Exercise


May 14'23

Answer

Key: C

[[math]] p(k) =\frac{2}{k} p(k-1) = (0 + \frac{2}{k})p(k-1) [[/math]]

Thus an (a, b, 0) distribution with a = 0, b = 2.

Thus Poisson with [math]\lambda = 2 [/math].

[[math]] p(4) = \frac{e^{-2}2^4}{4!} = 0.09 [[/math]]

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

00