Checking and Repairing a Proposed Probability Model

Why Check a Proposed Model?

A finite probability model assigns probabilities to outcomes and, from them, to events. Before using the model to answer a question, check that its assignments could actually be probabilities. Two basic failures are negative point probabilities and a total probability other than one. A model can also fail because its event probabilities contradict one another, even when each number lies between zero and one.

Checking and repairing are different tasks. A check asks whether all stated assignments can hold together. A repair changes one or more assignments to obtain a valid model, while retaining the conditions that the problem says must be preserved. Without such conditions, there may be many possible repairs rather than one uniquely correct answer.

Checking Point Probabilities

The two checks address different requirements. Nonnegativity must hold outcome by outcome; a positive total cannot offset a negative point probability. The total-mass check asks whether the probabilities of all outcomes together equal one. If either requirement fails, the proposed list does not define a probability model.

Why These Two Checks Are Sufficient

For a finite sample space, checking the point probabilities is not merely a preliminary screening rule. Nonnegative point probabilities with total one determine a probability for every event, and those event probabilities satisfy the probability axioms.

Proof. First suppose the numbers satisfy nonnegativity and sum to one. Define \(P(A)\) by the stated finite sum for each \(A\subseteq\Omega\). Every such sum is nonnegative, so \(P(A)\geq0\), and \(P(\Omega)=\sum_{i=1}^m p_i=1\). If \(A\) and \(B\) are disjoint, each outcome in their union appears in exactly one of the two sums. Therefore \(P(A\cup B)=P(A)+P(B)\). The assignment is thus a probability measure on the finite space.

Conversely, suppose \(P\) is a probability measure on the subsets of \(\Omega\), and set \(p_i=P(\{\omega_i\})\). Nonnegativity of a probability measure gives \(p_i\geq0\). The singleton events partition \(\Omega\), so finite additivity gives \(1=P(\Omega)=\sum_{i=1}^mP(\{\omega_i\})=\sum_{i=1}^mp_i\). Both conditions are necessary, completing the proof. \(\square\)

This result provides a complete check when the proposal gives point probabilities for every outcome. If the proposal instead gives probabilities for events, those assignments must also agree with one another and with some valid set of point probabilities.

Checking Assignments to Overlapping Events

Event probabilities cannot generally be checked one at a time. For example, knowing \(P(A)\) and \(P(B)\) does not mean their probabilities can simply be added if the events overlap. The overlap must be counted only once in the union. Decomposing the events into disjoint parts gives the required consistency rule.

Derivation. The sets \(A\setminus B\), \(B\setminus A\), and \(A\cap B\) are disjoint, and their union is \(A\cup B\). Also, \(A\) is the disjoint union of \(A\setminus B\) and \(A\cap B\), while \(B\) is the disjoint union of \(B\setminus A\) and \(A\cap B\). Finite additivity therefore gives \(P(A\cup B)=P(A\setminus B)+P(B\setminus A)+P(A\cap B)\), \(P(A)=P(A\setminus B)+P(A\cap B)\), and \(P(B)=P(B\setminus A)+P(A\cap B)\). Adding the last two equalities counts the intersection twice; subtracting it once gives the stated formula. \(\square\)

Repairing a Finite Model Under Constraints

A practical repair method is to express the unknown point probabilities as variables. Each required event probability becomes a sum of the variables for outcomes in that event. Add the conditions that all point probabilities are nonnegative and that their total is one. Solve these equations while retaining the stated constraints, then check the resulting values.

The original list happened to sum to one, but it failed the constraints: \(0.25\ne0.15\), \(0.45\ne2(0.15)\), and its probability for \(\{a,b\}\) was \(0.25+0.15=0.40\). This illustrates why a total of one is necessary but not enough when a problem specifies additional event relationships.

Common Pitfall: Normalizing Every List

A tempting response to a total other than one is to divide every proposed probability by the total. This produces nonnegative values summing to one when the original values are nonnegative and their sum is positive. But it changes every point probability, so it may violate event probabilities or other constraints that were supposed to remain fixed. Normalization is appropriate when the numbers are explicitly weights to be converted into probabilities; it is not a universal repair for a model with fixed assignments.

For example, in the first worked example, normalizing \(0.25,0.50,0.35\) by \(1.10\) gives approximately \(0.2273,0.4545,0.3182\). These values sum to one, but they do not preserve the specified assignments \(p_a=0.25\) and \(p_b=0.50\). Under those constraints the correct repair was \(p_c=0.25\).

A second common error is to assume that any collection of event probabilities in the interval from zero to one is consistent. The union calculation above disproves that assumption: event assignments must satisfy the relationships implied by the sets themselves. When constraints are insufficient to determine every point probability, more than one repair may work; state the constraints and verify the chosen solution rather than claiming it is uniquely forced.