Example Problems on Probability Measures
ST779 Homework 4 on Probability Measures
1
Let Ω be a set of F be a field on it. A function P:F→[0,1] with P(Ω)=1 is called finitely additive on F for any A1,…,An∈F disjoint, n≥1, P(∪ni=1Ai)=∑ni=1P(Ai).
If Pn, n=1,2,…, are probability measures on Ω,A and F is a field on Ω, F⊂A, such that for all A∈F, Pn(A) converges to a limit, to be called P(A), then show P:F→[0,1] is finitely additive.
Is P also countable additive? Prove or give a counterexample.
We can show finite additivity.
P(m⋃i=1Ai)=limHowever, we do not have infinite additivity. For example, take \Omega = \overline{ \mathbb R } and \overline{ \mathcal A }. Let’s look at the interval [-n,n]. Take A_1, \dots, A_{2n} to be a partition over [-n,n] where each A_i has length \frac{ 1 }{ 2n }. Looking at the finite case,
1 = P \left( \bigcup_{i=1}^{2n} A_i \right) = \sum_{i=1}^{2n} P(A_i) = \sum_{i=1}^{2n}\frac{ 1 }{ 2n } = 1.However, if we take n \rightarrow \infty, notice that the length of each A_i \rightarrow 0, but their union covers the all of \Omega.
\sum_{i=1}^\infty P(A_i) = \sum_{i=1}^\infty 0 = 0 \neq 1 = P(\Omega) = P\left( \bigcup_{i=1}^\infty A_i \right)2
Let \Omega be a set and \mathcal A be a \sigma-field on it. P: \mathcal A\rightarrow [0,1] with P(\Omega) = 1 be finitely additive and satisfy the following property: For all A_1, A_2, \dot , \in \mathcal A disjoint such that \bigcup_{n=1}^\infty A_n = \Omega, we have that \sum_{n=1}^\infty P(A_n) = 1. Then show that P is a (countable additive) probability measure on (\Omega, \mathcal A).
We need to show that P is countably additive. Take B_1, B_2, \dots \in \mathcal A to be disjoint sets. Then \Omega \setminus \left( \bigcup_{i=1}^\infty B_i\right) is disjoint from all B_i and
\bigcup_{i=1}^\infty B_i \cup \left( \Omega \setminus \left( \bigcup_{i=1}^\infty B_i\right) \right) = \Omega.Then,
\begin{align} 1 & = P(\Omega) \\ & = P\left(\bigcup_{i=1}^\infty B_i \cup \left( \Omega \setminus \left( \bigcup_{i=1}^\infty B_i\right) \right)\right) \\ & = \sum_{j=1}^\infty \left[ P\left(B_j \right) + P\left( \Omega \setminus \left( \bigcup_{i=1}^\infty B_i\right) \right)\right] \\ & = \sum_{j=1}^\infty \left[P(B_j) \right] + P\left(\Omega \setminus \left( \bigcup_{i=1}^\infty B_i\right)\right) \\ \sum_{j=1}^\infty P(B_j) & = 1 - P\left(\Omega \setminus \left( \bigcup_{i=1}^\infty B_i\right)\right) \\ & = 1 - \left( P(\Omega) - P\left(\bigcup_{i=1}^\infty B_i\right) \right) \\ & = P\left(\bigcup_{i=1}^\infty B_i\right) \end{align}3
Let P be countable additive on a semifield \mathcal C on \Omega (i.e. whenever C_n \in \mathcal C disjoint and \bigcup_{n=1}^\infty C_n \in \mathcal C, we have P\left( \bigcup_{n=1}^\infty C_n \right) = \sum_{n=1}^\infty P(C_n).) Condisere that Caratheodory extension
P^*(A) = \inf \left\{ \sum_{i=1}^\infty P(C_i): C_1, C_2, \dots \in \mathcal C, A \in \bigcup_{i=1}^\infty C_i, \right\}for any A \subset \Omega. Show that there exists B \in \sigma \langle \mathcal C \rangle containing A such that P^*(A) = P^*(B).
By the definition of infimum
\sum_{i=1}^\infty P(C_{i,k}) < P^*(A) + 1/k.Then \forall k, A \subset \bigcup_{i=1}^\infty C_{i,k}. And define B
A \subset \bigcap_{k=1}^\infty \left( \bigcup_{i=1}^\infty C_{i,k} \right) = B \Rightarrow P^*(A) < P^*(B).But also B \in \sigma \langle \mathcal C \rangle and B = \bigcup_{k=1}^\infty B_k. Then,
P^*(B) \leq P^*(B_k) \leq \sum_{i=1}^\infty P(C_{i,k}) < P^*(A) + 1/k, k > 0.Thus, P^*(A) = P^*(B).
4
Let A be a Borel subset of [0,1] such that its Lebsegue measure 0 < \lambda(A) < 1. Let 0 < \alpha < 1. Show that there exists a Borel subset B of A such that \lambda(B) = \alpha \lambda(A).
