From: radford@cs.toronto.edu (Radford Neal)
Newsgroups: sci.math,sci.philosophy.tech,comp.theory,sci.math.stat
Subject: Re: Combining probabilities?
Date: 20 Nov 92 20:33:40 GMT
Organization: Department of Computer Science, University of Toronto

In article <1992Nov20.172443.17984@njitgw.njit.edu> aaron@funcity.njit.edu (Aaron Watters) writes:

>I'm interested in the question of combining probabilities and any
>mathematical foundations thereof.
>
>   If sensor A says that object X is at position
>       P with 45% probability
>       Q with 54% probability
>       R with 1% probability and
>      sensor B says that object X is at position
>       P with 1% probability
>       Q with 36% probability
>       R with 63% probability

Your problem is that sensors are inherently incapable of saying things
like this. Sensors should be characterized by "likelihoods", not
"probabilities". (As technical terms, these are NOT synonyms.)  

To find the probability that X is at P, you need to know everything,
such as whether whoever owns the land at P is willing to let X be
there, or whether they patrol the place with a shotgun and shoo them
off immediately if they show up (in which case, any sensor saying
they're there is probably wrong). Likelihoods are different, they just
say how likely the sensor is to give various readings in various
circumstances, without any comittment to how likely these circumstances 
might be.

Suppose sensor A produces some reading, call it R_A, and sensor B
produces some reading, R_B. The designer of the sensor can say how
likely these reading are to occur given that the object is really in
certain positions. Consulting the manuals, you may find that:

  Probability (Sensor A reads R_A GIVEN THAT X is at P) = 0.30
  Probability (Sensor A reads R_A GIVEN THAT X is at Q) = 0.50
  Probability (Sensor A reads R_A GIVEN THAT X is at R) = 0.01

  Probability (Sensor B reads R_B GIVEN THAT X is at P) = 0.01
  Probability (Sensor B reads R_B GIVEN THAT X is at Q) = 0.40
  Probability (Sensor B reads R_B GIVEN THAT X is at R) = 0.70

Note that likelihoods don't have to sum up to anything in particular.

Now, IF you can assume that the behaviour of the two sensors is 
independent, given a particular position for X, you can conclude 
that 

  Prob (A reads R_A AND B reads R_B GIVEN X is at P) = 0.30*0.01 = 0.003
  Prob (A reads R_A AND B reads R_B GIVEN X is at Q) = 0.50*0.40 = 0.090
  Prob (A reads R_A AND B reads R_B GIVEN X is at R) = 0.01*0.70 = 0.007

I.e. you can combine the likelihoods from the two sensors. The results
are NOT probabilities - for one thing they don't add up to one. To get
probabilities, you need to put in your PRIOR probabilities for X being
at P, Q, or R. Suppose that

  Prob (X is at P) = 0.3
  Prob (X is at Q) = 0.2      (Note: these are the probabilities before
  Prob (X is at R) = 0.5             you know the sensor readings)

Then to get the probabilities after you know the sensor readings, you
multiply these prior probabilities by the combined likelihoods, and
rescale so they add up to one:

  Prob (X is at P GIVEN A reads R_A AND B reads R_B) = K*0.3*0.003 = K*0.0009
  Prob (X is at Q GIVEN A reads R_A AND B reads R_B) = K*0.2*0.090 = K*0.0180
  Prob (X is at R GIVEN A reads R_A AND B reads R_B) = K*0.5*0.007 = K*0.0035

To get these to sum to one, we need K = 0.0009+0.0180+0.0035, giving,
approximately:

  Prob (X is at P GIVEN A reads R_A AND B reads R_B) = 0.04
  Prob (X is at Q GIVEN A reads R_A AND B reads R_B) = 0.80
  Prob (X is at R GIVEN A reads R_A AND B reads R_B) = 0.16

This procedure is an application of "Bayes' Theorem".

Hope that helps,

      Radford Neal
Newsgroups: sci.math