Question

Let \(X\) be a random variable with space \(\mathcal{D}\). For \(D \subset \mathcal{D}\), recall that the probability induced by \(X\) is \(P_{X}(D)=P[\\{c: X(c) \in D\\}] .\) Show that \(P_{X}(D)\) is a probability by showing the following: (a) \(P_{X}(\mathcal{D})=1\). (b) \(P_{X}(D) \geq 0\). (c) For a sequence of sets \(\left\\{D_{n}\right\\}\) in \(\mathcal{D}\), show that $$ \left\\{c: X(c) \in \cup_{n} D_{n}\right\\}=\cup_{n}\left\\{c: X(c) \in D_{n}\right\\} $$ (d) Use part (c) to show that if \(\left\\{D_{n}\right\\}\) is sequence of mutually exclusive events, then $$ P_{X}\left(\cup_{n=1}^{\infty} D_{n}\right)=\sum_{n=1}^{\infty} P_{X}\left(D_{n}\right) $$

Short answer

To prove \(P_{X}(D)\) to be a valid probability, the following steps were undertaken: Proved \(P_{X}(\mathcal{D}) = 1\); Proved \(P_{X}(D) \geq 0\); Proved that \(\{c: X(c) \in \cup_{n} D_{n}\} = \cup_{n}\{c: X(c) \in D_{n}\}\); Proved \(P_{X}(\cup_{n=1}^{\infty} D_{n}) = \sum_{n=1}^{\infty} P_{X}(D_{n})\). Each step used basic properties of probabilities and set operations.

Step-by-step solution

Prove total probability equals 1

To prove \(P_{X}(\mathcal{D}) = 1\), note that the set \(\{c: X(c) \in \mathcal{D}\}\) is the entire sample space \(\Omega\) since all outcomes of \(X\) are contained in \(\mathcal{D}\). Therefore, \(P_{X}(\mathcal{D}) = P(\Omega) = 1\) by definition of the probability of the entire sample space.

Prove probability is nonnegative

To prove \(P_{X}(D) \geq 0\), note that \(P_{X}(D)\) is a probability measure by definition, and one of the properties of a probability measure is that it is always nonnegative. Thus, \(P_{X}(D) \geq 0\).

Assessing set equality

To assess set equality, we want to show that \(\{c: X(c) \in \cup_{n} D_{n}\} = \cup_{n}\{c: X(c) \in D_{n}\}\). To see why this is true, consider the left-hand side (LHS). The LHS contains all outcomes \(c\) such that the corresponding value of \(X\) belongs to at least one of the sets in the sequence \(\{D_n\}\). This set of outcomes is the same as the union of the sets of outcomes in which the corresponding value of \(X\) belongs to each \(D_n\), which is exactly the right-hand side (RHS). Hence, the two sets are equal.

Prove addition rule holds

To prove \(P_{X}(\cup_{n=1}^{\infty} D_{n}) = \sum_{n=1}^{\infty} P_{X}(D_{n})\), note that if \(\{D_n\}\) is a sequence of mutually exclusive events, then by definition of probability measures, the probability of the union of these events is equal to the sum of their individual probabilities. Since we've shown in step 3 that the set equalities hold, we can directly use the properties of probabilities to state the result, thereby completing the proof.

Key concepts

Random Variable

A random variable is a fundamental concept in statistics and probability theory. It's a mathematical function that assigns a numerical value to each outcome of a random experiment. For example, when you toss a coin, you might define a random variable, let's call it 'C,' where C is 1 if the coin lands on heads and 0 if it lands on tails.

The set of all possible values that the random variable can take is called its 'sample space.' In our coin example, the sample space for C is {0, 1}. Random variables can be discrete, taking on distinct or countable values (like the coin toss), or continuous, taking on any value within a range. By converting random outcomes into numerical values, random variables allow us to calculate probabilities and build statistical models.

Probability Induced by a Random Variable

The probability induced by a random variable is a way to quantify the likelihood of different outcomes. It uses the function of the random variable to assign probabilities to subsets of its sample space. In the technical talk, for a given random variable \(X\) and a subset \(D\) of its domain, \(\mathcal{D}\), the probability induced by \(X\) on \(D\) is denoted by \(P_X(D)\). This notation represents the probability that \(X\) takes a value within the set \(D\).

The rule that \(P_X(\mathcal{D}) = 1\) underscores that the total probability over the entire set of possible outcomes is always 1—essentially, the random variable will take some value within its domain with certainty. The fact that \(P_X(D) \geq 0\) for all subsets \(D\) reflects the non-negativity of probabilities; \(0\) meaning the event won't occur and \(1\) meaning it's certain—nothing in between can be less than \(0\).

Sequence of Mutually Exclusive Events

A sequence of mutually exclusive events is a string of outcomes where the occurrence of one event prevents any of the others from occurring at the same time. Think of it as booking a single room in a hotel—once you've booked it for a specific date, no one else can book that same room for the same date. Each event is like a separate booking; they don't overlap.

In probability, if you have a sequence of such events, the probabilities of these events don't interfere with one another. This is why the probability of the union of these mutually exclusive events equals the sum of their individual probabilities. It's like adding up the probabilities of booking each room separately on different dates. This property is part of the axioms of probability and helps calculate the likelihood of any one of several possible non-overlapping events occurring.