# Kolmogorov probability axioms without non-negativity condition

What is a minimal consistent modification of probability axioms to include negative values? Is it enough to use a minimal modification of axioms obtained by formal exclusion of non-negativity requirement, i.e.:

There are sample space $\Omega$, event space $F$ ($\sigma$-algebra of subsets of $\Omega$) and a function $P$ satisfying axioms:

(A1) $P(E) \in {\mathbb R}$, $\forall E \in F$

(A2) $P(\Omega) = 1$

(A3) Any countable sequence of pairwise disjoint elements $E_k \in F$ satisfies

$$P(\bigcup_k E_k) = \sum_k P(E_k).$$

So, the only difference with standard probability axioms is lack of condition $P(E) \geq 0$ in A1.

I am not quite sure, is it necessary to add yet another axiom?:

(A0) $P(\emptyset) = 0$

From the one hand, the axiom A0 together with A1 and A3 define a signed measure and so known to be consistent. Definition above could be shortly rewritten as: Extended ("negative") probability space $\Omega =(\Omega,F,P)$ is signed measure space with $P(\Omega)=1$.

On the other hand in usual probability theory A0 is consequence of other axioms and I suppose that here it also follows from application of A3 to formal expression $P(E \cup \emptyset) = P(E)$ with arbitrary $E$.

It is possible to prove that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ and for complementary event $P(\overline{E}) \equiv P(\Omega \setminus E) = 1 - P(E)$.

Yet, monotonicity does not follow from the axioms, i.e., for $A \subseteq B$ it is not necessary $P(A) \leq P(B)$.

The question: if axioms A1, A2, A3 (maybe, together with A0) define minimal logically consistent model of "extended" probabilities? If yes, that is the possible caveats (e.g. some essential theorems are not valid or useful tools do not work - cf lack of monotonicity).

Note: My interest was inspired by an application to geometrical probability, but below is suggested a more elementary example (it may be omitted, I wrote that due to a reasonable question about interpretation of extended probabilities).

There is a family with father, mother, son and daughter. The family is poor and may only buy Xmas gift for single person. So each year son or daughter may have a present with equal probabilities and (in average) we get a distribution of gifts: father: 0, mother: 0, son: 0.5 and daughter 0.5

But let's suppose, that parents do not want to upset both children and after buying a gift they also search for gifts received during own childhood to present one to second child. So both children have gifts, but one of parents lost his own old gift. Now distribution of gifts may be formally written: father: -0.5, mother: -0.5, son: 1 and daughter: 1

In fact, the example shows, that $P > 1$ may cause even more objections against probabilistic interpretation, than $P < 0$. Here we have distribution of gifts: parents: -1, children: 2. Indeed, $P(\overline{E}) > 1$ is inevitable consequence of $P(E) < 0$ due to axiom A2.

[EDIT 2-Dec-2011] With taking into account comments of Andreas Blass and Emil Jeřáb,

I could suggest some clarification:

The question: if axioms A1, A2, A3 (maybe, together with A0) - are logically consistent system for "extended" probabilities? If yes, which are the possible caveats (e.g. some essential theorems are not valid or useful tools do not work - cf lack of monotonicity).

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The question, whether the proposed axioms "define minimal logically consistent model of 'extended probabilities'," is not very clear. The axioms are logically consistent, but (like most consistent axiom systems) they have many models, so I wouldn't say they define a model at all. Nor is there a clear ordering relation between models, with which to make sense of "minimal." Finally, since "extended probabilities" are only described by an example, it's not clear what would be required of a "model" of them. –  Andreas Blass Dec 1 '11 at 13:54
@Andreas Blass: I agree, it may be not very clear - I added a note about "caveats" partially due to that... –  Alex 'qubeat' Dec 1 '11 at 14:00
I concur with Andreas that the question is quite unclear. However, I note that in reasonable logic, if a set of axioms is consistent, then every its subset is also consistent. Thus, the one and only minimal consistent axiom system is the one with no axioms at all. –  Emil Jeřábek Dec 1 '11 at 14:10
Emil's comment about the minimal consistent axiom system is correct, and it is reasonable to suppose that Alex meant "axiom system" where he wrote "model", so that "logically consistent model" becomes sensible. But then I'd suspect that he didn't really mean "minimal" (not only because of what Emil pointed out but also because of "minimal consistent modification" in the first sentence of the question). Unfortunately, taking that formulation seriously, one finds modifications that replace $P(E)\geq0$ with $P(E)\geq-\epsilon$, so minimality looks hopeless. –  Andreas Blass Dec 1 '11 at 16:22
@Andreas Blass: Yes, I meant "minimal modification", not minimal axiom system. Yet, seems, due to A3 $\epsilon+\epsilon = \epsilon$, so either $\epsilon = 0$ or $\epsilon = \infty$. –  Alex 'qubeat' Dec 1 '11 at 16:36

Your question is discussed in detail (including possible physical interpretations etc) in the book by A. Khrennikov, Interpretations of probability, VSP, Utrecht, 1999 (2nd edition, 2003).

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Thank you for the note, but the problem for me - is using of $p$-adic framework there. –  Alex 'qubeat' Dec 2 '11 at 16:19

Just a technical point, be sure to include absolute convergence of the sum in your axiom A3. This is automatic when P is positive, and I'm sure it was understood in you formulation. (This is not an answer but I can't post a comment).

The problem with non absolute convergence of convergent sum is a famous remark that one can make them converge to any real value or even diverge (by oscillation or to both infinities) just by permuting their terms.

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But the left hand sight in the axiom $P(\bigcup_k E_k)= \sum_k P(E_k)$ is independent of the order and so is the right hand sight which implies absolute convergence. –  Jochen Wengenroth Feb 14 '12 at 7:59
Thank you for reminder about that. @Jochen Wengenroth: If the problem may be really resolved in such a simple way? What about possibility of some analogue of Banach-Tarski paradox? –  Alex 'qubeat' Feb 14 '12 at 11:13
Let us consider sequence of elements $S_k$, $k = 1,\ldots,\infty$ with positive $P(S_k) = 1/(\ln(2) 2k (2k-1))$, $\sum_k P(S_k) = 1$. Now let us split each element on pair $S_k = E_{2k-1} \cup E_{2k}$, $P(E_k) = (-1)^{k+1}/(\ln(2) k)$, $P(S_k) = P(E_{2k-1})+P(E_{2k})$ using the gifts" model above. Then $\sum_k |P(E_k)| = \infty$, but why we should exclude such a model from consideration? –  Alex 'qubeat' Feb 14 '12 at 13:35

There is a wikipedia article with some references:

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Yes, I read that before post the question –  Alex 'qubeat' Feb 14 '12 at 11:16