Exercise


ABy Admin
May 06'23

Answer

Solution: A

[[math]] \begin{align*} \operatorname{Cov}( C_1 , C_2 ) &= \operatorname{Cov}( X + Y , X + 1.2Y ) \\ &= \operatorname{Cov}( X , X ) + \operatorname{Cov}(Y , X ) + \operatorname{Cov}( X ,1.2Y ) + \operatorname{Cov}( Y,1.2Y ) \\ &= \operatorname{Var}(X) + \operatorname{Cov}( X , Y ) + 1.2\operatorname{Cov}( X , Y ) + 1.2\operatorname{Var}(Y) \\ &= \operatorname{Var}(X) + 2.2 \operatorname{Cov}( X , Y ) + 1.2\operatorname{Var}(Y) \end{align*} [[/math]]

[[math]] \operatorname{Var}(X) = \operatorname{E}( X^2 ) − ( \operatorname{E}( X ) )^2 = 27.4 − 5^2 = 2.4 [[/math]]

[[math]] \operatorname{Var}(Y) = \operatorname{E}( Y^2 ) − ( \operatorname{E}( Y ) )^2 = 27.4 − 5^2 = 2.4 [[/math]]

[[math]] \operatorname{Var}(X+Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) + 2 \operatorname{Cov}( X , Y ) [[/math]]

[[math]] \operatorname{Cov}( C_1 , C_2 ) = 2.4 + 2.2 (1.6 ) + 1.2 ( 2.4 ) = 8.8 [[/math]]

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

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