Let [math]X[/math] be a random variable that takes on the values –1, 0, and 1 with equal probabilities. Let [math]Y = X^2 [/math] . Which of the following is true?

• Cov(X, Y) > 0; the random variables X and Y are dependent
• Cov(X, Y) > 0; the random variables X and Y are independent
• Cov(X, Y) = 0; the random variables X and Y are dependent
• Cov(X, Y) = 0; the random variables X and Y are independent
• Cov(X, Y) < 0; the random variables X and Y are dependent

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: C

[[math]] \operatorname{Cov}( X , Y ) = E ( XY ) − E ( X ) E (Y ) - \operatorname{E}( X^3 ) − \operatorname{E}( X )\operatorname{E}( X^2 ) [[/math]]

[[math]] \operatorname{E}( X ) = \operatorname{E}(X^3) = (1/3)(−1 + 0 +1) = 2/3 [[/math]]

[[math]] \operatorname{E}( X^2 ) = (1/3)(1+0+1) = 2/3 [[/math]]

[[math]] \operatorname{Cov}( X , Y ) = 0 − 0(2 / 3) = 0. [[/math]]

They are dependent, because

[[math]] \operatorname{P}( X= 0, Y= 0)= \operatorname{P}( X= 0, X^2 =0)= \operatorname{P}( X= 0)= 1/ 3 [[/math]]

but

[[math]] \operatorname{P}( X= 0) \operatorname{P}(Y= 0)= (1/ 3)(1/ 3)= 1/ 9 ≠ 1/ 3. [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.