A piece of equipment is being insured against early failure. The time from purchase until failure of the equipment is exponentially distributed with mean 10 years. The insurance will pay an amount x if the equipment fails during the first year, and it will pay 0.5x if failure occurs during the second or third year. If failure occurs after the first three years, no payment will be made.

Calculate x such that the expected payment made under this insurance is 1000.

• 3858
• 4449
• 5382
• 5644
• 7235

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

Solution: D

We want to find [math]x[/math] such that

[[math]] \begin{align*} 1000 = \operatorname{E}[P] &= \int_0^1 \frac{x}{10} e^{-t/10} dt + \int_1^3 \frac{x}{2} \frac{1}{10} e^{-t/10} dt \\ &= 1000 \int_0^1 x(0.1)e^{-t/10} dt + \int_1^30.5x(0.1)e^{-t/10} dt \\ &= -xe^{-t/10} \Big |_0^1 - 0.5xe^{-t/10} \Big |_1^3 \\ &= -xe^{-1/10} + x - 0.5xe^{-3/10} + 0.5xe^{-1/10} + 0.5x^{-1/10} \\ &= 0.1722x. \end{align*} [[/math]]

Thus [math]x = 5644 [/math].

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