3 corrected spelling

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, that is which are the possible caveats (e.g. some essential theorems are not valid or useful tools do not work - cf lack of monotonicity).

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, that is the possible caveats (e.g. some essential theorems are not valid or useful tools do not work - cf lack of monotonicity).

1

# 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.