Question

For a fixed event \(B\), show that the collection \(P(A \mid B)\), defined for all events \(A\), satisfies the three conditions for a probability. Conclude from this that $$ P(A \mid B)=P(A \mid B C) P(C \mid B)+P\left(A \mid B C^{c}\right) P\left(C^{C} \mid B\right) $$ Then directly verify the preceding equation.

Short answer

In summary, we have shown that the conditional probability \(P(A|B)\) satisfies the three axioms of probability: non-negativity, normalization, and additivity. We then derived the given equation: \(P(A \mid B)=P(A \mid B C) P(C \mid B)+P\left(A \mid B C^{c}\right) P\left(C^{C} \mid B\right)\) by expressing \(P(A\cap B)\) using the law of total probability and applying the definition of conditional probability. We then directly verified the derived equation by multiplying the terms together and relating them back to \(P(A|B)\).

Step-by-step solution

Show that the conditional probability satisfies the three axioms of probability

The three axioms of probability are: 1. Non-negativity: \(P(A) \geq 0\) for any event A. 2. Normalization: \(P(S) = 1\), where S is the sample space. 3. Additivity: If \(A_1, A_2, A_3, \dots\) are mutually exclusive events, then \(P\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} P(A_i)\) To prove that \(P(A|B)\) satisfies these conditions, we'll use the definition of conditional probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\) given that \(P(B) > 0\) Axiom 1: Non-negativity Since \(P(A \cap B) \geq 0\) and \(P(B) > 0\), then it is clear that \(P(A|B) \geq 0\). Axiom 2: Normalization We need to show that \(P(S|B) = 1\). Since \(S \cap B = B\), then \(P(S|B) = \frac{P(B)}{P(B)} = 1\). Axiom 3: Additivity Let \(A_1, A_2, A_3, \dots\) be mutually exclusive events. We need to show that: \(P\left(\bigcup_{i=1}^{\infty} A_i | B\right) = \sum_{i=1}^{\infty} P(A_i|B)\) By the definition of conditional probability, we get: \(\frac{P\left((\bigcup_{i=1}^{\infty} A_i) \cap B\right)}{P(B)} = \sum_{i=1}^{\infty} \frac{P(A_i \cap B)}{P(B)}\) Since \(A_1, A_2, A_3, \dots\) are mutually exclusive, their intersections with B are also mutually exclusive. This means we can rewrite the left side of the equation as: \(\frac{\sum_{i=1}^{\infty} P(A_i \cap B)}{P(B)}\) Comparing both sides of the equation, we find they are equal: \(\frac{\sum_{i=1}^{\infty} P(A_i \cap B)}{P(B)} = \sum_{i=1}^{\infty} \frac{P(A_i \cap B)}{P(B)}\) -----------

Derive the given equation

To derive the equation: \(P(A \mid B)=P(A \mid B C) P(C \mid B)+P\left(A \mid B C^{c}\right) P\left(C^{C} \mid B\right)\) Let's first examine the term \(P(A \cap B)\). Using the law of total probability, we can write it as: \(P(A \cap B) = P(A \cap B \cap C) + P(A \cap B \cap C^c)\) Now, let's divide both sides of the equation by \(P(B)\): \(\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B \cap C)}{P(B)} + \frac{P(A \cap B \cap C^c)}{P(B)}\) Using the definition of conditional probability, we can rewrite this as: \(P(A|B) = P(A \cap C | B)P(C | B) + P(A \cap C^c | B)P(C^c | B)\) This is the equation we sought to derive. -----------

Verify the derived equation

Now, let's directly verify that our derived equation is true: \(P(A | B) = P(A | BC)P(C|B) + P(A | BC^c)P(C^c | B)\) By definition, \(P(A | BC) = \frac{P(A \cap B \cap C)}{P(B \cap C)}\) and \(P(A | BC^c) = \frac{P(A \cap B \cap C^c)}{P(B \cap C^c)}\) Using the definition of conditional probability for the other terms, we have: \(P(C|B) = \frac{P(C \cap B)}{P(B)}\) and \(P(C^c|B) = \frac{P(C^c \cap B)}{P(B)}\) Now, multiplying the terms together, and using the fact that \(P(A \cap B \cap C) = P(A|BC)P(B \cap C)\) and \(P(A \cap B \cap C^c) = P(A|BC^c)P(B \cap C^c)\), we get: \(\frac{P(A \cap B \cap C)}{P(B)} + \frac{P(A \cap B \cap C^c)}{P(B)}\) This is equal to: \(\frac{P(A \cap B \cap C) + P(A \cap B \cap C^c)}{P(B)} = \frac{P(A \cap B)}{P(B)}\) Using the definition of conditional probability, we find that this is equal to \(P(A|B)\), which verifies the derived equation directly.

Key concepts

Axioms of Probability

Probability serves as a mathematical framework for quantifying uncertainty, and at its core lie the axioms of probability—fundamental principles that govern how we assign probabilities to events.

Non-negativity

Firstly, non-negativity dictates that the probability of any event is never negative, reflecting the idea that an event cannot occur less than never.

Normalization

Normalization, the second axiom, requires that the probability of the entire sample space equals 1, akin to saying that something will definitely happen.

Additivity

Lastly, additivity concerns itself with mutually exclusive events—those which cannot occur simultaneously. When dealing with such distinct events, the probability of any one of them occurring is the sum of their separate probabilities. These axioms ensure a consistent and rigorous approach to assessing probabilities.

Normalization

Normalization is the axiom that anchors probabilities to a scale of certainty by asserting that the probability of the entire sample space is 1. In the context of conditional probability, normalization is evidenced by how the probability of the entire sample space, given any event B, is still 1. For instance, if event B is certain, then the sample space of possible outcomes restricted by B, by definition, contains B exclusively, and the probability of B happening given B has occurred is indeed 1.

Additivity

Additivity is the cornerstone of compound events, simplifying the complexity of overlapping possibilities. It states that if you have a collection of events that can't happen at the same time—termed mutually exclusive—the probability of any one of these events occurring is merely the sum of their individual probabilities. This easy addition holds up even if the number of events stretches into infinity, but two events can't interfere with one another; they are independent in their occurrence.

Non-Negativity

Non-negativity is an axiom that reassures us of the sane nature of probability - you cannot have less than no chance of something happening. It sets the lower boundary of probability at zero. This reflects the simple truth that an event either happens or it doesn't; there is no realm of occurrence less likely than an impossibility.

Law of Total Probability

The law of total probability is like a unifying thread that brings together separate scenarios underneath a single comprehensive outcome. It dictates that if you divide the sample space into exhaustive and mutually exclusive parts, the total probability of any event is the sum of the probabilities of that event occurring within each of those parts. This law proves invaluable when we tackle problems with conditional probabilities, where the event of interest is considered over different complementary scenarios, ensuring we account for all possible ways an event can occur.

Mutually Exclusive Events

Imagine two events with no chance of a shared moment; these events are known as mutually exclusive. The concept of mutually exclusive events is critical when applying the axiom of additivity, as these events cannot occur at the same moment, making their probabilities neatly summable. When analyzing conditional probabilities, recognizing events that are mutually exclusive is crucial, as this acknowledgment guides us through accurate calculations and helps avoid the pitfall of double-counting probabilities.