Discussion

Let \(S = \{1, 2, 3, \ldots\}\) and suppose that every subset of \(S\) is an event. Let \(\mathcal{P}(S)\) denote the power set of \(S\). Which of the following statements is/are true?

A.There exists a probability function \(P: \mathcal{P}(S) \to [0, \infty)\) such that \(P(\{n\}) = \frac{1}{n+1},\ n \geq 1\)
B.There exists a probability function \(P: \mathcal{P}(S) \to [0, \infty)\) such that \(P(A) = 0\) if \(A\) is a finite set and \(P(A) = 1\) if \(A\) is an infinite set
C.There exists a probability function \(P: \mathcal{P}(S) \to [0, \infty)\) such that \(P(\{1, 2, \ldots, n\}) = \int_{1}^{n} \frac{1}{x}\,dx,\ n \geq 1\)
D.There exists a probability function \(P: \mathcal{P}(S) \to [0, \infty)\) such that \(P(\{1, 2, \ldots, n\}) = \int_{0}^{n} e^{-x}\,dx,\ n \geq 1\)

Answer: D

The only true statement is the one with \(P(\{1, 2, \ldots, n\}) = \int_{0}^{n} e^{-x}\,dx\).

A probability function must be non-negative, give \(P(S) = 1\), and be countably additive. Check each proposal against these:

\(P(\{n\}) = \frac{1}{n+1}\): countable additivity forces \(P(S) = \sum \frac{1}{n+1}\), a divergent harmonic series, not 1. Impossible.
Finite sets 0, infinite sets 1: \(S\) is the countable union of the singletons \(\{n\}\), each of probability 0, so countable additivity gives \(P(S) = 0\), not 1. Impossible.
\(P(\{1, \ldots, n\}) = \int_{1}^{n} \frac{dx}{x} = \log_{e} n\): this exceeds 1 once \(n \geq 3\), which no probability can. Impossible.
\(P(\{1, \ldots, n\}) = 1 - e^{-n}\): set \(P(\{1\}) = 1 - e^{-1}\) and \(P(\{n\}) = e^{-(n-1)} - e^{-n} \gt 0\) for \(n \geq 2\). These are positive and sum to \(\lim(1 - e^{-n}) = 1\), so they define a valid probability.

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