Finite Probability Models With Weighted Outcomes

From Outcome Weights to Event Probabilities

A finite sample space makes probability calculations explicit: an event is a set of outcomes, and its probability is obtained by adding the weights of the outcomes it contains. The weights need not be equal. What matters is that they are nonnegative and that their total is one. This construction gives a probability measure and makes clear exactly when a uniform model is appropriate.

Throughout, \(\Omega\) is a finite, nonempty sample space, \(\mathcal F=2^\Omega\) is the sigma-algebra of all subsets of \(\Omega\), and \(\mathbb P\) is a probability measure on \((\Omega,\mathcal F)\). Thus \(\mathbb P\) is a nonnegative, countably additive set function with \(\mathbb P(\Omega)=1\). The elements \(\omega\in\Omega\) are outcomes, and the sets \(A\in\mathcal F\) are events. A weight assignment is a function \(p:\Omega\to[0,\infty)\). Since \(\Omega\) is finite, all sums over outcomes below are finite.

A weight \(p(\omega)=0\) is allowed: it means that the singleton event \(\{\omega\}\) has probability zero in this model. Nonnegative weights are essential, while strict positivity is not. Normalization is also essential: without a total of one, the formula still defines a nonnegative set function, but not a probability measure.

Why the Weight Formula Defines a Probability Measure

Proof. For every \(A\subseteq\Omega\), \(\mathbb P(A)\ge0\), because it is a sum of nonnegative weights. The empty sum gives \(\mathbb P(\varnothing)=0\), and normalization gives \(\mathbb P(\Omega)=1\). To check countable additivity, let \(A_1,A_2,\ldots\) be pairwise disjoint subsets of \(\Omega\). Only finitely many of these sets can be nonempty: each nonempty \(A_j\) contains at least one point of the finite set \(\Omega\), and disjoint nonempty sets must contain different points. The union is therefore a finite disjoint union, and each outcome in \(\bigcup_j A_j\) occurs in exactly one of the sums over \(A_j\). Rearranging these finite sums gives \(\mathbb P(\bigcup_j A_j)=\sum_j\mathbb P(A_j)\); all terms from empty \(A_j\) are zero. Thus \(\mathbb P\) satisfies the probability-measure axioms. \(\square\)

The construction also describes every probability measure on a finite sample space when all subsets are measurable. If \(\mathbb P\) is already a probability measure on \((\Omega,2^\Omega)\), define the weight of an outcome by \(p(\omega)=\mathbb P(\{\omega\})\). These weights are nonnegative. Since \(\Omega\) is the disjoint union of its finitely many singleton sets, finite additivity gives \(\sum_{\omega\in\Omega}p(\omega)=\mathbb P(\Omega)=1\). For any event \(A\subseteq\Omega\), its singleton decomposition gives \(\mathbb P(A)=\sum_{\omega\in A}p(\omega)\). So on a finite space with all subsets measurable, specifying the probability measure and specifying normalized outcome weights are equivalent.

Often the given numbers are raw weights rather than probabilities. If \(w(\omega)\ge0\) and \(W=\sum_{\omega\in\Omega}w(\omega)>0\), set \(p(\omega)=w(\omega)/W\). Then \(\sum_{\omega\in\Omega}p(\omega)=W/W=1\), and the resulting event probability is \(\mathbb P(A)=\sum_{\omega\in A}w(\omega)/W\). The requirement \(W>0\) matters: if every raw weight is zero, there is no total by which to normalize.

Worked Models

Checks, Complements, and a Common Pitfall

For any \(A\subseteq\Omega\) in a finite probability space, the event and its complement partition \(\Omega\), so finite additivity gives \(\mathbb P(A)+\mathbb P(A^c)=1\). Also, if \(A\subseteq B\), then \(B\) is the disjoint union of \(A\) and \(B\setminus A\); nonnegativity gives \(\mathbb P(B)=\mathbb P(A)+\mathbb P(B\setminus A)\ge\mathbb P(A)\). These provide useful checks on calculations: event probabilities must be between zero and one, and a subset cannot have greater probability than the set containing it.

Common mistake: treating raw weights as probabilities. In the weighted-die example, the raw weights of faces \(1,2,3\) add to \(1+1+2=4\). Calling \(4\) the event probability is wrong: these are raw weights, whose total over all faces is \(12\), not \(1\). The correction is to divide by the total, giving \(4/12=1/3\). A related error is to use \(|A|/|\Omega|\) in a nonuniform model. For that same event there are three outcomes among six, so this shortcut gives \(3/6=1/2\); it is the probability only in the uniform model, whereas the specified weights give \(1/3\).

The finite construction is useful because it separates two decisions. First choose the possible outcomes and the measurable events; here every subset is an event. Then assign nonnegative weights and normalize them. Once these choices are made, an event probability is always the sum over the outcomes it contains. In particular, no assumption about equal likelihood should be introduced unless equal weights were actually assigned.