Question

(a) Note that (3.4) assumes \(P(A) \neq 0\) since \(P_{A}(B)\) is meaningless if \(P(A)=0\) Assuming both \(P(A) \neq 0\) and \(P(B) \neq 0,\) show that if (3.4) is true, then \(P(A)=P_{B}(A) ;\) that is if \(B\) is independent of \(A,\) then \(A\) is independent of \(B\) If either \(P(A)\) or \(P(B)\) is zero, then we use (3.5) to define independence. (b) When is an event \(E\) independent of itself? When is \(E\) independent of "not \(E " ?\)

Short answer

If B is independent of A, then A is independent of B. An event E is independent of itself if \( P(E) = 0 \) or \( P(E) = 1 \). E is independent of \( \text{not } E \) if \( P(E) = \frac{1}{2} \).

Step-by-step solution

Understand the problem

We need to show that if event B is independent of event A, then event A is also independent of event B. This will involve some known probabilities and conditional probability formulas.

Recall the definition of independence

Two events A and B are independent if and only if \[ P(A \text{ and } B) = P(A)P(B) \]

Use the conditional probability formula

The formula for conditional probability is given by \[ P(B|A) = \frac{P(A \text{ and } B)}{P(A)} \]

Substitute in the independence condition

By the definition of independence, if B is independent of A, then \[ P(B|A) = P(B) \].Using the conditional probability formula, we get \[ P(B) = \frac{P(A \text{ and } B)}{P(A)} \].

Rearrange the equation

Rearranging the equation we get \[ P(A \text{ and } B) = P(A)P(B) \].This implies that A and B are independent.

Show independence in the reverse direction

We need to show that if B is independent of A, then A is independent of B. We already have \[ P(A \text{ and } B) = P(A)P(B) \].Now consider \[ P(A|B) = \frac{P(A \text{ and } B)}{P(B)} \], substituting for \[ P(A \text{ and } B) \], we get \[ P(A|B) = \frac{P(A)P(B)}{P(B)} \] which simplifies to \[ P(A|B) = P(A) \].Hence, A is independent of B.

Discuss the independence condition on probabilities

If either \( P(A) \) or \( P(B) \) is zero, the independence definition simplifies since an event with probability zero does not affect the probability of any other event.

Answer part (b)

An event E is independent of itself if \( P(E) = 1 \) or \( P(E) = 0 \). An event E is independent of not E (\( E^c \)) if \( P(E) = \frac{1}{2} \).

Key concepts

Conditional Probability

Conditional probability is the likelihood of an event occurring, given that another event has already occurred. The formula for conditional probability is:
\[ P(B|A) = \frac{P(A \text{ and } B)}{P(A)} \]
This means we are interested in the probability of event B happening given that event A has already happened. Essentially, it's updating our probability assessment in light of new information.
In the given exercise, understanding conditional probability is essential to demonstrate that if event B is independent of event A, then event A is also independent of event B.

Probability Theory

Probability theory forms the mathematical basis for analyzing and interpreting random events. The theory includes understanding fundamental concepts like independent events and conditional probability.
An important aspect of probability theory is defining how events are independent. Two events A and B are considered to be independent if the occurrence of one does not affect the probability of the other. Mathematically, this is expressed as:
\[ P(A \text{ and } B) = P(A)P(B) \]
In probability theory, the independence of events simplifies calculations. The exercise illustrates this by showing the forward (if B is independent of A) and reverse (if A is independent of B) conditions.

Mathematical Proof

A mathematical proof is a logical process that demonstrates the truth of a statement, based entirely on previously established statements such as theorems, axioms, and lemmas.
In the exercise, the proof involves showing that if event B is independent of event A, then event A is also independent of event B.
Steps of the proof in the exercise:

• Start by recalling the definition of independence: \[ P(A \text{ and } B) = P(A)P(B) \]
• Use the conditional probability formula: \[ P(B|A) = \frac{P(A \text{ and } B)}{P(A)} \]
• Substitute in the condition for independence: \[ P(B|A) = P(B) \]
• Rearrange to show the independence of A and B going in both directions.

In doing so, the proof confirms the bidirectional nature of independence.

Probability Distributions

Probability distributions describe how probabilities are distributed over the values of a random variable. It can be expressed in terms of a probability mass function (for discrete variables) or a probability density function (for continuous variables).
In the context of the exercise, we deal with probabilities associated with events A and B. Independence of these events implies that the probability distribution of one does not affect the probability distribution of the other.
Key concepts related to probability distributions include:

• Mean: the average value.
• Variance: the measure of how much the values differ from the mean.
• Standard Deviation: the square root of the variance.

Understanding how these distributions work can help solve more complex probability problems by breaking down events into simpler, independent parts.