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Motivating the Gaifman condition on first-order probabilities

0

$begingroup$

I've recently been looking at the literature on assigning probabilities to first-order formulae (not just propositional formulae). Following the treatment here, let $L$ be a first-order language, let $D$ be some domain, and let $L(D)$ be $L$ expanded by enough constants $a_1, a_2, dots$ to name all elements of $D$. Gaifman showed in 1964 that if we are given a probability map from the quantifier-free sentences of $L(D)$ to $[0, 1]$, then there is a unique extension of that to a probability map from the sentences of $L(D)$ to $[0, 1]$ satisfying the following condition:
$$ p(exists x phi(x)) = mathrmsup p(phi(a_i_1) lor cdots lor phi(a_i_k)) $$
where the supremum is taken over all finite sets of constants.

My question is this: what is the motivation for imposing the Gaifman condition?

If I'm reading it right, then it identifies the probability of an existentially quantified claim with the maximal probability assigned to some disjunction of substitution-instances.

But don't we want to allow that the probability of an existentially quantified statement might be greater than the probability of any given finite disjunction (for, roughly, the same reasons that we allow that the probability of a disjunction might be higher than the probabilities of either of its disjuncts)?

probability-theory first-order-logic predicate-logic

share|cite|improve this question

edited yesterday

Lee David Chung Lin

4,50841342

asked Mar 26 at 9:59

anygivenpointanygivenpoint

1305

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not clear... He uses $text sup$; this means that the existential statement will have a prob Greater-or-equal to...
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 26 at 10:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh that's a very good point (which I'm now embarrassed to have missed)!
  $endgroup$
  – anygivenpoint
  Mar 26 at 10:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Basically, I imagine that if there is no... we want that its prob will be $0$, and thus every finite disjunct must be False (otherwise at least one of them will have a positive prob and the sup will be $ne 0$). On the other end, if there is some, at least one finite disjunct will be True, and its prob will be Greater than $0$ (I imagine).
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 26 at 10:52
  

  
  
  
  

add a comment | 

0

$begingroup$

I've recently been looking at the literature on assigning probabilities to first-order formulae (not just propositional formulae). Following the treatment here, let $L$ be a first-order language, let $D$ be some domain, and let $L(D)$ be $L$ expanded by enough constants $a_1, a_2, dots$ to name all elements of $D$. Gaifman showed in 1964 that if we are given a probability map from the quantifier-free sentences of $L(D)$ to $[0, 1]$, then there is a unique extension of that to a probability map from the sentences of $L(D)$ to $[0, 1]$ satisfying the following condition:
$$ p(exists x phi(x)) = mathrmsup p(phi(a_i_1) lor cdots lor phi(a_i_k)) $$
where the supremum is taken over all finite sets of constants.

My question is this: what is the motivation for imposing the Gaifman condition?

If I'm reading it right, then it identifies the probability of an existentially quantified claim with the maximal probability assigned to some disjunction of substitution-instances.

But don't we want to allow that the probability of an existentially quantified statement might be greater than the probability of any given finite disjunction (for, roughly, the same reasons that we allow that the probability of a disjunction might be higher than the probabilities of either of its disjuncts)?

probability-theory first-order-logic predicate-logic

share|cite|improve this question

edited yesterday

Lee David Chung Lin

4,50841342

asked Mar 26 at 9:59

anygivenpointanygivenpoint

1305

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not clear... He uses $text sup$; this means that the existential statement will have a prob Greater-or-equal to...
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 26 at 10:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh that's a very good point (which I'm now embarrassed to have missed)!
  $endgroup$
  – anygivenpoint
  Mar 26 at 10:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Basically, I imagine that if there is no... we want that its prob will be $0$, and thus every finite disjunct must be False (otherwise at least one of them will have a positive prob and the sup will be $ne 0$). On the other end, if there is some, at least one finite disjunct will be True, and its prob will be Greater than $0$ (I imagine).
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 26 at 10:52
  

  
  
  
  

add a comment | 

$begingroup$

I've recently been looking at the literature on assigning probabilities to first-order formulae (not just propositional formulae). Following the treatment here, let $L$ be a first-order language, let $D$ be some domain, and let $L(D)$ be $L$ expanded by enough constants $a_1, a_2, dots$ to name all elements of $D$. Gaifman showed in 1964 that if we are given a probability map from the quantifier-free sentences of $L(D)$ to $[0, 1]$, then there is a unique extension of that to a probability map from the sentences of $L(D)$ to $[0, 1]$ satisfying the following condition:
$$ p(exists x phi(x)) = mathrmsup p(phi(a_i_1) lor cdots lor phi(a_i_k)) $$
where the supremum is taken over all finite sets of constants.

My question is this: what is the motivation for imposing the Gaifman condition?

If I'm reading it right, then it identifies the probability of an existentially quantified claim with the maximal probability assigned to some disjunction of substitution-instances.

But don't we want to allow that the probability of an existentially quantified statement might be greater than the probability of any given finite disjunction (for, roughly, the same reasons that we allow that the probability of a disjunction might be higher than the probabilities of either of its disjuncts)?

probability-theory first-order-logic predicate-logic

share|cite|improve this question

edited yesterday

Lee David Chung Lin

4,50841342

asked Mar 26 at 9:59

anygivenpointanygivenpoint

1305

$endgroup$

share|cite|improve this question

edited yesterday

Lee David Chung Lin

4,50841342

asked Mar 26 at 9:59

anygivenpointanygivenpoint

1305

share|cite|improve this question

edited yesterday

Lee David Chung Lin

4,50841342

asked Mar 26 at 9:59

anygivenpointanygivenpoint

1305

edited yesterday

Lee David Chung Lin

4,50841342

edited yesterday

Lee David Chung Lin

4,50841342

asked Mar 26 at 9:59

anygivenpointanygivenpoint

1305

asked Mar 26 at 9:59

anygivenpointanygivenpoint

1305