Probability induced by weight function

Consider a finite, non-empty set \(E\) and a function: \(w : E \to \mathbb{R}^{+}\). Then, there exists a unique function \(P\) that makes the triplet \((E, \mathcal{P}(E), P)\) a probability space such that, for any \(a, b \in E\):

\[w(b) P(a) = w(a) P(b)\]

The function \(w\) can be thought as assigning a "weight" to each value in \(E\). The larger its weight, the larger its probability.

Proof

To prove this, let \(n = |E| > 0\). We can then list the elements in \(E\):

\[E = \{ e_1, ..., e_n \}\]

Since \(E\) is finite, \(P\) is determined by its image for single-outcome events. Let \(k \in \{1, ..., n\}\) be an arbitrary index from \(E\). Then:

\[P(e_k) = 1 - P(E \setminus \{e_k\})\]

The probability of any given event is the sum of the probabilities of all the outcomes in the event. Therefore:

\[P(e_k) = 1 - P(E \setminus \{e_k\}) = 1 - \sum_{i \in \{1, ..., n\} \setminus \{k\}} P(e_i)\]

Now we use the property relating probabilities to weights in order to change \(P(e_i)\) into \(P(e_k)\):

\[P(e_k) = 1 - \sum_{i \in \{1, ..., n\} \setminus \{k\}} \frac{w(e_i)}{w(e_k)} P(e_k)\]

The terms without \(i\) can be factored out from the sum:

\[P(e_k) = 1 - \frac{P(e_k)}{w(e_k)} \sum_{i \in \{1, ..., n\} \setminus \{k\}} w(e_i)\]

Now we have \(P(e_k)\) at both sides of the equation. If we solve for it:

\[P(e_k) = \frac{1}{1 + \frac{\sum_{i \in \{1, ..., n\} \setminus \{k\}} w(e_i)}{w(e_k)}}\]

We could stop here for our proof, but let's simplify this expression.

\[P(e_k) = \frac{1}{\frac{w(e_k)}{w(e_k)} + \frac{\sum_{i \in \{1, ..., n\} \setminus \{k\}} w(e_i)}{w(e_k)}}\]

We can now complete the sum:

\[P(e_k) = \frac{1}{\frac{\sum_{i \in \{1, ..., n\}} w(e_i)}{w(e_k)}}\]

Moving \(w(e_k)\) to the numerator gives us the final form of the probability formula for single-outcome events:

\[P(e_k) = \frac{w(e_k)}{\sum_{i \in \{1, ..., n\}} w(e_i)}\]

And since \(k\) was chosen arbitrarily, this completely and uniquely determines \(P\).

Final remark

In the proof, we used the fact that \(P(e_k) \neq 0\). This follows from the equation:

\[w(b) P(a) = w(a) P(b)\]

Indeed, given \(a \in E\), were \(P(a)\) to be 0, since \(w(a) > 0\), it would follow that \(P(b) = 0\) for any other \(b \in E\). This would contradict the fact that \((E, \mathcal{P}(E), P)\) is a probability space.