Question

Show that $$ P\left(\bigcup_{i=1}^{n} E_{i}\right) \leqslant \sum_{i=1}^{n} P\left(E_{i}\right) $$ This is known as Boole's inequality. Hint: Either use Equation (1.2) and mathematical induction, or else show that \(\bigcup_{i=1}^{n} E_{i}=\bigcup_{i=1}^{n} F_{i}\), where \(F_{1}=E_{1}, F_{i}=E_{i} \bigcap_{j=1}^{i-1} E_{j}^{c}\), and use property (iii) of a probability.

Short answer

We defined new sets of events \(F_i\), where \(F_1 = E_1\) and for \(i = 2, 3, \ldots, n\), \(F_i = E_i \cap \left(\bigcap_{j=1}^{i-1} E_j^c\right)\). We proved that these \(F_i\) events are mutually exclusive and that \(\bigcup_{i=1}^n E_i = \bigcup_{i=1}^n F_i\). Since \(F_i\) events are mutually exclusive, we have \(P\left(\bigcup_{i=1}^n F_i\right) = \sum_{i=1}^n P(F_i)\) and since \(F_i \subseteq E_i\), we get \(P(F_i) \leq P(E_i)\) for all \(i\). Therefore, \(P\left(\bigcup_{i=1}^n E_i\right) = \sum_{i=1}^n P(F_i) \leq \sum_{i=1}^n P(E_i)\), proving Boole's inequality.

Step-by-step solution

Defining the Fi events

For this problem, we will define a new set of events \(F_i\) such that: \[ F_1 = E_1 \] \[ F_i = E_i \cap \left(\bigcap_{j=1}^{i-1} E_j^c\right) \: \text{ for } \: i = 2, 3, \ldots, n \]

Proving that the Fi events are mutually exclusive

To prove that the \(F_i\) events are mutually exclusive, we need to show that the intersection between any two events is the empty set, i.e., \(F_i \cap F_j = \emptyset\) for all \(i \neq j\). Let's take the intersection of \(F_i\) and \(F_j\) with \(i < j\): \[ F_i \cap F_j = \left(E_i \cap \left(\bigcap_{k=1}^{i-1} E_k^c\right)\right) \cap \left(E_j \cap \left(\bigcap_{k=1}^{j-1} E_k^c\right)\right) \] Applying the distributive property, we get: \[ F_i \cap F_j = E_j \cap E_i^c \cap \left(\bigcap_{k=1}^{i-1} E_k^c\right) \cap \left(\bigcap_{k=1}^{j-1} E_k^c\right) \] Since \(i < j\), it is clear that \(E_i^c\) will be part of both intersections. Thus, the intersection of \(F_i\) and \(F_j\) will always contain \(E_j \cap E_i^c \ne \emptyset\), resulting in \(F_i \cap F_j = \emptyset\). Therefore, we have proven that the \(F_i\) events are mutually exclusive.

Proving the union of the Ei and Fi events are equal

We want to show that: \[ \bigcup_{i=1}^n E_i = \bigcup_{i=1}^n F_i \] To prove this, we will use the fact that an element \(x\) is in the union of the \(E_i\) events if and only if it is in the union of the \(F_i\) events. Let \(x \in \bigcup_{i=1}^n E_i\). This means that \(x\) is in at least one of the \(E_i\) events. Let \(j\) be the first event that \(x \in E_j\). Since \(x\) is not in any \(E_i\) with \(i < j\), \(x \in \bigcap_{k=1}^{j-1} E_k^c\). Therefore, \(x \in F_j\) and \(x \in \bigcup_{i=1}^n F_i\). Now, let \(x \in \bigcup_{i=1}^n F_i\). This means that \(x\) is in at least one of the \(F_i\) events. By definition, if \(x \in F_i\), then \(x \in E_i\). Therefore, \(x \in \bigcup_{i=1}^n E_i\). We have shown that \(\bigcup_{i=1}^n E_i = \bigcup_{i=1}^n F_i\).

Applying probability properties to prove Boole's inequality

Since \(F_i\) events are mutually exclusive, we have: \[ P\left(\bigcup_{i=1}^n F_i\right) = \sum_{i=1}^n P(F_i) \] Recall that \(P\left(\bigcup_{i=1}^n E_i\right) = P\left(\bigcup_{i=1}^n F_i\right)\). So, we have: \[ P\left(\bigcup_{i=1}^n E_i\right) = \sum_{i=1}^n P(F_i) \] Since \(F_i \subseteq E_i\) for all \(i\), we have \(P(F_i) \leq P(E_i)\) for all \(i\). Therefore, we get: \[ P\left(\bigcup_{i=1}^n E_i\right) = \sum_{i=1}^n P(F_i) \leq \sum_{i=1}^n P(E_i) \] Thus, we have proven Boole's inequality, completing the exercise.

Key concepts

Probability Theory

At its core, probability theory is a branch of mathematics concerned with the analysis of random phenomena. The central concept is the probability, which quantifies the likelihood of events occurring within a set framework, often expressed as a number between 0 and 1.

In probability theory, we deal with various kinds of events, from simple ones like flipping a coin to complex series of occurrences with multiple possible outcomes. One of the seminal results within this domain is Boole's inequality, which gives an upper bound on the probability of the union of several events.

Boole's inequality is especially useful when events have complex interrelations and cannot easily be considered independent or mutually exclusive. Understanding probability theory allows one to navigate this uncertainty, make predictions, and quantify risk in many disciplines including science, engineering, finance, and social sciences.

Mathematical Induction

Mathematical induction is a powerful technique for proving statements or formulas which are asserted for all natural numbers. It operates by leveraging a domino effect, establishing the validity of a base case and then proving that, if any one case is true, it implies the next case is also true.

When using mathematical induction in probability, such as in the proof of Boole's inequality, one proves the base case for the smallest set, often when the number of events is 1. Subsequently, one assumes the statement to be true for a certain number of events, say n, and then demonstrates that it remains true when another event is added, making the total n+1 events. With mathematical induction, you can effectively create a 'proof for all numbers' one step at a time.

Mutually Exclusive Events

Mutually exclusive events are those that cannot happen at the same time. An intuitive example of mutually exclusive events is tossing a single coin; it cannot land both heads and tails simultaneously.

When events are mutually exclusive, the probability of their union is equal to the sum of their individual probabilities. However, in real-world scenarios, not all events are mutually exclusive. This is where Boole’s inequality becomes an important tool, as it provides an upper bound of the probability for the union of several events, not necessarily mutually exclusive.

Understanding whether events are mutually exclusive is crucial when applying probability theories, as it informs the techniques and formulas we use, such as Boole's inequality, to accurately calculate probabilities in more complex situations.