An actuary analyzes a company’s annual personal auto claims, [math]M[/math], and annual commercial auto claims, [math]N[/math]. The analysis reveals that [math]\operatorname{\operatorname{Var}}(M) = 1600 [/math], [math]\operatorname{Var}(N) = 900 [/math], and the correlation between [math]M[/math] and [math]N[/math] is 0.64.

Calculate [math]\operatorname{\operatorname{Var}}(M + N)[/math].

• 768
• 2500
• 3268
• 4036
• 4420

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

Solution: D

[[math]] \begin{align*} 0.64 &= \rho = \frac{\operatorname{Cov}(M,N)}{\sqrt{\operatorname{Var}(M) \operatorname{Var}(N)}} \\ \operatorname{Cov}(M,N) &= 0.64 \sqrt{1600(900)} = 768 \\ \operatorname{Var}(M+N) &= \operatorname{Var}(M) + \operatorname{Var}(N) + 2\operatorname{Cov}(M,N) = 1600 + 900 + 2(768) = 4036 \end{align*} [[/math]]

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