A motorist makes three driving errors, each independently resulting in an accident with probability 0.25. Each accident results in a loss that is exponentially distributed with mean 0.80. Losses are mutually independent and independent of the number of accidents. The motorist’s insurer reimburses 70% of each loss due to an accident.

Calculate the variance of the total unreimbursed loss the motorist experiences due to accidents resulting from these driving errors.

• 0.0432
• 0.0756
• 0.1782
• 0.2520
• 0.4116

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

Solution: B

Let N denote the number of accidents, which is binomial with parameters 3 and 0.25 and thus has mean 3(0.25) = 0.75 and variance 3(0.25)(0.75) = 0.5625.

Let X i denote the unreimbursed loss due to the ith accident, which is 0.3 times an exponentially distributed random variable with mean 0.8 and therefore varianc\operatorname{E}(0.8)2 = 0.64. Thus, X i has mean 0.8(0.3) = 0.24 and variance 0.64(0.3) 2 = 0.0576 .

Let X denote the total unreimbursed loss due to the N accidents.

This problem can be solved using the conditional variance formula. Note that independence is used to write the variance of a sum as the sum of the variances.

[[math]] \begin{align*} \operatorname{Var}(X ) &= \operatorname{Var}[ \operatorname{E}( X | N )] + \operatorname{E}[\operatorname{Var}( X | N )] \\ &= \operatorname{Var}[ \operatorname{E}( X_1 + \cdots + X_N )] + \operatorname{E}[\operatorname{Var}( X 1 + \cdots + X_N )] \\ &= \operatorname{Var}[ N\operatorname{E}( X_1 )] + \operatorname{E}[ N\operatorname{Var}( X_1 )] \\ &= \operatorname{Var}(0.24 N ) + \operatorname{E}(0.0576 N ) \\ &= 0.242 \operatorname{Var}( N ) + 0.0576 \operatorname{E}( N ) \\ &= 0.0576(0.5625) + 0.0576(0.75)= 0.0756. \end{align*} [[/math]]

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