Which Kolmogorov axioms must a probability assignment satisfy to be valid?

## The three axioms A probability model is a sample space `S` together with a function `P` assigning a number to each event. Kolmogorov's axioms are the minimal conditions that make `P` a probability: 1. **Non-negativity** — `P(A) ≥ 0` for every event `A`. 2. **Normalisation** — `P(S) = 1`, where `S` is the entire sample space. 3. **Countable additivity** — if `A1, A2, A3, ...` are pairwise disjoint (no two share an outcome), then `P(A1 ∪ A2 ∪ ...) = P(A1) + P(A2) + ...` That is the whole foundation. Notice what is *not* an axiom: there is no axiom saying probabilities are at most 1, no axiom about complements, and nothing at all about independence or conditional probability. Those are either consequences or separate definitions. ## What follows from them - **The empty event has probability 0.** The empty set is disjoint from itself, and additivity forces `P(empty) = 0`. - **Complement rule**: `A` and `A^c` are disjoint and their union is `S`, so `P(A) + P(A^c) = 1`, giving `P(A^c) = 1 - P(A)`. - **Upper bound**: since `P(A^c) ≥ 0`, the complement rule gives `P(A) ≤ 1`. The familiar "probabilities live between 0 and 1" is half axiom and half theorem. - **Monotonicity**: if every outcome of `A` is in `B`, write `B` as the disjoint union of `A` and `B` minus `A`; both pieces are non-negative, so `P(A) ≤ P(B)`. - **Addition rule**: `P(A ∪ B) = P(A) + P(B) - P(A ∩ B)`, derived by cutting the union into disjoint pieces. ## Checking a proposed assignment The practical version of this question is: someone hands you numbers, and you say whether they can be probabilities. **A negative value.** Suppose a model outputs `-0.05` for one category. That is not a rounding curiosity; it violates non-negativity outright, and the model is not a probability model. This happens in practice when numbers come from a fitting procedure that was never constrained to the simplex. **Values summing above 1.** Suppose three churn reasons are described as mutually exclusive and exhaustive, with probabilities 0.5, 0.4 and 0.3. Because they are disjoint, additivity says their union has probability `0.5 + 0.4 + 0.3 = 1.2`. Because they are exhaustive, that union is `S`, so normalisation demands 1. The assignment is contradictory and must be rejected. **Values summing below 1.** If those same three exhaustive reasons totalled 0.85, the model implicitly leaves 0.15 of the probability unassigned, which means the categories are not really exhaustive: some outcome is missing, often an "other" or "unknown" case. **Values summing above 1 that are perfectly fine.** This is the nuance that separates a rehearsed answer from an understood one. If three *overlapping* labels have probabilities 0.6, 0.5 and 0.3, the total 1.4 breaks nothing. A user can carry several labels at once, so additivity does not apply and there is no contradiction. The sum-to-one check is only meaningful for a set of events that are pairwise disjoint and cover the sample space. Before declaring numbers invalid, verify that claim about the events. ## What the axioms deliberately do not do The axioms are a consistency contract, not a source of numbers. They cannot tell you whether a coin's probability of heads is 0.5 or 0.6; both assignments satisfy all three axioms. Where the numbers come from is a separate question, answered by symmetry arguments, by observed long-run frequencies, or by a considered subjective judgment. A candidate who says "the axioms tell you the probability" has confused the rules of arithmetic with the measurement. The axioms also say nothing about independence. `P(A ∩ B) = P(A) P(B)` is the *definition* of independence, an extra property a particular model may or may not have, and it is a common error to recite it as a fourth axiom. ## Interview framing Expect this to arrive disguised: "here are the outputs of a scoring system, is anything wrong with them?" The strong answer names the axiom that is violated, states whether the events involved are actually disjoint and exhaustive, and distinguishes a genuine contradiction from a legitimate overlapping-tag total. If nothing is violated, say the assignment is internally consistent and then ask the separate question of whether it is *well calibrated*, which the axioms cannot judge. ## Traps to avoid - Listing "probabilities are between 0 and 1" as an axiom. The upper bound is derived. - Adding independence as a fourth axiom. - Declaring any set of numbers that sums above 1 invalid without checking that the events are disjoint. - Treating a sum below 1 over supposedly exhaustive categories as a rounding issue rather than a missing outcome.