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Chapter 8: Problem 4

Let \(A \subset \mathbb{R}^{n}\) be a Borel measurable set of finite Lebesgue measure \(\lambda(A) \in(0, \infty)\) and let \(X\) be uniformly distributed on \(A\) (see Example 1.75). Let \(B \subset A\) be measurable with \(\lambda(B)>0\). Show that the conditional distribution of \(X\) given \(\\{X \in B\\}\) is the uniform distribution on \(B\).

Short Answer

Expert verified
Therefore, the conditional distribution of \(X\) given \(X \in B\) is the uniform distribution on \(B\). This is because the conditional distribution function equals the distribution that it would over a uniform space as \(B\).

Step by step solution

01

Study of Uniform Distribution

Since \(X\) is uniformly distributed on \(A\), the probability density function (\(pdf\)) of \(X\), \(f_{X}(x)\), is given by \(f_{X}(x) = 1/ \lambda(A)\) if \(x \in A\) and 0 otherwise.
02

Studying the Conditional Probability

We need to find the conditional distribution of \(X\) given \(X \in B\), which can be written as \(f_{X|X \in B}(x)\). According to the definition of the conditional probability, we have \(f_{X|X \in B}(x) = f_{X}(x) / Pr(X \in B)\) if \(x \in B\) and 0 otherwise.
03

Applying definitions

Since \(X\) is uniformly distributed on \(A\), and \(B\) is a subset of \(A\), the probability of \(X\) being in \(B\) is the measure of \(B\) divided by the measure of \(A\), i.e., \(\lambda(B) / \lambda(A)\). Therefore, we have \(f_{X|X \in B}(x) = 1/ \lambda(A) / (\lambda(B) / \lambda(A)) = 1/ \lambda(B)\) if \(x \in B\) and 0 otherwise.

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Key Concepts

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

Uniform Distribution
The idea of a uniform distribution is a friendly concept in probability. Imagine a perfectly shuffled deck of cards where each card has the same chance of being drawn. This is analogous to a uniform distribution.
In our context, we talk about a set, like the Borel measurable set \(A\), where a random variable \(X\) is uniformly distributed. This means every point in \(A\) has an equal chance of being selected.
A uniform distribution over a set \(A\) can be described by the probability density function (pdf), where if \(X\) takes a value \(x\) inside \(A\), the pdf \(f_X(x) = \frac{1}{\lambda(A)}\), and it's zero otherwise.
  • \(A\): A set where the distribution occurs.
  • \(\lambda(A)\): The size of \(A\) or its total measure.
  • Uniformity: Equal chance for every element within \(A\).

This paths us to understanding that uniform distributions are all about equal likelihood over a well-defined space.
Borel Measurable Set
A Borel measurable set is a fundamental concept in measure theory. If you've heard of open or closed intervals on a number line, Borel sets help generalize these notions.
Borel measurable sets, like our set \(A\) in \(\mathbb{R}^n\), include all possible constructs you can make starting from open intervals using operations such as union, intersection, and complement. This includes what we might consider normal shapes or figures on a plane or higher dimensions.
Why is this important? It ensures that certain operations and measurements, like probability, are well defined.
  • Inclusion of all open intervals.
  • Closure under countable union, intersection, and complement operations.
  • Essential for well-formulated probability measures.

Ultimately, recognizing that a set is Borel measurable assures us that we can apply various mathematical tools and theorems effectively.
Probability Density Function
The probability density function, or pdf, is a concept that describes how the probabilities are distributed over the values of a continuous random variable. Think of it as a curve that tells you the likelihood of various outcomes. But remember, for continuous variables, the value of the pdf itself does not give probability directly; it's the area under the curve that counts.
In simpler terms, if you can imagine pouring water over a graph and measuring the area it covers over a section, that represents the probability for that range. For our exercise, the pdf helps us understand the distribution of \(X\) over \(A\) and further when conditioned on subset \(B\).
  • The pdf \(f_X(x)\) gives the shape of the distribution.
  • Integral of pdf over a particular range gives the probability of \(X\) falling within that range.
  • Key to defining uniform distribution, which assigns equal probabilities across an interval.

Grasping the essence of a pdf empowers you to predict and analyze the behavior of random variables in continuous domains.

