An insurance policy pays a total medical benefit consisting of two parts for each claim. Let [math]X[/math] represent the part of the benefit that is paid to the surgeon, and let [math]Y[/math] represent the part that is paid to the hospital. The variance of [math]X[/math] is 5000, the variance of [math]Y[/math] is 10,000, and the variance of the total benefit, [math]X[/math] + [math]Y[/math], is 17,000. Due to increasing medical costs, the company that issues the policy decides to increase [math]X[/math] by a flat amount of 100 per claim and to increase [math]Y[/math] by 10% per claim.

Calculate the variance of the total benefit after these revisions have been made.

• 18,200
• 18,800
• 19,300
• 19,520
• 20,670

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

Solution: C

We use the relationships

[[math]] \operatorname{\operatorname{Var}}(aX +b) = a^2 \operatorname{\operatorname{Var}}(X), \, \operatorname{Cov}(aX,bY) = ab \operatorname{Cov}(X,Y) [[/math]]

and

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

First we observe

[[math]] 17,000 = \operatorname{Var} ( X + Y ) = 5000 + 10, 000 + 2 \operatorname{Cov}( X , Y ) [[/math]]

and so [math]\operatorname{Cov}(X,Y) = 1000 [/math]. We want to find

[[math]] \begin{align*} \operatorname{Var} [( X + 100 ) + 1.1Y] = \operatorname{Var}[( X + 1.1Y ) + 100 ] &= \operatorname{Var} [ X + 1.1Y ] \\ & = \operatorname{Var} X + \operatorname{Var}[(1.1) Y ] + 2 \operatorname{Cov}( X ,1.1Y )\\ &= \operatorname{Var} X + (1.1)^2 \operatorname{Var} Y + 2 (1.1) \operatorname{Cov}( X , Y ) \\ &= 5000 + 12,100 + 2200 \\ &= 19,300. \end{align*} [[/math]]

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