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Chapter 12: Problem 1

Suppose \((\Omega, \mathcal{F}, P)\) is a probability space and \(A \in \mathcal{F}\). Prove that \(A\) and \(\Omega \backslash A\) are independent if and only if \(P(A)=0\) or \(P(A)=1\)

Short Answer

Expert verified
To prove that \(A\) and \(\Omega \backslash A\) are independent if and only if \(P(A)=0\) or \(P(A)=1\), we must show that \(P(A \cap (\Omega \backslash A)) = P(A)P(\Omega \backslash A)\). Observing that the intersection is empty yields \(P(A \cap (\Omega \backslash A)) = 0\). Since \(P(\Omega \backslash A) = 1 - P(A)\), we have the equation \(0 = P(A)(1 - P(A))\), which has two possible solutions: \(P(A)=0\) or \(P(A)=1\). Thus, \(A\) and \(\Omega \backslash A\) are independent events if and only if \(P(A)=0\) or \(P(A)=1\).

Step by step solution

01

Define independence

In order for two events \(A\) and \(B\) to be independent, the following condition must be satisfied: \(P(A \cap B) = P(A)P(B)\)
02

Compute the intersection of A and the complement of A

Let's first observe the intersection of \(A\) and its complement \(\Omega \backslash A\). Since \(A \cap (\Omega \backslash A)\) represents the intersection of elements that are in both \(A\) and not in \(A\), this intersection is empty and \(P(A \cap (\Omega \backslash A)) = 0\).
03

Compute the probability of the complement of A

By definition of a probability space, the probability of the sample space \(\Omega\) is 1. Also, since \(\Omega\) can be considered the union of set \(A\) and its complement \(\Omega \backslash A\), we have \(P(\Omega) = P(A\cup(\Omega\backslash A))\). Since \(A\) and \(\Omega \backslash A\) are disjoint, we have \(P(A) + P(\Omega\backslash A) = 1\), and thus, \(P(\Omega \backslash A) = 1 - P(A)\).
04

Apply the definition of independence

Now that we have found the probability of the intersection \(A \cap (\Omega \backslash A)\) and the probability of the complement of \(A\), we can apply the definition of independence: \(P(A \cap (\Omega \backslash A)) = P(A)P(\Omega \backslash A)\) \(0 = P(A)(1 - P(A))\)
05

Solve for P(A)

In order to satisfy the condition for independence, we must solve the equation \(0 = P(A)(1 - P(A))\): There are two possible solutions for this equation: 1. \(P(A) = 0\) 2. \(P(A) = 1\)
06

Conclusion

Based on our analysis, we can conclude that \(A\) and \(\Omega \backslash A\) are independent events if and only if \(P(A)=0\) or \(P(A)=1\).

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Key Concepts

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

Probability Space
Understanding the concept of a probability space is fundamental to the study of probability theory. A probability space鈥攄enoted as \( (\Omega, \mathcal{F}, P) \)鈥攊s essentially the mathematical framework that lays the foundation for defining probabilities of various events. It is formed of three elements:\
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  • \Sample space (\Omega):\ This is the set of all possible outcomes of an experiment. Imagine it as a universal set that contains everything that could happen when an experiment is conducted.
    • \
  • \\sigma-algebra (\mathcal{F}):\ A collection of subsets of the sample space, including the sample space itself and the empty set, that satisfies certain mathematical properties. It organizes these subsets into events that we can assign probabilities to.
    • \
  • \Probability measure (P):\ A function that assigns a probability to the events in the \sigma-algebra, adhering to the axioms of probability. It quantifies how likely an event is to happen, with 0 indicating impossibility and 1 indicating certainty.
    • \
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A proper appreciation of the probability space concept helps you make sense of complex events and their probabilities within the framework of mathematical probability.
Independent Events
In probability theory, the independence of events is a critical concept that guides our understanding of how one event affects鈥攐r doesn't affect鈥攖he likelihood of another event. Two events, \(A\) and \(B\), are considered \independent\ if the occurrence of one does not change the probability of occurrence of the other. This can be mathematically expressed as \(P(A \cap B) = P(A)P(B)\), where \(P(A \cap B)\) is the probability of both events occurring together.
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The Role of Independence in Complex Probability

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Independence is essential when dealing with multiple events. For example, flipping a fair coin, each toss is independent of the previous one鈥攖he result of one flip does not affect the next. If the events are not independent, we would need to consider additional information, and calculations would become more complex, as they would involve conditional probabilities.
Complement of Events
The complement of an event is a crucial idea that is often used alongside the concepts of probability spaces and independent events. If \(A\) is an event in a probability space, its complement鈥攄enoted as \(\Omega \backslash A\) or sometimes simply \(A^c\)鈥攔epresents all outcomes of the sample space that are \not\ in event \(A\). One of the fundamental rules involving complements is that the probability of an event and its complement always add up to 1, i.e., \(P(A) + P(\Omega \backslash A) = 1\).
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Understanding Through Contrasts

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Complementary events are helpful in assessing probabilities in a more intuitive way. For instance, if it is challenging to calculate the probability of a complex event directly, we might find it easier to calculate the probability of it not happening (the complement) and then use the rule \(P(A) = 1 - P(\Omega \backslash A)\) to determine \(P(A)\). This approach is especially useful in scenarios where direct calculation can be exhaustive or when events have a large number of outcomes.

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Most popular questions from this chapter

Give an example of a probability space \((\Omega, \mathcal{F}, P)\) and events \(A, B_{1}, B_{2}\) such that \(A\) and \(B_{1}\) are independent, \(A\) and \(B_{2}\) are independent, but \(A\) and \(B_{1} \cup B_{2}\) are not independent.

Give an example of a probability space \((\Omega, \mathcal{F}, P)\) and \(X, Y \in \mathcal{L}^{2}(P)\) such that \(\sigma^{2}(X+Y)=\sigma^{2}(X)+\sigma^{2}(Y)\) but \(X\) and \(Y\) are not independent random variables.

Suppose \((\Omega, \mathcal{F}, P)\) is a probability space and \(A \in \mathcal{F}\) is an event such that \(0

Suppose \((\Omega, \mathcal{F}, P)\) is a probability space, \(A \in \mathcal{F},\) and \(B_{1} \subset B_{2} \subset \cdots\) is an increasing sequence of events such that \(A\) and \(B_{n}\) are independent events for each \(n \in \mathbf{Z}^{+}\). Show that \(A\) and \(\bigcup_{n=1}^{\infty} B_{n}\) are independent.

Suppose \((\Omega, \mathcal{F}, P)\) is a probability space and \(\left\\{A_{k}\right\\}_{k \in \Gamma}\) is a family of events. Prove the family \(\left\\{A_{k}\right\\}_{k \in \Gamma}\) is independent if and only if the family \(\left\\{\Omega \backslash A_{k}\right\\}_{k \in \Gamma}\) is independent.
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