Let [math]X[/math] denote the loss amount sustained by an insurance company’s policyholder in an auto collision. Let [math]Z[/math] denote the portion of [math]X[/math] that the insurance company will have to pay. An actuary determines that [math]X[/math] and [math]Z[/math] are independent with respective density and probability functions

[[math]] f(x) = \begin{cases} (1/8)e^{-x/8}, \, x \gt0 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

and

[[math]] \operatorname{P}[Z = z] = \begin{cases} 0.45, \, z= 1 \\ 0.55, \, z=0 \end{cases} [[/math]]

Calculate the variance of the insurance company’s claim payment [math]ZX[/math].

• 13.0
• 15.8
• 28.8
• 35.2
• 44.6

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

Solution: E

X has an Exponential distribution with mean 8 and variance 64. The second moment is 128. The mean and second moment of Z are both 0.45. Then (using the independence of X and Z),

[[math]] \begin{align*} \operatorname{E}( ZX ) = \operatorname{E}( Z ) \operatorname{E}( X ) = 0.45(8) \\ \operatorname{E}[( ZX )^ 2 ] = \operatorname{E}( Z ^2) \operatorname{E}( X^2 ) 0.45(128) = 57.6 \\ \operatorname{Var}(ZX) = 57.6 − 3.62 = 44.64. \end{align*} [[/math]]

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