Consider the two statements. \(S_1 :\) There exist random variables \(X\) and…
2021
Consider the two statements.
\(S_1 :\) There exist random variables \(X\) and \(Y\) such that
\(\left(\mathbb E[(X-\mathbb E(X))(Y-\mathbb E(Y))]\right)^2>\textsf{Var}[X]\textsf{Var}[Y]\)
\(S_2 :\) For all random variables \(X\) and \(Y\),
\(Y, \textsf{Cov}[X,Y]=\mathbb E \left[|X-\mathbb E[X]|\;|Y-\mathbb E[Y]|\right ]\)
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.
Attempted by 33 students.
Show answer & explanation
Correct answer: D
Answer: Both statements are false.
Reason for the first statement:
By definition, Cov(X,Y) = E[(X - E[X])(Y - E[Y])]. The Cauchy–Schwarz inequality applied to the random variables X - E[X] and Y - E[Y] gives |Cov(X,Y)| ≤ sqrt(Var(X) Var(Y)), hence
Cov(X,Y)^2 ≤ Var(X) Var(Y).
Therefore a strict inequality Cov(X,Y)^2 > Var(X) Var(Y) cannot occur. Equality holds exactly when X - E[X] and Y - E[Y] are linearly dependent almost surely (for example, when Y = aX + b a.s.).
Reason for the second statement:
The proposed identity asserts Cov(X,Y) = E[|X - E[X]| |Y - E[Y]|]. This cannot hold in general because the right-hand side is always nonnegative, while the left-hand side can be zero or negative.
Counterexample:
Let X and Y be independent centered Rademacher variables: P(X=1)=P(X=-1)=1/2 and similarly for Y. Then E[X]=E[Y]=0, so Cov(X,Y)=E[XY]=0.
However, |X - E[X]| = |X| = 1 and |Y - E[Y]| = |Y| = 1 almost surely, so E[|X - E[X]| |Y - E[Y]|] = 1.
Thus Cov(X,Y) = 0 while E[|X - E[X]| |Y - E[Y]|] = 1, showing the claimed equality is false in general.
Conclusion: Both S1 and S2 are false.