The distribution of the number of claims, N, is a member of the (a, b, 0) class. You are given:

1. [math]\operatorname{P}(N=k) = p_k[/math]
2. [math]\frac{p_6}{p_4} = 0.5 [/math] and [math]\frac{p_5}{p_4} = 0.8[/math]

A zero-modified distribution, [math]N^M[/math], associated with [math]N[/math] has [math]\operatorname{P}(N^M = 0) = 0.1 [/math].

Calculate [math]\operatorname{E}(N^M) [/math].

• 3.64
• 3.73
• 3.85
• 4.00
• 4.05

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

Key: A

[[math]] \begin{aligned} &\frac{p_k}{p_{k-1}} = a + \frac{b}{k} \Rightarrow 0.8 = a + \frac{b}{5} \quad \textrm{and} \frac{0.5}{0.8} = 0.625 = a + \frac{b}{6} \\ &\Rightarrow 0.8 − 0.625 = b(\frac{1}{5} - \frac{1}{6} ) \Rightarrow b = 5.25 \quad \textrm{and} \, a = -0.25. \\ &\Rightarrow \textrm{ N is binomial with } \, \frac{-q}{1-q} = -025 \quad \textrm{and} \, (m+1) \frac{q}{1-q} = 5.25 \Rightarrow m = 20 \quad \textrm{and} \, q = 0.2 \\ &\Rightarrow \operatorname{E}(N^M) = (1-p_0^M) \frac{mq}{1-(1-q)^m} = 3.64. \end{aligned} [[/math]]

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