ABy Admin
May 07'23

Exercise

Under an insurance policy, a maximum of five claims may be filed per year by a policyholder. Let [math]p(n)[/math] be the probability that a policyholder files n claims during a given year, where [math]n = 0,1,2,3,4,5 [/math]. An actuary makes the following observations:

  1. [math]p(n) ≥ p (n + 1) [/math] for [math]n = 0, 1, 2, 3, 4 [/math] .
  2. The difference between [math]p (n) [/math] and [math]p(n + 1)[/math] is the same for [math] n = 0,1,2,3,4 [/math].
  3. Exactly 40% of policyholders file fewer than two claims during a given year.

Calculate the probability that a random policyholder will file more than three claims during a given year.

  • 0.14
  • 0.16
  • 0.27
  • 0.29
  • 0.33

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

ABy Admin
May 07'23

Solution: C

Due to the equal spacing of probabilities, [math]p= p_0 − nc[/math] for [math]c = 1, 2, 3, 4, 5.[/math] Also,

[[math]]0.4 = p_0 + p_1 = p_0 + p_0 − c = 2 p_0 − c.[[/math]]

Because the probabilities must sum to 1,

[[math]]1 = p_0 + p_0 − c + p_0 − 2c + p_0 − 3c + p_0 − 4c + p_0 − 5c = 6 p_0 − 15c. [[/math]]

This provides two equations in two unknowns. Multiplying the first equation by 15 gives [math]6 =30 p_0 − 15c[/math]. Subtracting the second equation gives [math]5= 24 p_0 ⇒ p_0= 5 / 24 .[/math]

Inserting this in the first equation gives [math]c = 1/60.[/math] The requested probability is

[[math]] p_4 + p_5= 5 / 24 − 4 / 60 + 5 / 24 − 5 / 60= 32 /120= 0.267. [[/math]]

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

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