Question

Consider the probability space $(\Omega, \mathcal{G}, P)$, where $\Omega = [0,2]$ and $\mathcal{G} = \{\emptyset, \Omega, [0,1], (1,2]\}$. Let $X$ and $Y$ be two functions on $\Omega$ defined as} \[ X(\omega) = \begin{cases} 1 & \text{if } \omega \in [0,1] \\ 2 & \text{if } \omega \in (1,2] \end{cases} \] and \[ Y(\omega) = \begin{cases} 2 & \text{if } \omega \in [0,1.5] \\ 3 & \text{if } \omega \in (1.5,2] \end{cases} \] Then which one of the following statements is true?

• A. $X$ is a random variable with respect to $\mathcal{G}$, but $Y$ is not a random variable with respect to $\mathcal{G}$
• B. $Y$ is a random variable with respect to $\mathcal{G}$, but $X$ is not a random variable with respect to $\mathcal{G}$
• C. Neither $X$ nor $Y$ is a random variable with respect to $\mathcal{G}$
• D. Both $X$ and $Y$ are random variables with respect to $\mathcal{G}$

Hint:

- A random variable is defined as a function whose pre-images of Borel sets belong to the sigma-algebra $\mathcal{G}$.
- If a function's set values or their pre-images do not match the elements of the sigma-algebra, it is not a random variable.

Answer 1

The Correct Option is A

1) Random Variable Definition:
A function \( f: \Omega \to \mathbb{R} \) is a random variable with respect to a sigma-algebra \( \mathcal{G} \) if the pre-image of any Borel set in \( \mathbb{R} \) is in \( \mathcal{G} \).

2) Analyzing \( X \):
The function \( X \) is defined as: \[ X(\omega) = \begin{cases} 1 & \text{if } \omega \in [0,1] \\ 2 & \text{if } \omega \in (1,2] \end{cases} \] The pre-images of the sets \( \{1\} \) and \( \{2\} \), which are \( \{[0,1]\} \) and \( \{(1,2]\} \), are both elements of the sigma-algebra \( \mathcal{G} \), hence \( X \) is a random variable with respect to \( \mathcal{G} \).

3) Analyzing \( Y \):
The function \( Y \) is defined as: \[ Y(\omega) = \begin{cases} 2 & \text{if } \omega \in [0,1.5] \\ 3 & \text{if } \omega \in (1.5,2] \end{cases} \] The pre-images of the sets \( \{2\} \) and \( \{3\} \) are \( \{[0,1.5]\} \) and \( \{(1.5,2]\} \), but \( \{[0,1.5]\} \) is not in \( \mathcal{G} \) (since \( [0,1.5] \) is not one of the sets in the sigma-algebra \( \mathcal{G} \)). Therefore, \( Y \) is not a random variable with respect to \( \mathcal{G} \).

Hence, the correct answer is (A).