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Most popular questions from this chapter

Let \(E\) be a countable set and let \(P\) and \(Q\) be probability measures on \(E .\) Assume there is a \(c>0\) with $$ f(e):=\frac{Q(\\{e\\})}{P(\\{e\\})} \leq c \quad \text { for all } e \in E \text { with } P(\\{e\\})>0 $$ Let \(X_{1}, X_{2}, \ldots\) be independent random variables with distribution \(P\). Let \(U_{1}, U_{2}, \ldots\) be i.i.d. random variables that are independent of \(X_{1}, X_{2}, \ldots\) and that are uniformly distributed on \([0,1] .\) Let \(N\) be the smallest (random) nonnegative integer \(n\) such that \(U_{n} \leq f\left(X_{n}\right) / c\) and define \(Y:=X_{N}\) Show that \(Y\) has distribution \(Q\) Remark. This method for generating random variables with a given distribution \(Q\) is called rejection sampling, as it can also be described as follows. The random variable \(X_{1}\) is a proposal for the value of \(Y\). This proposal is accepted with probability \(f\left(X_{1}\right) / c\) and is rejected otherwise. If the first proposal is rejected, the game starts afresh with proposal \(X_{2}\) and so on.

Let \(X\) and \(Y\) be real random variables with joint density \(f\) and let \(h: \mathbb{R} \rightarrow \mathbb{R}\) be measurable with \(\mathbf{E}[|h(X)|]<\infty .\) Denote by \(\lambda\) the Lebesgue measure on \(\mathbb{R}\). (i) Show that almost surely $$ \mathbf{E}[h(X) \mid Y]=\frac{\int h(x) f(x, Y) \lambda(d x)}{\int f(x, Y) \lambda(d x)}. $$ (ii) Let \(X\) and \(Y\) be independent and \(\exp _{\theta}\)-distributed for some \(\theta>0\). Compute \(\mathbf{E}[X \mid X+Y]\) and \(\mathbf{P}[X \leq x \mid X+Y]\) for \(x \geq 0\).

Consider a theatre with \(n\) seats that is fully booked for this evening. Each of the \(n\) people entering the theatre (one by one) has a seat reservation. However, the first person is absent-minded and takes a seat at random. Any subsequent person takes his or her reserved seat if it is free and otherwise picks a free seat at random. (i) What is the probability that the last person gets his or her reserved seat? (ii) What is the probability that the \(k\) th person gets his or her reserved seat?

Consider the earth as a ball (as widely accepted nowadays). Let \(X\) be a random point that is uniformly distributed on the surface. Let \(\Theta\) be the longitude and let \(\Phi\) be the latitude of \(X .\) A little differently from the usual convention, assume that \(\Theta\) takes values in \([0, \pi)\) and \(\Phi\) in \([-\pi, \pi) .\) Hence, for fixed \(\Theta\), a complete great circle is described when \(\Phi\) runs through its domain. Now, given \(\Theta\), is \(\Phi\) uniformly distributed on \([-\pi, \pi) ?\) One could conjecture that any point on the great circle is equally likely. However, this is not the case! If we thicken the great circle slightly such that its longitudes range from \(\Theta\) to \(\Theta+\varepsilon\) (for a small \(\varepsilon\) ), on the equator it is thicker (measured in metres) than at the poles. If we let \(\varepsilon \rightarrow 0\), intuitively we should get the conditional probabilities as proportional to the thickness (in metres). (i) Show that \(\mathbf{P}[\\{\Phi \in \cdot\\} \mid \Theta=\theta]\) for almost all \(\theta\) has the density \(\frac{1}{4}|\cos (\phi)|\) for \(\phi \in[-\pi, \pi)\) (ii) Show that \(\mathbf{P}[\\{\Theta \in \cdot\\} \mid \Phi=\phi]=\mathcal{U}[0, \pi)\) for almost all \(\phi\). Hint: Show that \(\Theta\) and \(\Phi\) are independent, and compute the distributions of \(\Theta\) and \(\Phi\).

Show that \((\mathbb{R}, \mathcal{B}(\mathbb{R}))\) and \(\left(\mathbb{R}^{n}, \mathcal{B}\left(\mathbb{R}^{n}\right)\right)\) are isomorphic. Conclude that every Borel set \(B \in \mathcal{B}\left(\mathbb{R}^{n}\right)\) is a Borel space.
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