Don't know as I can help much, but a few questions occur to mind. 
It strikes me that there may be some constraints on the system that you 
haven't mentioned.  For example:
 1.  Are features L and R mutually exclusive, so that if L is found 
        at position i, R cannot be found at that position?
 2.  Are L and R exhaustive, so that P(Li) = 1 - P(Ri)?  If not, is there 
        some other identifiable constraint, perhaps of the form
        P(Li) + P(Ri) = k < 1  (or <=, or >=, instead of =)?
 3.  Are R and L geographically constrained, so to speak -- e.g., when L 
        is at postion i, P(Rj) = 0 for j = i-2,...,i+2 ?  (Or, 
        alternatively, sum(P(Rj)) = 1 for j in some neighborhood of i.) 
 4.  In vector F, for features L and R, are the numbers of L's and R's 
        constrained in any way?  E.g., L and R both occur exactly once, 
        or at least twice, or at most five times, or ...
5.  Can either feature occur more than once at any position i ?

On Thu, 15 Jun 2000, Mark Everingham wrote:

> Any help, hints, direction etc. on this appreciated:
> 
> I have a vector of measurements F, and two feature models L and R which
> give the pdf of a measurement Fi for that model i.e. P(Fi|Li) and
> P(Fi|Ri). Denoting the probability that there is a feature L at position
> i in the vector as P(Li) and similarly feature R at position j as P(Rj)
> I am interested in obtaining
> 
> P(Li,Rj|F)
> 
> i.e. the probability that there is feature L at position i and feature R
> at position j, given the vector F. The assumptions I will make are
> 
> P(Li|F)=P(Li|Fi) and P(Ri|F)=P(Ri|Fi)
> 
> i.e. the feature depends only on local value of the vector. What I will
> *not* assume is that P(Li) and P(Rj) are independent. i.e. the position
> of feature L influences that position of feature R.
> 
> What I am looking for is an expression for P(Li,Rj|F) derived only from
> P(Fi|Li), P(Fj|Rj) and P(Li|Rj) or similar.
> 
> P(Li,Rj|F)
> =P(Li|Rj,F)P(Ri,F)
> =P(Li|Rj,F)P(Ri,Fi)
> ...?

Should not the second factor read P(Rj|F) rather than P(Ri,F) ?
 
> Knowing that P(Li|F)=P(Li|Fi) can I simplify P(Li|Rj,F) further?

Sorry I can't contribute any more;  I'm woefully out of practice on 
this sort of probability calculus.
                                        -- DFB.
 ------------------------------------------------------------------------
 Donald F. Burrill                                 [EMAIL PROTECTED]
 348 Hyde Hall, Plymouth State College,          [EMAIL PROTECTED]
 MSC #29, Plymouth, NH 03264                                 603-535-2597
 184 Nashua Road, Bedford, NH 03110                          603-471-7128  


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