Question

Consider the two statements: 
\[ S_1:\ \exists \text{ random variables } X \text{ and } Y \text{ such that } \big( \mathbb{E}[(X-\mathbb{E}(X))(Y-\mathbb{E}(Y))] \big)^2 > \mathrm{Var}[X]\mathrm{Var}[Y] \] \[ S_2:\ \text{For all random variables } X \text{ and } Y,\ \mathrm{Cov}[X,Y] = \mathbb{E}\big[\,|X-\mathbb{E}[X]|\,|Y-\mathbb{E}[Y]|\,\big] \] Which one of the following choices is correct?

• A. Both \( S_1 \) and \( S_2 \) are true.
• B. \( S_1 \) is true, but \( S_2 \) is false.
• C. \( S_1 \) is false, but \( S_2 \) is true.
• D. Both \( S_1 \) and \( S_2 \) are false.

Hint:

Cauchy–Schwarz inequality provides an upper bound on covariance in terms of variances.

Answer 1

The Correct Option is D

Step 1: Analysis of Statement \( S_1 \).
By the Cauchy–Schwarz inequality, \[ \big( \mathbb{E}[(X-\mathbb{E}X)(Y-\mathbb{E}Y)] \big)^2 \le \mathrm{Var}[X]\mathrm{Var}[Y]. \] Hence, the inequality in \( S_1 \) can never hold. Therefore, \( S_1 \) is false.

Step 2: Analysis of Statement \( S_2 \).
In general, \[ \mathrm{Cov}[X,Y] = \mathbb{E}[(X-\mathbb{E}X)(Y-\mathbb{E}Y)], \] which is not equal to the expectation of the product of absolute deviations. Thus, \( S_2 \) is also false.

Step 3: Conclusion.
Since both statements are false, the correct option is (D).