Question

Suppose \(\mathscr{B}\) is the \(\sigma\)-field generated by some random variable \(Y\) (having, then, at most a denumerable number of possible values). Show that a random variable \(X\) is \(\mathscr{B}\)-measurable if and only if \(X=f(Y)\) for some real-valued function \(f\).

Short answer

In conclusion, a random variable \(X\) is \(\mathscr{B}\)-measurable if and only if \(X=f(Y)\) for some real-valued function \(f\). This is based on showing that if \(X\) is \(\mathscr{B}\)-measurable, then \(X=f(Y)\), and conversely, if \(X=f(Y)\), then \(X\) is \(\mathscr{B}\)-measurable. The proof involves analyzing the properties of the \(\sigma\)-field generated by \(Y\), and verifying that \(X^{-1}(A) \in \mathscr{B}\) for all \(A \in \mathscr{B}(\mathbb{R})\).

Step-by-step solution

Understanding \(\mathscr{B}\)-measurable random variables

A random variable \(X\) is said to be \(\mathscr{B}\)-measurable if for any set \(A\) in the Borel \(\sigma\)-field \(\mathscr{B}(\mathbb{R})\) (i.e., the smallest \(\sigma\)-field containing all open intervals), its preimage \(X^{-1}(A)\) belongs to \(\mathscr{B}\). In other words, \(X\) is \(\mathscr{B}\)-measurable if: \[X^{-1}(A) \in \mathscr{B} \, \, \text{for all} \, \, A \in \mathscr{B}(\mathbb{R})\]

Define \(\mathscr{B}\), the \(\sigma\)-field generated by \(Y\)

The \(\sigma\)-field \(\mathscr{B}\) is generated by the random variable \(Y\). It is the smallest \(\sigma\)-field containing all sets of the form \(\{Y \in A\}\), where \(A \in \mathscr{B}(\mathbb{R})\). Since \(Y\) has at most a denumerable number of possible values, we can write the set of all possible values of \(Y\) as \(\{y_1, y_2, \dots\}\). The sets \(\{Y \in A\}\) are those sets where \(Y\) takes a value from the set \(A\).

If \(X\) is \(\mathscr{B}\)-measurable, show that \(X=f(Y)\) for some function \(f\)

We are given that \(X\) is \(\mathscr{B}\)-measurable, which means \(X^{-1}(A) \in \mathscr{B}\) for all \(A \in \mathscr{B}(\mathbb{R})\). Let \(E\) be any element of \(\mathscr{B}\). Now consider the following set: \[A = \{x \in \mathbb{R} : f(x) \in E\}\] Since \(\mathscr{B}\) is the \(\sigma\)-field generated by \(Y\), there exists a set \(B \in \mathscr{B}(\mathbb{R})\) such that \(Y^{-1}(B) = E\). Thus, we have: \[X^{-1}(A) = \{Y \in B\}\] Since \(X^{-1}(A) \in \mathscr{B}\), we must have that \(X\) is a function of \(Y\). Specifically, \(X=f(Y)\) for some function \(f\).

If \(X=f(Y)\) for some function \(f\), show that \(X\) is \(\mathscr{B}\)-measurable

Let \(X=f(Y)\) for some function \(f\). We want to show that \(X^{-1}(A) \in \mathscr{B}\) for all \(A \in \mathscr{B}(\mathbb{R})\). Let \(A\) be any set in \(\mathscr{B}(\mathbb{R})\). Since \(X=f(Y)\), we have: \[X^{-1}(A) = Y^{-1}(f^{-1}(A))\] Since \(f^{-1}(A)\) is a set in \(\mathscr{B}(\mathbb{R})\), and \(Y\) is a random variable that defines the \(\sigma\)-field \(\mathscr{B}\), it follows that \(Y^{-1}(f^{-1}(A)) \in \mathscr{B}\). Therefore, we have shown that \(X^{-1}(A) \in \mathscr{B}\) for all \(A \in \mathscr{B}(\mathbb{R})\), which means that \(X\) is \(\mathscr{B}\)-measurable. In conclusion, we have shown that a random variable \(X\) is \(\mathscr{B}\)-measurable if and only if it is a function of \(Y\). That is, \(X=f(Y)\) for some real-valued function \(f\).