Probability Foundations – Problem (36/365)

I did say it would be one post a day and it already looks like i’ll only achieve it as an expectation of posts every day. So, let me catch up first by solving more problems. This time from chapter 2 of the book: Mathematical Foundations of Probability Theory.

This chapter introduces us to how we can extend the probability framework we had for finite sample spaces. The key problem we face is that in the finite case we were simply able to assign a probability to each \omega \in \Omega and therefore get P(X \subseteq \Omega) = \sum_{x \in X} p(x). But we can no longer follow this approach for an infinite sample space.

Anyway, the problem asks the following. Let \Omega be the set of rational numbers in [0,1]. Let \mathcal{A} be the algebra of sets where each set takes on one of these forms: \{r : a < r < b \}, \{r : a \le r < b \}, \{r : a < r \le b \}, \{r : a \le r \le b \} and P(A) = b - a. Show that P(A) is a finitely additive set function but not countably additive.

Let A < B \in \mathcal{A} be disjoint sets. Then, we see that P(\cdot) is finitely additive.

\displaystyle  \text{We can write } P(A) = b - a = \sup A - \inf A \\ P(A \cup B) = \sup (A \cup B) - \inf (A \cup B) \\ = \sup B - \inf A \\ = (\sup A + P(B)) - \inf A \\ = P(A) + P(B)

To show that P(\cdot) is not countably additve we need to show that we can come up with an infinite sequence of disjoint sets whose sum of probabilitites is not equal to the probability of its union. This should bring back memories of converging sequences. Consider the sets (\frac{1}{2},1], (\frac{1}{3}, \frac{1}{2}], \dots, (\frac{1}{n+1}, \frac{1}{n}], \dots. It is clear that the union of these sets is [0,1]. But

\displaystyle  \sum_{i=1}^\infty P(( \frac{1}{n+1}, \frac{1}{n} ]) \\ = \sum_{i=1}^\infty \frac{1}{n} - \frac{1}{n+1} \\ = \sum_{i=1}^\infty \frac{1}{n(n+1)} \\ = \frac{1}{1(2)} + \frac{1}{2(3)} + \frac{1}{3(4)} + \frac{1}{4(5)} + \frac{1}{5(6)} + \frac{1}{6(7)} + \frac{1}{7(8)} + \dots \\  > \frac{1}{2(2)} + \frac{1}{4(4)} + \frac{1}{4(4)} + \frac{1}{8(8)} + \frac{1}{8(8)} + \frac{1}{8(8)} + \frac{1}{8(8)} + \dots \\ = \frac{1}{4} + 2 \frac{1}{4(4)} + 4 \frac{1}{8(8)} + \dots \\ = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \dots \\ = \infty

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