Company XYZ provides a warranty on a product that it produces. Each year, the number of warranty claims follows a Poisson distribution with mean [math]c[/math]. The probability that no warranty claims are received in any given year is 0.60. Company XYZ purchases an insurance policy that will reduce its overall warranty claim payment costs. The insurance policy will pay nothing for the first warranty claim received and 5000 for each claim thereafter until the end of the year.

Calculate the expected amount of annual insurance policy payments to Company XYZ.

• 554
• 872
• 1022
• 1354
• 1612

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

Solution: A

Let N denote the number of warranty claims received. Then,

[[math]] 0.6 =\operatorname{P}( N =0) =e^{−c} ⇒ c =− \ln(0.6) =0.5108. [[/math]]

The expected yearly insurance payments are:

[[math]] \begin{align*} &5000[ \operatorname{P}( N = 2) + 2 \operatorname{P}( N = 3) + 3\operatorname{P}( N = 4) + \cdots] \\ &= 5000[ \operatorname{P}( N = 1)+ 2 \operatorname{P}( N = 2)+ 3\operatorname{P}( N = 3) + \cdots] − 5000[ \operatorname{P}( N = 1) + \operatorname{P}( N =+ 2) \operatorname{P}( N = 3) + \cdots] \\ &=5000 \operatorname{E}( N ) − 5000[1 − \operatorname{P}( N =0)] =5000(0.5108) − 5000(1 − 0.6) =554. \end{align*} [[/math]]

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