Probability Help: Bayes

Cantab.

Greetings,

I can follow line 1 to line 2, but cannot follow line 2 to 3 and line 3 to 4 for the life of me.

It's straight out of a well-regarded paper so I'm sure there's no mistake.

Can anyone help?

Cantab.

I think I've figured out line 2 to 3, based on the following:



Could someone explain this equation to me?

Fremen

The equation you posted in your second comment is known as the law of total probability. It's a bit easier to understand if you ignore the Y term initially. Once you've got your head around that, you can just condition everything on Y.

The idea follows from Bayes' theorem. I'll use "n" for intersection since I'm not bothered typesetting into a pic and linking (bloody boards :rolleyes: )

P(A) = P(A n + P(A n B compliment)

and Bayes gives expressions for the probability of both of these intersections. This is true more generally for disjoint sets whose union has probability 1, which is exactly what those elements z in Z are.

Seems like you're missing a "b" in P(o | sprime, a) just before the summation symbol in the third equality. Is this a mistake?

The notation in the last line is not standard. Does he give any clues in the paper? Have you got a link to the paper? I could take a look if you want.

Cantab.

Fremen wrote: »

The equation you posted in your second comment is known as the law of total probability. It's a bit easier to understand if you ignore the Y term initially. Once you've got your head around that, you can just condition everything on Y.

The idea follows from Bayes' theorem. I'll use "n" for intersection since I'm not bothered typesetting into a pic and linking (bloody boards :rolleyes: )

P(A) = P(A n + P(A n B compliment)

and Bayes gives expressions for the probability of both of these intersections. This is true more generally for disjoint sets whose union has probability 1, which is exactly what those elements z in Z are.

Seems like you're missing a "b" in P(o | sprime, a) just before the summation symbol in the third equality. Is this a mistake?

The notation in the last line is not standard. Does he give any clues in the paper? Have you got a link to the paper? I could take a look if you want.

Hi Freman,

Thanks for the message. Just knowing the phrase "law of total probability" makes my search so much easier. I didn't get what you said with:

This is true more generally for disjoint sets whose union has probability 1, which is exactly what those elements z in Z are.

You're right about the missing "b". I believe I can ignore this because it so happens that the probability doesn't depend on b and can be safely dropped.

You're right about the last line: it's non-standard notation.

You're probably dying to know the paper:
http://www.cis.upenn.edu/~mkearns/papers/barbados/klc-pomdp.pdf

Thanks again.

Fremen

Sorry, should have tried to make that clearer. I meant, it holds if

P(z1 U z2 U ... U zn) = 1

and for any two distinct choices of the z's, say z2 and z6,

P(z3 n z6) = 0

In other words, the calculation holds for a set of distinct events, providing one and only one of the events can happen.

Fremen

If nothing depends on b, then he's just defining

O(sprime,a o) = P(o | a, sprime)

T(s,a,sprime) = P(sprime | a,b,s)

b(s) = P(s | a,b)