Notice that for a measure \mu and S \subset T,
\begin{align} T & = (T \setminus S) \cup (T \cap S) \\ & = (T \setminus S) \cup S \\ \mu(T) & = \mu(T \setminus S) \cup \mu(S) \\ \mu(T \setminus S) & = \mu(T) - \mu(S). \end{align}Recall the intermediate value theorem, that is for a continuous function f(x) on [a,b] with minimum and maximum \min and \max. Then \forall l such that \min < l < \max, \exists c such that f(c) = l.
In this case a = 0 and b=1 and f(x) = \lambda \left( A \cap [0,x] \right). And our min and max are
\begin{align} f(0) & = \lambda \left( A \cap 0\right) = \begin{cases} \lambda(0) \\ \lambda(\varnothing) \end{cases} = 0 \\ f(1) & = \lambda \left( A \cap [0,1] \right) = \lambda \left( A \right) \end{align}Now we need to show that f(x) is continuous. That is, \forall \epsilon > 0 \exists \delta such that \mid x - y \mid < \delta \Rightarrow \mid f(x) - f(y) \mid < \epsilon. Without loss of generality, take y \geq x so A \cap [0,y] \subset A \cap [0,y].
\begin{align} f(x) - f(y) & = \lambda\left( A \cap [0,x] \right) - \lambda \left( A \cap [0,y]\right) \\ & = \lambda \left( \left( A \cap [0,x] \right) \setminus \left( A \cap [0,y] \right) \right) \\ & = \lambda \left( \left( A \cap [0,x] \right) \cap \left( A \cap [0,y] \right)^C \right) \\ & = \lambda \left( \left( A \cap [0,x] \right) \cap \left( A^C \cup (y, 1] \right) \right) \\ & = \lambda \left( \left( A \cap [0,x] \cap A^C \right) \cup \left( A \cap [0,x] \cap (y, 1] \right) \right) \\ & = \lambda \left( \left( \varnothing \right) \cup \left( A \cap [0,x] \cap (y, 1] \right) \right) \\ & = \lambda \left( A \cap [0,x] \cap (y, 1] \right) \\ & < \epsilon \end{align}Then take l = \alpha \lambda(A). By the intermediate value theorem \exists z such that
f(z) = l = \alpha \lambda(A) = \lambda(\underbrace{A \cap [0,z]}_{B}).Then B \subset A and is a Borel subset.
5
Let \alpha > 0. For \delta > 0 and A \in \mathbb R, let
H_\delta^\alpha(A) = \inf \left\{ \sum_{n=1}^\infty \mid I_n \mid^\alpha: \cup_{n=1}^\infty I_n \supset A, I_n \text{ are intervals of length } \mid I_n \mid < \delta \right\}.Note that H_\delta^\alpha(A) increases as \delta \downarrow 0. Define H^\alpha(A) = \lim_{\delta\rightarrow 0} H_\delta^\alpha(A), called the \alpha-Hausdorff outer measure of A. It can be shown, using the Caratheodory extension theorem that this is a measure on the Borel \sigma-field \mathcal R. Note that H^1 is the Lebesgue measure.
Note the following: if H^\alpha(A) = \infty and \beta < \alpha, then H^\beta(A) = \infty. Define the Hausdorff dimension of a Borel A set by
\text{Hdim}(A) = \inf \left\{ \alpha > 0: H^\alpha(A) < \infty \right\} = \sup \left\{ \alpha >0 : H^\alpha(A) = \infty \right\}.Show that the Hausdorff dimension of the Cantor set C is \log(2) / \log(3).
Recall that C = \bigcap_{n=1}^\infty C_n and that C_n has 2^n disjoint elements each of length (1/3)^n with \lambda(C_n) = (2/3)^n. We want to take a small \delta = (1/3)^n + \epsilon for \epsilon > 0. Then we can take each I_m to be one of the elements of C_n. Then,
\sum_{m=1}^{2^n}\mid I_m \mid ^\alpha = \sum_{m=1}^{2^n} (1/3)^{n\alpha} = \left( \frac{ 2 }{ 3^\alpha } \right)^n.This holds for all \alpha, so take \alpha = 0.001 Then,
\left(\frac{ 2 }{ 3^\alpha } \right)^n \approx 2^n.And since C = \bigcap_{n=1}^\infty C_n we know C \subset C_n \forall n and H^\alpha(C) < H^\alpha(C_n). Now we can take the limit of n.
\lim_{n\rightarrow \infty} 2^n \rightarrow \inftyAnd if we take \alpha large, such as \alpha = 1
\lim_{n \rightarrow \infty} (2/3)^n \rightarrow 0.We want to find \alpha such that H^\alpha(C) = 1.
2/3^\alpha = 1 \rightarrow \alpha = \frac{ \log(2) }{ \log(3) }