Measurable Spaces – Problem (42/365)

The chapter on measurable spaces introduces a \sigma-algebra over the real numbers R = (-\infty, \infty). The Borel algebra, \mathcal{B}(R), is the smallest \sigma-algebra \sigma(\mathcal{A}) where \mathcal{A} is the algebra generated by finite disjoint sums of intervals of the form (a,b]. By the direct product of algebras we also get algebras over higher dimensions \mathcal{B}(R^n) = \mathcal{B}(R) \otimes \dots \otimes \mathcal{B}(R). We also get a legal \sigma-algebra for the infinite direct product \mathcal{B}(R^\infty) = \mathcal{B}(R) \otimes \mathcal{B}(R) \otimes \dots.

The book asks to show that certain sets are members of \mathcal{B}(R^\infty). Show that the following are Borel sets.

\displaystyle  \{ x \in R^\infty : \sup \inf x_n > a \} \\ \{ x \in R^\infty : \inf \sup x_n \le a \}

Take the first case. Note that \sup \inf x_n > a is not satisfied if for every i, \inf \{ x_k : k \ge i \} \le a. This can only happen if there are an infinite number of coordinates whose value is \le a. Let

\displaystyle  \text{let } I \subset \mathbb{N} \text{ finite subset of natural numbers} \\ S(I) = \cup_{i=1}^\infty \{ x \in \mathcal{B}(R^\infty) : x_i \le a \text{ if } i \in I \text{ and } x_i > a \text{ otherwise} \}

The set S(I) is a Borel set since we have constructed it as a countable union of Borel sets. Therefore, \{ x \in R^\infty : \sup \inf x_n > a \} = \cup_{I} S(I) (this is also a countable union) is a Borel set. A similar argument is made for the other.

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Measurable Spaces – Problem (41/365)

Let \mathcal{D} = \{ D_1, D_2, \dots \} be a countable decomposition of \mathcal{\Omega} and \mathcal{A} = \sigma(\mathcal{D}) be the \sigma-algebra generated by \mathcal{D}. Are there only countably many sets in \mathcal{A}?

The answer is no. We can show this by showing that the natural numbers is a strict subset of \mathcal{A}. Every natural number n can be written as \sum_{i=0}^\infty a_i 2^i where a_i \in \{0,1\} and only a finite number of a_i = 1 (because if an infinite number of a_i = 1 the sum is \infty).

Note that since \mathcal{D} is a decomposition we can write every set in \mathcal{A} as a countable (because this is a \sigma-algebra) union of a subset of \mathcal{D}. This means we can encode every set in \mathcal{A} as \cup_{i=1}^\infty (A_i \cap D_i) where A_i \in \{ \emptyset, \Omega \}. First, since \mathcal{D} is countable there is a bijection between the natural numbers and \mathcal{D}, however, \mathcal{A} is not countable since we can have a countable number of A_i = \Omega.

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Measurable Spaces – Problem (40/365)

The past few problems looked at how probability works when we have an infinite sample space. It didn’t cover how one can actually assign probabilities to such spaces. That will be the next task. Before that, the book covers the topic of \sigma-algebras which form the algebra of events on top of which we can assign a measure.

Given a set \Omega, and a set of subsets \mathcal{A}, we say that \mathcal{A} is an algebra if \Omega \in \mathcal{A} and is closed under unions and complementation. A \sigma-algebra adds to that the requirement that it also be closed under countable unions. The pair (\Omega, \mathcal{A}) is called a measureable space.

Let \mathcal{A}_1, \mathcal{A}_2 be \sigma-algebras of \Omega. Are the following systems of sets \sigma-algebras?

\displaystyle  \mathcal{A}_1 \cap \mathcal{A}_2 = \{ A : A \in \mathcal{A}_1 \text{ and } A \in \mathcal{A}_2 \} \\ \mathcal{A}_1 \cup \mathcal{A}_2 = \{ A : A \in \mathcal{A}_1 \text{ or } A \in \mathcal{A}_2 \}

The intersection of \sigma-algebras is also a \sigma-algebra because \Omega \in \mathcal{A}_1 \cap \mathcal{A}_2, and A_1 \cup A_2 \cup \dots \in in the intersection is contained in both \mathcal{A}_1 and \mathcal{A}_2.

