Consider the two statements.
S1: There exist random variables X and Y such that
S2: For all random variables X and
Which one of the following choices is correct?
(A) Both S1 and S2 are true
(B) S1 is true, but S2 is false
(C) S1 is false, but S2 is true
(D) Both S1 and S2 are false
Answer: (D)
Explanation: Theorem : Square of Covariance is Less Than or Equal to Product of Variances.
Let X and Y be random variables.
Let the variances of X and Y exist and be finite.
Then:
(cov(X,Y))2 ≤ var(X)var(Y)
where cov(X,Y) denotes the covariance of X and Y.
Proof :
We have, by the definition of variance, that both:
E((X−E(X))2)
and:
E((Y−E(Y))2)
exist and are finite.
Therefore:
(cov(X,Y))2 = (E((X−E(X))(Y−E(Y))))2
Definition of Covariance
(cov(X,Y))2 ≤ E((X−E(X))2)E((Y−E(Y))2)
Square of Expectation of Product is Less Than or Equal to Product of Expectation of Squares
(cov(X,Y))2 = var(X)var(Y)
Definition of Variance.
Hence, both given statement (S1) and (S2) are false.
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