GATE CS 2021 Set 1 — Question 45

To determine the correctness of the statements $S_1$ and $S_2$, we analyze them individually using the principles of probability theory.---

Analysis of Statement $S_1$

The term inside the square on the left-hand side of the inequality is the covariance of $X$ and $Y$:$\text{Cov}[X, Y] = \text{E}[(X - \text{E}[X])(Y - \text{E}[Y])]$Thus, the inequality in $S_1$ can be written as:$(\text{Cov}[X, Y])^2 > \text{Var}[X] \text{Var}[Y]$By the Cauchy-Schwarz inequality for random variables, for any two real-valued random variables $U$ and $V$ with finite second moments:$(\text{E}[UV])^2 \le \text{E}[U^2] \text{E}[V^2]$Setting $U = X - \text{E}[X]$ and $V = Y - \text{E}[Y]$, we get:$(\text{E}[(X - \text{E}[X])(Y - \text{E}[Y])])^2 \le \text{E}[(X - \text{E}[X])^2] \text{E}[(Y - \text{E}[Y])^2]$Since $\text{Var}[X] = \text{E}[(X - \text{E}[X])^2]$ and $\text{Var}[Y] = \text{E}[(Y - \text{E}[Y])^2]$, this simplifies to:$(\text{Cov}[X, Y])^2 \le \text{Var}[X] \text{Var}[Y]$This inequality holds universally for all random variables $X$ and $Y$. Therefore, it is impossible for any random variables $X$ and $Y$ to satisfy:$(\text{Cov}[X, Y])^2 > \text{Var}[X] \text{Var}[Y]$Thus, $S_1$ is false.---

Analysis of Statement $S_2$

Statement $S_2$ asserts that for all random variables $X$ and $Y$:$\text{Cov}[X, Y] = \text{E}[|X - \text{E}[X]| |Y - \text{E}[Y]|]$Let us test this statement with a counterexample. Let $X$ and $Y$ be two random variables such that $Y = -X$, where $X$ is a non-constant random variable. For simplicity, let $X$ be a standard normal random variable, $X \sim \mathcal{N}(0, 1)$. Then:

• $\text{E}[X] = 0$ and $\text{E}[Y] = 0$.
• $X - \text{E}[X] = X$ and $Y - \text{E}[Y] = -X$.

Now, we compute both sides of the equation in $S_2$: $ \text{Cov}[X, Y] = \text{E}[XY] = \text{E}[X(-X)] = -\text{E}[X^2] = -1$ $ \text{E}[|X - \text{E}[X]| |Y - \text{E}[Y]|] = \text{E}[|X| |-X|] = \text{E}[|X|^2] = \text{E}[X^2] = 1$Since $\text{LHS} = -1$ and $\text{RHS} = 1$, we have $\text{LHS} \neq \text{RHS}$. Thus, the equality does not hold for all random variables, which means $S_2$ is false.---

Conclusion

Both statements $S_1$ and $S_2$ are false.Correct Option: D