Question

Let \(x\) and \(y\) be random variables, not necessarily independent, that take real values in the interval \([0, 1]\). Let \(z = xy\) and let the mean values of \(x, y, z\) be \(\bar{x}, \bar{y}, \bar{z}\), respectively. Which one of the following statements is TRUE?

• A. \(\bar{z} = \bar{x} \bar{y}\)
• B. \(\bar{z} \leq \bar{x} \bar{y}\)
• C. \(\bar{z} \geq \bar{x} \bar{y}\)
• D. \(\bar{z} \leq \bar{x}\)

Hint:

When working with random variables in bounded intervals, remember that products of variables are constrained by their individual ranges.

Answer 1

The Correct Option is D

The random variables \(x\) and \(y\) are in the interval \([0, 1]\), and \(z = xy\). For any pair of real values \(x, y \in [0, 1]\), the value of \(xy\) is always less than or equal to \(x\) (since \(y \leq 1\)). Therefore, the mean of \(z\), \(\bar{z}\), satisfies: \[ \bar{z} \leq \bar{x}. \] This is always true, regardless of the dependence or independence of \(x\) and \(y\). Analysis of the other options:
Option (A): \(\bar{z} = \bar{x} \bar{y}\) is true only when \(x\) and \(y\) are independent. This is not guaranteed here. Hence, (A) is incorrect.
Option (B): \(\bar{z} \leq \bar{x} \bar{y}\) is true due to the general inequality \(E[xy] \leq E[x]E[y]\) (Cauchy-Schwarz inequality), but this is weaker than the stricter constraint \(\bar{z} \leq \bar{x}\). Hence, (B) is not the strongest statement.
Option (C): \(\bar{z} \geq \bar{x} \bar{y}\) is false because the inequality \(E[xy] \leq E[x]E[y]\) holds.
Option (D): \(\bar{z} \leq \bar{x}\) is always true and the strongest statement in this case. Hence, (D) is correct. Final Answer: \[ \boxed{\text{(D)}} \]