Problem 6 - A Computational Introduction to Number Theory and Algebra

Question

Let \(\mathcal{B}\) be an event, and let \(\left\\{\mathcal{B}_{i}\right\\}_{i \in I}\) be a finite, pairwise disjoint family of events whose union is \(B\). Generalizing the law of total probability (equations (8.9) and (8.10)\(),\) show that for every event \(\mathcal{A},\) we have \(\mathrm{P}[\mathcal{A} \cap \mathcal{B}]=\) \(\sum_{i \in I} \mathrm{P}\left[\mathcal{A} \cap \mathcal{B}_{i}\right],\) and if \(\mathrm{P}[B] \neq 0\) and \(I^{*}:=\left\\{i \in I: \mathrm{P}\left[\mathcal{B}_{i}\right] \neq 0\right\\},\) then $$ \mathrm{P}[\mathcal{A} \mid \mathcal{B}] \mathrm{P}[\mathcal{B}]=\sum_{i \in I^{*}} \mathrm{P}\left[\mathcal{A} \mid \mathcal{B}_{i}\right] \mathrm{P}\left[\mathcal{B}_{i}\right] $$ Also show that if \(\mathrm{P}\left[\mathcal{A} \mid \mathcal{B}_{i}\right] \leq \alpha\) for each \(i \in I^{*},\) then \(\mathrm{P}[\mathcal{A} \mid \mathcal{B}] \leq \alpha\)

Step-by-Step Solution

Answer

Question: Given a finite, pairwise disjoint family of events $\left\{\mathcal{B}_{i}\right\}_{i \in I}$, show that if $\mathrm{P}\left[\mathcal{A} \mid \mathcal{B}_{i}\right] \leq \alpha$ for each $i \in I^{*}$, then $\mathrm{P}[\mathcal{A} \mid \mathcal{B}] \leq \alpha$. Answer: To prove this statement, we can use the generalized law of total probability for events $\mathcal{A}$ and $\mathcal{B}$, where $\mathcal{B}$ is the union of the family of events $\left\{\mathcal{B}_{i}\right\}_{i \in I}$. Since $\mathrm{P}\left[\mathcal{A} \mid \mathcal{B}_{i}\right] \leq \alpha$ for each $i \in I^{*}$, we can apply the result that $\mathrm{P}[\mathcal{A} \mid \mathcal{B}] \mathrm{P}[\mathcal{B}] \leq \alpha \mathrm{P}[\mathcal{B}]$ by the law of total probability. Dividing both sides by $\mathrm{P}[\mathcal{B}]\) (which is nonzero), we obtain $\mathrm{P}[\mathcal{A} \mid \mathcal{B}] \leq \alpha$ as desired.

Step 1: Prove the first statement

Using the fact that \(\mathcal{B}_{i}\) are pairwise disjoint, we know that $$\mathcal{A} \cap \mathcal{B} = \mathcal{A} \cap \bigcup_{i \in I} \mathcal{B}_{i} = \bigcup_{i \in I} (\mathcal{A} \cap \mathcal{B}_{i})$$ Thus, by the axiom of probability, $$\mathrm{P}[\mathcal{A} \cap \mathcal{B}] = \mathrm{P}\left[\bigcup_{i \in I} (\mathcal{A} \cap \mathcal{B}_{i})\right]$$ Applying the axiom for the disjoint union of events, we get: $$\mathrm{P}[\mathcal{A} \cap \mathcal{B}] = \sum_{i \in I} \mathrm{P}\left[\mathcal{A} \cap \mathcal{B}_{i}\right]$$

Step 2: Prove the second statement

Assuming \(\mathrm{P}[\mathcal{B}] \neq 0\), we must show that: $$\mathrm{P}[\mathcal{A} \mid \mathcal{B}] \mathrm{P}[\mathcal{B}] = \sum_{i \in I^{*}} \mathrm{P}\left[\mathcal{A} \mid \mathcal{B}_{i}\right] \mathrm{P}\left[\mathcal{B}_{i}\right]$$ Now, notice that $$\mathrm{P}[\mathcal{A} \mid \mathcal{B}] \mathrm{P}[\mathcal{B}] = \mathrm{P}[\mathcal{A} \cap \mathcal{B}] = \sum_{i \in I} \mathrm{P}\left[\mathcal{A} \cap \mathcal{B}_{i}\right]$$ For \(i \in I^{*}\), we have \(\mathrm{P}\left[\mathcal{B}_{i}\right] \neq 0\), and so we can rewrite the summation as: $$\sum_{i \in I^{*}} \mathrm{P}\left[\mathcal{A} \cap \mathcal{B}_{i}\right] = \sum_{i \in I^{*}} \mathrm{P}\left[\mathcal{A} \mid \mathcal{B}_{i}\right] \mathrm{P}\left[\mathcal{B}_{i}\right]$$ which is the desired result.

