An auto insurance company insures an automobile worth 15,000 for one year under a policy with a 1,000 deductible. During the policy year there is a 0.04 chance of partial damage to the car and a 0.02 chance of a total loss of the car. If there is partial damage to the car, the amount [math]X[/math] of damage (in thousands) follows a distribution with density function

[[math]] f(x) = \begin{cases} 0.5003e^{− x /2}, \, 0 \lt x \lt 15 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the expected claim payment.

• 320
• 328
• 352
• 380
• 540

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

Solution: B

Let [math]Y[/math] denote the claim payment made by the insurance company. Then

[[math]] Y = \begin{cases} 0 \quad \textrm{with probability 0.94} \\ \max(0,x-1) \quad \textrm{with probability 0.04} \\ 14 \quad \textrm{with probability 0.02} \end{cases} [[/math]]

and

[[math]] \begin{align*} \operatorname{E}[Y] &= ( 0.94 )( 0 ) + ( 0.04 )( 0.5003)\int_1^{15} (x-1)e^{-x/2} dx + (0.02)(14) \\ &= 0.28 + ( 0.020012 ) \left [-2x^{-x/2} \Big |_1^{15} + 2 \int_1^{15}e^{-x/2} dx - \int_1^{15}e^{-x/2} dx \right ] \\ &= 0.28 + ( 0.020012 ) \left [-30e^{-7.5} + 2e^{-0.5} -2e^{-x/2} \Big |_1^{15} \right ] \\ &= 0.28 + ( 0.020012 ) \left [-30e^{-7.5} + 2e^{-0.5} -2e^{-7.5} + 2e^{-0.5} \right ] \\ &= 0.28 + ( 0.020012 ) \left [-32e^{-7.5} + 4e^{-0.5} \right ] \\ &= 0.28 + ( 0.020012 )( 2.408 ) \\ &= 0.328 \quad \textrm{(in thousands)} \end{align*} [[/math]]

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