wanted to get my final form is that I only have a distribution for P(H), P(J|H), P(G|L,H), and P(L|H) and there is a conditional independence assumption that P(J|H)=P(J|G,L,H).

So, I need to find P(H|J, G, L) in terms of those.

If I were to follow your steps to solve for P(H|J, G, L), would it look something like:

P(H|J, G, L) = P(H, J, G, L) / ((P(H|J, G, L)*P(J|G, L)*P(G|L)*P(L))

I'm not sure how to proceed at that point..

Thanks for the help, I hope I can wrap my head around this.

4. Oct 9, 2011

### Stephen Tashi

I think you need more information to calculate P(H| J, G, L). You need to be able to calculate P(J,G,L) without the condition that the event H is true. If you know the problem has an answer, think about whether there is other known information.