Path: utzoo!mnetor!uunet!husc6!bbn!oberon!ll-xn!vanhove
From: vanhove@XN.LL.MIT.EDU (Patrick Van Hove)
Newsgroups: comp.ai
Subject: Re: Probability Bounds from Bayes Theory: (A Problem).
Message-ID: <838@xn.LL.MIT.EDU>
Date: 24 Dec 87 16:32:56 GMT
References: <1578@brahma.cs.hw.ac.uk>
Organization: MIT Lincoln Laboratory, Lexington, MA
Lines: 133
In-reply-to: tom@cs.hw.ac.uk's message of 11 Dec 87 14:10:21 GMT



	Here are my 2 pennies' worth

LITTLE FLAME
------------

	First, a few words about the problem statement. The data
is well explained, but very few words are written about what one is looking
for. In addition, the data "p(~l, ~t) = 0.95"
which appears in the "PRIORS" doesn't seem to appear anywhere in the "PROBLEM".
Now, this piece of data is crucial to the solution, because it tells us that, although
3% have tb and 4% have cancer, these overlap in such a way that the total is only 5%.

MY SOLUTION
-----------

	One can approach this problem in the following way: there are
three events: t, l, pv. The combination of these events results in 8
possible outcomes for every person. These can be represented, for example,
by a Karnaugh map. The complete statistical description of these events
is represented by the probabilities of these 8 events. The following constraints
apply:

The probabilities sum to 1.		(1 =ity constraint)
There are 3 known priors		(3 =ity constraints)
There are 3 known conditional		(3 =ity constraints)
All probabilities are non-negative	(8 ~=ity constraints)

Since there is no redundancy in the equality constraints (proof left to reader),
there is a set of solutions depending on one parameter, and this set is
not empty, as the inequalities can all be satisfied simultaneously.


In a problem of this type, it may always be necessary to determine all the
probabilities to find the desired solution, but in this particular case, it is.

The solution is underdetermined, but the following conclusions can be obtained
by grinding through the algebra (algebra given at the end of this message):

The upper bound of p(t,l|pv) is attained when all the cases of t which are undetected
correspond to patients with t and ~l. 	I found that

	 upper bound [ p(t,l|pv) ] = .1802..

In that case, by the way, p(pv|t,l) is 1.0!

The lower bound of p(t,l|pv) is attained when all the cases where t is undetected
correspond to t and l , whereas all cases of t and ~l are properly detected. I found

	 lower bound [ p(t,l|pv) ] = .1644..

In that case, p(pv|t,l) is 0.925!

It is interesting to see that the extreme bounds for the requested probability
are pretty tight. This is only possible because the prior knowledge that

	p(~l, ~t) = 0.95
	
implies a serious overlap between t and l. As both are detected "most of the time",
their combination must be detected "most of the time" too.
If there was no overlap between t and l, then the answer would be undetermined.
As a consequence, if the data did not guarantee an overlap between t and l,
the bounds on the desired probability would be very weak.

An other interesting aspect of this solution is that it was done based only
on probability theory; it is liekly that, knowing the physics of the real 
problem , one can formulate valid additional constraints.

For example, it might be reasonable to assume that p(pv|l and t)
is at least as large as p(pv|l and ~t) and as p(pv|~l and t);
with this assumption, you would get tighter bounds.
But thus assumption was not part of the problem statement.

Finally, the algebraic statement of the problems has 8 variables
and 7 equality constraints; if you give any additional linear equality constraint
which is not redundant with the other 7, then you reduce to set of
possibilities to a single solution, or none if the inequalities
can't be met. 


THE SOLUTION BY ALGEBRA

Notation: 

x1 = p( l, ~t, ~pv)
x2 = p( l, ~t,  pv)
x3 = p( l,  t, ~pv)
x4 = p( l,  t,  pv)
x5 = p(~l,  t, ~pv)
x6 = p(~l,  t,  pv)
x7 = p(~l, ~t,  pv)
x8 = p(~l, ~t, ~pv)

Constraints:

(1): all xi >= 0
(2): Sum all xi = 1
(3): x1+x2+x3+x4 = 0.04
(4): x3+x4+x5+x6 = 0.03
(5): x7+x8 = 0.95
(6): [x2+x4] / [x1+x2+x3+x4] = 0.9
(7): [x4+x6] / [x3+x4+x5+x6] = 0.95
(8): x7 / [x7+x8] = 0.07

From all the equalities, one can express all the probabilities as a function of x4.

x1 = -0.016 + x4
x2 =  0.036 - x4
x3 =  0.02  - x4
x5 = -0.0185 + x4
x6 =  0.0285 - x4
x7 =  0.0665
x8 =  0.8835

Since x1, x2, x3, x5, x6 must be non-negative, x4 must comply with the following bounds

lower bounds: 0.016 , 0.0185
upper bounds: 0.036, 0.02, 0.0285

As a consequence, we must have that

0.0185 < x4 < 0.02

p(t and l | pv) = p(t and l and pv) / p ( pv )
		= x4 / [ x2+x4+x6+x7 ]
		= x4 / ( 0.131 - x4 )

The bounds are:

0.1644.. < x4 < 0.1802..


	Patrick Van Hove