However, the union of \sigma-algebras is not always a \sigma-algebra. For instance, let \mathcal{A} = \{ A, \bar{A}, \emptyset, \Omega_1 \}, \mathcal{B} = \{ B, \bar{B}, \emptyset, \Omega_2 \}, then their union does not contain A \cup B.

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Probability Foundations – Problem (39/365)

Let \mu be a finite measure on algebra \mathcal{A}, A_n \in \mathcal{A} for n = 1,2,\dots and A = \lim_n A_n (i.e. A=\overline{\lim}A_{n}=\underline{\lim}A_{n}). Show that \mu(A) = \lim_n \mu(A_n.

\displaystyle  \overline{\lim}A_{n} = \cap_{n=1}^{\infty}\cup_{k=n}^{\infty}A_{k} \\ = \left(\cup_{k=1}^{\infty}A_{k}\right)\cap\left(\cup_{k=2}^{\infty}A_{k}\right)\cap\dots \\ = \left(\cup_{k=2}^{\infty}A_{k}\right)\cap\dots \\ = \lim_{n}\left(\cup_{k=n}^{\infty}A_{k}\right) \\ \underline{\lim}A_{n} =	\cup_{n=1}^{\infty}\cap_{k=n}^{\infty}A_{k} \\ = \left(\cap_{k=1}^{\infty}A_{k}\right)\cup\left(\cap_{k=2}^{\infty}A_{k}\right)\cup\dots \\ = \left(\cap_{k=2}^{\infty}A_{k}\right)\cup\dots \\ = \lim_{n}\left(\cap_{k=n}^{\infty}A_{k}\right)

Since A=\overline{\lim}A_{n}=\underline{\lim}A_{n},

\displaystyle  \lim_{n}\left(\cup_{k=n}^{\infty}A_{k}\right) =	\lim_{n}\left(\cap_{k=n}^{\infty}A_{k}\right) \\ \text{implies limit is } \lim_n A_n \text{ because the union and intersection have to agree}\\ \text{so } \mu(A) = \lim_n \mu(A_n)

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Probability Foundations – Problem (38/365)

Let \mu be a finitely additive measure on an algebra \mathcal{A}, and let A_1, A_2, \dots \in \mathcal{A} be pairwise disjoint and satisfy A = \cup_{i=1}^\infty A_i \in \mathcal{A}. Then show that \mu(A) \ge \sum_{i=1}^\infty \mu(A_i).

\displaystyle  \text{Since } A \in \mathcal{A}, A - A_1 \in \mathcal{A} \text{, and } A - \cup_{i=1}^k A_i \in \mathcal{A} \\ \text{Let } B_i = \cup_{k=i}^\infty A_k \\ \sum_{i=1}^\infty \mu(A_i) \\ = \sum_{i=1}^\infty \mu(B_i - B_{i+1}) \\ = \sum_{i=1}^\infty \mu(B_i) - \mu(B_{i+1}) \\ = \mu(B_1) - \mu(B_2) + \mu(B_2) - \mu(B_3) + \mu(B_4) - \mu(B_5) + \dots \\ = \mu(B_1) - \lim_n B_n \\ = \mu(A) - \lim_n B_n \\ \le \mu(A)

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Probability Foundations – Problem (37/365)

A problem similar to the previous post. Let \Omega be a countable set and \mathcal{A} a collection of all its subsets. Put \mu(A) = 0 if A is finite and \mu(A) = \infty if A is inifinite. Show that the set function \mu is finitely additive but not countable additive.

To see that it is finitely additive, let A,B \in \mathcal{A} be disjoint.

\displaystyle  \mu(A \cup B) = \mu(A) + \mu(B) = 0 + 0 = 0 \text{ if both finite} \\ \mu(A \cup B) = \mu(A) + \mu(B) = \infty \text{ if either one infinite}

To show that it is not countably additive, consider the case where \Omega = \mathbb{N} is the set of natural numbers. Then

\displaystyle  \mu(\{ 1, 2, 3, \dots \}) = \infty \\ \text{But,} \sum_{i=1}^\infty \mu(i) = 0

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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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