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: $$\boxed{\text{(D)}}$$