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Chapter 1: Problem 5

Let \(S\) be the support of a distribution on a Euclidean space \((\mathcal{X}, \mathcal{A})\). Then, \((i) S\) is closed; (ii) \(P(S)=1 ;\) (iii) \(S\) is the intersection of all closed sets \(C\) with \(P(C)=1\).

Short Answer

Expert verified
The support \( S \) is closed, satisfies \( P(S)=1 \), and is the intersection of all closed sets \( C \) with \( P(C)=1 \).

Step by step solution

01

Define the support

The support of the distribution, denoted by \( S \), is the smallest closed set with probability 1. This means for any point \( x \in \mathcal{X} \setminus S \), there exists a neighborhood around \( x \) with probability zero.
02

Prove (i) S is closed

To demonstrate that \( S \) is closed, recall that the closure of a set includes the limit points. By definition, the support \( S \) cannot be a smaller set, since removing any point would lower its probability below 1. Therefore, all limit points must be within \( S \), making it closed.
03

Prove (ii) P(S)=1

Since \( S \) is the smallest closed set with probability 1 in the definition of support, by definition we have \( P(S) = 1 \). This is intrinsic to the definition of support in any probability distribution.
04

Prove (iii) S as intersection of closed sets

Let \( C \) be any closed set with \( P(C) = 1 \). The support \( S \) is the smallest such set, therefore, \( S \) is included in \( C \). Thus, \( S \) can be expressed as the intersection of all such closed sets \( C \), because taking their intersection still maintains probability 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Closed Set
In the context of this exercise, we need to understand what a closed set is. **A closed set** in a Euclidean space is a set that contains all its boundary or limit points. Here's a simpler way to understand it:
  • If you have a closed interval like [0, 1], it includes its endpoints 0 and 1. In other words, there's no 'missing' points on the edge.
  • If you think of a closed set like a sealed jar, everything that wants to be inside has to fit exactly within the boundaries without even the tiniest gap.
In our problem, when we call the support of a distribution closed, it means this set includes every point that can potentially 'reach' (or be close) from within the set itself. You can't have a point just outside the set trying to be part of it without actually including it in the closed set.
Probability Measure
A **probability measure** is a function that assigns a probability to different possible outcomes in a space. When we talk about the subset of this space, we are checking how likely it is to 'end up' in that subset.
  • Imagine a fair dice, each number from 1 to 6 has a probability of \(\frac{1}{6}\) because each outcome is equally likely.
  • For the support of a distribution, the condition is that the closed set, like you could see in the dice example, sums up to probability 1. This means the whole equation equals complete certainty, covering all possible outcomes.
In the case of our exercise, the support is the smallest closed set that accounts for all 'probability mass' of the distribution. This is why we say for support \( P(S) = 1 \). Any slight removal from support would mean a probability loss, which isn't allowed here.
Euclidean Space
A **Euclidean space** is what mathematicians call the kind of space where conventional geometry rules can be applied. It refers to the 2D and 3D geometry most of us are familiar with, where:
  • Distances are computed using the Pythagorean theorem.
  • Shapes like squares and triangles reside in a flat plane environment, or extended into a 3D world.
  • There's a coordinate system to help place points, lines, and shapes clearly and mathematically.
In the exercise, the term Euclidean space defines the 'playing field' for our distribution. This is crucial because it implies any mathematical rules or definitions applied are subject to the familiar rules of Euclidean geometry. With this backdrop, the concepts of support being a closed set or probability measures maintaining their values operate under common spatial rules.

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Most popular questions from this chapter

Suppose \(k_{n}^{\prime} / k_{n} \rightarrow \infty\). (a) If \(R_{n}=0\left(1 / k_{n}\right)\) and \(R_{n}^{\prime}=0\left(1 / k_{n}^{\prime}\right)\), then \(R_{n}+R_{n}^{\prime}=0\left(1 / k_{n}\right)\). (b) If \(R_{n}=o\left(1 / k_{n}\right)\) and \(R_{n}^{\prime}=o\left(1 / k_{n}^{\prime}\right)\), then \(R_{n}+R_{n}^{\prime}=o\left(1 / k_{n}\right)\).

Show that the \(k\)-dimensional sphere \(\Sigma_{i=1}^{k} x_{1}^{2} \leq c\) is convex.

Show that (a) If \(A_{1} \subset A_{2} \subset \cdots\), then \(\underline{A}=\bar{A}=\cup A_{n}\). (b) If \(A_{1} \supset A_{2} \supset \cdots\), then \(\underline{A}=\bar{A}=\cap A_{n}\).

Give an example showing that a convex function need not be continuous on a closed interval.

Let \(U\) be uniformly distributed on \((0,1)\), and let \(F\) be a distribution function on the real line. (a) If \(F\) is continuous and strictly increasing, show that \(F^{-1}(U)\) has distribution function \(F\). (b) For arbitrary \(F\), show that \(F^{-1}(U)\) continues to have distribution function \(F\). [Hint: Take \(F^{-1}\) to be any nondecreasing function such that \(F^{-1}[F(x)]=x\) for all \(x\) for which there exists no \(x^{\prime} \neq x\) with \(F\left(x^{\prime}\right)=F(x)\). \(]\)
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