Step 3: Prove the third statement

We are given that for each \(i \in I^{*}\), \(\mathrm{P}\left[\mathcal{A} \mid \mathcal{B}_{i}\right] \leq \alpha\), and we need to prove that \(\mathrm{P}[\mathcal{A} \mid \mathcal{B}] \leq \alpha\). Using the second statement, we have: $$\mathrm{P}[\mathcal{A} \mid \mathcal{B}] \mathrm{P}[\mathcal{B}] = \sum_{i \in I^{*}} \mathrm{P}\left[\mathcal{A} \mid \mathcal{B}_{i}\right] \mathrm{P}\left[\mathcal{B}_{i}\right]$$ Since \(\mathrm{P}\left[\mathcal{A} \mid \mathcal{B}_{i}\right] \leq \alpha\) and \(\mathrm{P}\left[\mathcal{B}_{i}\right] \geq 0\) for each \(i \in I^{*}\), we have: $$\mathrm{P}[\mathcal{A} \mid \mathcal{B}] \mathrm{P}[\mathcal{B}] \leq \sum_{i \in I^{*}} \alpha \mathrm{P}\left[\mathcal{B}_{i}\right] = \alpha \sum_{i \in I^{*}} \mathrm{P}\left[\mathcal{B}_{i}\right] = \alpha \mathrm{P}[\mathcal{B}]$$ Dividing both sides by \(\mathrm{P}[\mathcal{B}]\) (which is nonzero), we get: $$\mathrm{P}[\mathcal{A} \mid \mathcal{B}] \leq \alpha$$

Key Concepts

Conditional Probability

Understanding conditional probability is crucial in determining the likelihood of an event, given that another event has already occurred. It is denoted as \( P(A \mid B) \) and read as 'the probability of A given B'. This concept allows us to update our beliefs about the occurrence of an event A in the light of new information that event B has occurred. When calculating \( P(A \mid B) \) we assume that \( P(B) eq 0 \), and the formula used is \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \).

In relation to the law of total probability, we see how conditional probability plays a role in breaking down complex events into simpler, conditional ones. If you know that an event B is composed of several disjoint events \( B_i \), then you can express the probability of A given B as a weighted sum of the probabilities of A given these disjoint events \( B_i \), which can provide a more detailed picture of your probabilistic landscape.

Disjoint Events

When we talk about disjoint events, we refer to events that cannot occur simultaneously. In other words, the intersection of two disjoint events, \( A \cap B \) for events A and B, is an empty set. This concept is vital for the law of total probability because it allows us to 'partition' a sample space into several non-overlapping events.

In the provided exercise, we deal with a family of events \( \{ \mathcal{B}_{i} \} \) that are pairwise disjoint, which means each pair of different events in the family doesn't intersect. This feature is a cornerstone when it comes to calculating probabilities of combined events, as it ensures that we are not double counting any outcomes when we sum the probabilities.

Probability Axioms

The probability axioms provide a mathematical foundation for the probability theory. There are three axioms: the non-negativity axiom, stating that the probability of any event is non-negative; the normalization axiom, asserting that the probability of the entire sample space is 1; and the additivity axiom, which means that if two events \( A \text{ and } B \) are disjoint, the probability of their union is the sum of their probabilities, \( P(A \cup B) = P(A) + P(B) \).

These axioms underpin the exercise we are discussing. They were used to derive the probability of the intersection of event \( \mathcal{A} \) and the union of the disjoint events \( \mathcal{B}_{i} \) in the first step, showcasing the essential role they play in structuring probability theory.

Union of Events

The union of two or more events represents the event that at least one of the considered events occurs. For example, the union of events A and B, denoted as \( A \cup B \), occurs if either event A, event B, or both happen. Handling unions of events is very straightforward when the events are disjoint, as the probability of their union is simply the sum of their individual probabilities.

However, when events are not disjoint, calculating their union's probability requires adjustments to avoid overcounting shared outcomes. The law of total probability, which considers a union of disjoint events, sidesteps this issue and allows for a clear-cut calculation, a concept heavily utilized in the given exercise to establish foundational relationships within conditional probabilities.