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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
The conditional distribution of \( X \) given \( X \in B \) is uniform on \( B \).

Step by step solution

01

Understand the Problem

We need to show that the conditional distribution of a random variable \( X \), given that \( X \) is in a subset \( B \) of \( A \), is uniform. \( X \) is uniformly distributed over \( A \) and \( B \) is a measurable subset with positive measure.
02

Identify the Conditional Distribution

The conditional distribution of \( X \) given \( \{ X \in B \} \) is the distribution of \( X \) with respect to the set \( B \). This requires us to find the probability \( P(X \in C \mid X \in B) \) for any measurable set \( C \subset B \).
03

Apply the Definition of Conditional Probability

The probability of \( X \in C \) given \( X \in B \) is given by \[ P(X \in C \mid X \in B) = \frac{P(X \in C \cap B)}{P(X \in B)}. \] Since \( C \subset B \), this simplifies to \( P(X \in C \mid X \in B) = \frac{P(X \in C)}{P(X \in B)}. \)
04

Use Uniform Distribution on \( A \)

Because \( X \) is uniformly distributed on \( A \), \( P(X \in C) = \frac{\lambda(C)}{\lambda(A)} \) for \( C \subset A \). Similarly, \( P(X \in B) = \frac{\lambda(B)}{\lambda(A)} \).
05

Substitute and Simplify

Substituting back, we get \[ P(X \in C \mid X \in B) = \frac{\frac{\lambda(C)}{\lambda(A)}}{\frac{\lambda(B)}{\lambda(A)}} = \frac{\lambda(C)}{\lambda(B)}. \] This demonstrates that \( X \) is uniformly distributed on \( B \), as this expression is the probability for a uniform distribution on the set \( B \).
06

Conclude

The conditional distribution of \( X \) given \( X \in B \) is uniform since the derived probability matches the form of a uniform distribution on the set \( B \).
07

Conclusion and Interpretation

Therefore, the conditional distribution of \( X \) given \( X \in B \) is indeed uniform on \( B \), matching the properties required by a uniform distribution for any measurable subset \( C \subset B \).

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Key Concepts

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

Uniform Distribution
A uniform distribution is a type of probability distribution where all outcomes are equally likely within a certain range. Think of it as slicing a cake where every slice is identical in size.
When we say that a random variable is uniformly distributed over a set, it simply means that each part of the set is equally probable to be selected.
  • For instance, if you're picking a number at random from 1 to 10 (inclusive), each number from 1 to 10 has an equal chance of being chosen.
  • In the case of the problem provided, our random variable is uniformly distributed over a set \( A \), which implies each element within \( A \) has the same likelihood of appearing.
Understanding this is key as it sets the foundation for examining how the distribution changes when we're focusing on a subset.
Borel Measurable Set
A Borel measurable set is a course-grained concept in measure theory, which is crucial for assigning probabilities. Any set that can be formed through operations like unions, intersections, and complements starting from open sets in the Euclidean space is Borel measurable.
This type of set is significant because it allows us to work with probabilities in a consistent manner and is foundational in defining events in probability spaces.
  • Imagine Borel sets like collections of events that we can actually measure in some finitary sense, using the tools of measure theory.
  • In practical terms, knowing a set is Borel measurable confirms that we can "measure" or assign a probability to it, as done with set \( A \) and \( B \) in the problem scenario.
This measurability enables us to apply the concepts of uniform distribution and conditional probability effectively.
Lebesgue Measure
The Lebesgue measure is a tool used to assign a volume or "size" to a large collection of sets, including all Borel measurable sets in the real number space. It extends the idea of length from intervals to more complex sets.
For a simple interval \((a, b)\) in the real line, the Lebesgue measure equals \( b - a \).
  • The importance of Lebesgue measure lies in its ability to "measure" sets which are potentially very complex in form.
  • In the exercise, sets \( A \) and \( B \) both have a finite Lebesgue measure, means they represent a finite "volume" in \( \mathbb{R}^n \).
Understanding and utilizing the Lebesgue measure allows us to compute and interpret probabilities for complex events.
Conditional Probability
Conditional probability is the probability of an event occurring given that another event has already occurred. It refines our expectation by incorporating more information.
In mathematical terms, the conditional probability of an event \(C\) given another event \(B\) is \( P(C \mid B) \), and can be calculated using \[ P(C \mid B) = \frac{P(C \cap B)}{P(B)}. \]
  • This formula allows us to re-focus our probability space to just \( B \), making \( B \) the new world of possibilities.
  • In the problem discussion, calculating the probability of \(X\) within \( C \) given \(X\) is in \( B \) utilizes this principle to prove uniform distribution over \( B \).
Conditional probability enables us to handle situations like the problem presented, where the context (being in \( B \)) alters the likelihood of certain outcomes.

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Most popular questions from this chapter

(Lack of memory of the exponential distribution) Let \(X>0\) be a strictly positive random variable and let \(\theta>0 .\) Show that \(X\) is exponentially distributed if and only if $$ \mathbf{P}[X>t+s \mid X>s]=\mathbf{P}[X>t] \quad \text { for all } s, t \geq 0 $$ In particular, \(X \sim \exp _{\theta}\) if and only if \(\mathbf{P}[X>t+s \mid X>s]=e^{-\theta t}\) for all \(s, t \geq 0\).

(Rejection sampling for generating random variables) 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.

(Borel's paradox) 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 meters) than at the poles. If we let \(\varepsilon \rightarrow 0\), intuitively we should get the conditional probabilities as proportional to the thickness (in meters). (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 \(\Phi\).

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 \(x \geq 0\)

Assume the random variable \((X, Y)\) is uniformly distributed on the disc \(B:=\left\\{(x, y) \in \mathbb{R}^{2}: x^{2}+y^{2} \leq 1\right\\}\) and on \([-1,1]^{2}\), respectively. (i) In both cases, determine the conditional distribution of \(Y\) given \(X=x\). (ii) Let \(R:=\sqrt{X^{2}+Y^{2}}\) and \(\Theta=\arctan (Y / X) .\) In both cases, determine the conditional distribution of \(\Theta\) given \(R=r\).
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