Question

Let \(\Omega\) consist of four points, each with probability \(\frac{1}{4}\). Find three events that are pairwise independent but not independent. Generalize.

Short answer

The events \( A = \{ \omega_1, \omega_2 \} \), \( B = \{ \omega_2, \omega_3 \} \), and \( C = \{ \omega_1, \omega_4 \} \) are pairwise independent but not independent.

Step-by-step solution

Define Sample Space and Probabilities

Consider a sample space \( \Omega = \{\omega_1, \omega_2, \omega_3, \omega_4\} \), where each point has a probability of \( \frac{1}{4} \). This means \( P(\omega_i) = \frac{1}{4} \) for each point \( \omega_i \in \Omega \).

Selection of Events

Select three events: \( A = \{ \omega_1, \omega_2 \} \), \( B = \{ \omega_2, \omega_3 \} \), and \( C = \{ \omega_1, \omega_4 \} \).

Calculate Individual Probabilities

Calculate the probability of each event: \( P(A) = P(\{ \omega_1, \omega_2 \}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \), \( P(B) = P(\{ \omega_2, \omega_3 \}) = \frac{1}{2} \), and \( P(C) = P(\{ \omega_1, \omega_4 \}) = \frac{1}{2} \).

Calculate Pairwise Intersection Probabilities

Compute the probabilities of the intersection of pairs: \( P(A \cap B) = P(\{ \omega_2 \}) = \frac{1}{4} \), \( P(A \cap C) = P(\{ \omega_1 \}) = \frac{1}{4} \), \( P(B \cap C) = P(\emptyset) = 0 \).

Determine Pairwise Independence

The events are pairwise independent if for each pair, \( P(X \cap Y) = P(X)P(Y) \). Check:- \( P(A \cap B) = \frac{1}{4} = \frac{1}{2} \times \frac{1}{2} \)- \( P(A \cap C) = \frac{1}{4} = \frac{1}{2} \times \frac{1}{2} \)- \( P(B \cap C) = 0 = \frac{1}{2} \times \frac{1}{2} \)Thus, they are pairwise independent.

Check Independence of All Three Events

For the events to be independent, the intersection of all three events should satisfy: \( P(A \cap B \cap C) = P(A)P(B)P(C) \). Calculate:- \( P(A \cap B \cap C) = P(\emptyset) = 0 \)- Compare with \( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \)Since \( 0 eq \frac{1}{8} \), the events are not independent.

Generalization

To generalize, consider a sample space of four points \( \Omega = \{\omega_1, \omega_2, \omega_3, \omega_4\} \), each with probability \( \frac{1}{4} \). Partition these into three pairwise independent events, ensuring that the intersection over three events is empty, making them dependent.

Key concepts

Pairwise Independence

Pairwise independence is a concept in probability theory where any two events are independent. To determine pairwise independence, we must look at two events at a time and calculate their intersection probabilities, comparing them to the product of their individual probabilities. Consider events A and B:

• Event A: Occurs based on one subset of outcomes.
• Event B: Occurs based on a different subset of outcomes.

Pairwise independence is confirmed if the probability of both events happening, the intersection, is equal to the product of their individual probabilities:\[ P(A \cap B) = P(A) \cdot P(B) \]In our example, events A, B, and C were each calculated separately respective of each other. They satisfy:

• \( P(A \cap B) = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} \)
• \( P(A \cap C) = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} \)
• \( P(B \cap C) = 0 = \frac{1}{2} \cdot \frac{1}{2} \)

These calculations confirm that our chosen events are pairwise independent.

Independence of Events

Independence of events involves more than just two events; it extends to a combination where each event does not influence the others. For a set of events to be entirely independent, the requirement is stricter. All permutations of event occurrences must independently conform to the condition where the probability of their collective intersection equals the product of their individual probabilities:\[ P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C) \]In this example, our events were not independent since:

• The calculated intersection probability \( P(A \cap B \cap C) = 0 \), did not match \( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \).

The lack of this equality indicates that despite pairwise independence, the trio of events does not have mutual global independence.

Sample Space

A sample space is the foundation of probability theory. It is a collection of all possible outcomes of an experiment. In our exercise, the sample space \( \Omega = \{ \omega_1, \omega_2, \omega_3, \omega_4 \} \) consists of four equally likely outcomes, each with a probability of \( \frac{1}{4} \). The importance of a sample space lies in its ability to frame the entire probability model. It helps us define events as subsets of outcomes within this space:

• Each point \( \omega_i \) represents a possible result.
• The sum of probabilities of all outcomes in the sample space equals 1.

Understanding the sample space is crucial because it breaks down how we view and calculate probabilities for events and intersections within that realm.

Intersection Probabilities

Intersection probabilities reveal how events overlap in the sample space. They reflect situations where multiple events occur simultaneously. To find these probabilities, we determine the common outcomes shared between events and their respective probabilities.In our problem, the intersection probabilities were analyzed across event pairs, such as \( A \cap B \). Calculations included:

• \( P(A \cap B) = P(\{\omega_2\}) = \frac{1}{4} \)
• \( P(A \cap C) = P(\{\omega_1\}) = \frac{1}{4} \)
• \( P(B \cap C) = P(\emptyset) = 0 \)

These intersections helped establish pairwise independence by showing how the event pairs relate in probability terms within the defined sample space. However, as seen, they also highlight why the complete set of events failed to achieve full independence due to differences in intersection proportions when considering all three events.