Re: Combining Conditional Probabilities

V wrote:

Let say I know the probabilities of:

p(x|a,b)
p(x|b,c)
p(a), p(b), p(c), p(x)

How can I use this to say anything about:

p(x|a,b,c)

Obviously there is not enough information to fully
determine p(x|a,b,c), but I only need a reasonable
approximation.

First of all, nothing in the information you give precludes P(abc) = 0, so P(x|abc) maynot even be defined. Supppose we can come up with a positive lower bound d for P(abc).

The best you can do is come up with bounds. There are fifteen degrees of freedom for this distribution; you have specified only six. Thus, you can come up with simultaneous equalities and inequalities.

For example, let us work in the following fifteen variables: Let ' indicate complement.
P(abc), P(abc'), P(ab'c), P(ab'c'), P(a'bc), P(a'bc'), P(a'b'c), [sic]
P(x|abc), P(x|abc'), P(x|ab'c), P(x|ab'c'), P(x|a'bc), P(x|a'bc'), P(x|a'b'c), P(x|a'b'c').
All of these must be nonnegative.

Then we have
P(abc) >= d
P(abc) + ... + P(a'b'c) <= 1
P(x|abc) P(abc) + ... + P(x|a'b'c) P(a'b'c) + P(x|a'b'c') {1 - [P(abc) + .... + P(a'b'c)]} = Px
P(abc) + P(abc') + P(ab'c) + P(ab'c') = Pa
P(abc) + P(abc') + P(a'bc) + P(a'bc') = Pb
P(abc) + P(ab'c) + P(a'bc) + P(a'b'c) = Pc
P(x|abc) P(abc) + P(x|abc') P(abc') = P(x|ab) [P(abc) + P(abc')]
P(x|abc) P(abc) + P(x|a'bc) P(a'bc) = P(x|bc) [P(abc) + P(a'bc)]

I think that's all we can say. Solve these to come up with bounds on P(x|a,b,c). Admittedly, this may not easy. What we have here is a nonlinear program - a linear objective function subject to linear and quadratic constraints. Alternatively, working instead with the variables P(xabc), P(xabc'),..., P(x'a'b'c), we could find the extrema of the nonlinear function
P(xabc) / [P(xabc) + P(x'abc)] subject to only linear constraints.

See also the thread with subject, "conditional probability," started by Tino on 21 July 2005.

--
Stephen J. Herschkorn                        sjherschko@xxxxxxxxxxxx
Math Tutor in Central New Jersey and Manhattan
.