Problem 18 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Suppose \(A\) and \(B\) are events with \(0<\mathrm{P}(A)<1\) and \(0<\mathrm{P}(B)<1\). a. If \(A\) and \(B\) are disjoint, can they be independent? b. If \(A\) and \(B\) are independent, can they be disjoint? c. If \(A \subset B, \operatorname{can} A\) and \(B\) be independent? d. If \(A\) and \(B\) are independent, can \(A\) and \(A \cup B\) be independent?

Step-by-Step Solution

Answer

a) Disjoint events cannot be independent. b) Independent events cannot be disjoint. c) A and B cannot be independent if A is a subset of B. d) A and A ∪ B are generally not independent.

Step 1: Understanding Independence and Disjoint Events

Two events, A and B, are independent if \( \mathrm{P}(A \cap B) = \mathrm{P}(A) \cdot \mathrm{P}(B) \). They are disjoint or mutually exclusive if \( \mathrm{P}(A \cap B) = 0 \).

Step a: Can Disjoint Events Be Independent?

If events A and B are disjoint, \( \mathrm{P}(A \cap B) = 0 \). For A and B to be independent, \( \mathrm{P}(A \cap B) \) should equal \( \mathrm{P}(A) \cdot \mathrm{P}(B) \). Since \( \mathrm{P}(A) > 0 \) and \( \mathrm{P}(B) > 0 \), \( \mathrm{P}(A) \cdot \mathrm{P}(B) > 0 \), which contradicts the condition for being disjoint. Thus, disjoint events cannot be independent.

Step b: Can Independent Events Be Disjoint?

Independent events satisfy \( \mathrm{P}(A \cap B) = \mathrm{P}(A) \cdot \mathrm{P}(B) \). If events A and B are also disjoint, \( \mathrm{P}(A \cap B) = 0 \). However, even when \( \mathrm{P}(A \cap B) = \mathrm{P}(A) \cdot \mathrm{P}(B) = 0 \), it implies one or both events have zero probability, contradicting our given \( 0 < \mathrm{P}(A), \mathrm{P}(B) < 1 \). Therefore, independent events cannot be disjoint.

Step c: Independence Given Subset Condition

If \( A \subset B \), then \( \mathrm{P}(A \cap B) = \mathrm{P}(A) \). For independence, \( \mathrm{P}(A \cap B) \) should be \( \mathrm{P}(A) \cdot \mathrm{P}(B) \). Thus, \( \mathrm{P}(A) = \mathrm{P}(A) \cdot \mathrm{P}(B) \), leading to \( 1 = \mathrm{P}(B) \). But \( \mathrm{P}(B) < 1 \), meaning A and B cannot be independent if \( A \subset B \) and \( \mathrm{P}(B) eq 1 \).

Step d: Independence Between Event and Union of Event

A and B are independent if \( \mathrm{P}(A \cap B) = \mathrm{P}(A) \cdot \mathrm{P}(B) \). Check if \( A \) and \( A \cup B \) are independent. \( \mathrm{P}(A \cap (A \cup B)) = \mathrm{P}(A) \). Independence would require \( \mathrm{P}(A) = \mathrm{P}(A) \cdot \mathrm{P}(A \cup B) \). Simultaneously, \( \mathrm{P}(A \cup B) = 1 - \mathrm{P}(A^c \cap B^c) \), which generally doesn't satisfy \( \mathrm{P}(A \cup B) = 1 \). Hence, A and \( A \cup B \) generally cannot be independent unless \( B = \emptyset \) leading to \( A = A \cup B \).

Key Concepts

Independent Events

In probability theory, two events A and B are called independent if the occurrence of one does not influence the occurrence of the other. Mathematically, this is expressed as \( \mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B) \). This formula indicates that the joint probability of A and B is equal to the product of their individual probabilities.

If events are independent, knowing that one event has occurred provides no additional information about the other. Consider flipping two coins: the outcome of the first flip does not affect the outcome of the second flip, making these events independent. It's important to remember that probabilities for independent events can range from greater than 0 to less than 1. This ensures that neither event is assured or impossible.

Disjoint Events

Events are considered disjoint, or mutually exclusive, when they cannot occur simultaneously. In mathematical terms, this means \( \mathrm{P}(A \cap B) = 0 \). An example of disjoint events is rolling a die and landing at either a 2 or a 5 — these outcomes cannot happen at the same time.

For disjoint events, if one event occurs, the other cannot. Therefore, it’s impossible for disjoint events to be independent because their joint probability is zero. For events to be independent, as mentioned, their joint probability should equal to the product of their individual probabilities which would be greater than zero if both events have non-zero chance. Thus, any consideration of independence for disjoint events leads to contradictions.

Subset Events

Events are in a subset relationship if every outcome of one event is also an outcome of another event. For instance, if \( A \subset B \), then any occurrence of A guarantees an occurrence of B.

If \( A \subset B \), \( \mathrm{P}(A \cap B) \) is equal to \( \mathrm{P}(A) \). Independence would imply \( \mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B) \), leading to \( \mathrm{P}(A) = \mathrm{P}(A) \times \mathrm{P}(B) \). This situation implies \( \mathrm{P}(B) = 1 \), which contradicts our assumption that \( \mathrm{P}(B) < 1 \).

• This suggests it is impossible for subsets to be independent unless \( B \) constitutes the entire sample space, thus \( \mathrm{P}(B) = 1 \).

Event Union

The union of events A and B, represented as \( A \cup B \), includes all outcomes that are in A, in B, or in both. This concept often relates to the probability that at least one of several events occurs.

When checking for independence between an event A and the union \( A \cup B \), you investigate if \( \mathrm{P}(A \cap (A \cup B)) = \mathrm{P}(A) \cdot \mathrm{P}(A \cup B) \). For A and \( A \cup B \) to be independent, the relationship \( \mathrm{P}(A) = \mathrm{P}(A) \cdot \mathrm{P}(A \cup B) \) must hold, which indicates different kinds of dependencies. Since typically \( \mathrm{P}(A \cup B) eq 1 \) unless B is an empty set, A and \( A \cup B \) cannot generally be independent.

• The special case when \( B = \emptyset \) makes \( A \cup B \) identical to A, nullifying independence considerations.