Let [math]N_1[/math] and [math]N_2[/math] represent the numbers of claims submitted to a life insurance company in April and May, respectively. The joint probability function of [math]N_1[/math] and [math]N_2[/math] is

[[math]] p(n_1,n_2) = \begin{cases} 2e^{-(x + 2y)}, \,\, n_1 = 1,2,3,\ldots \,\, n_2 = 1,2,3,\ldots \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the expected number of claims that will be submitted to the company in May, given that exactly 2 claims were submitted in April.

• [math]\frac{3}{16}(e^2 -1)[/math]
• [math]\frac{3}{16}e^2[/math]
• [math]\frac{3e}{4-e}[/math]
• [math]e^2 -1 [/math]
• [math]e^2[/math]

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

Solution: E

First, find the conditional probability function of [math]N_2[/math] given [math]N = n_1: p_{2|1}(n_2 | n_1) = \frac{p(n_1,n_2)}{p_1(n_1)}[/math] where [math]p_1(n_1) [/math] is the marginal probability function of [math]N_1[/math]. To find the latter, sum the joint probability function over all possible values of [math]N_2[/math] obtaining

[[math]] p_1(n_1) = \sum_{n_2 =1}^{\infty} p(n_1,n_2) = \frac{3}{4} \left (\frac{1}{4} \right)^{n_1-1} \sum_{n_2=1}^{\infty} (1-e^{-n_1})^{n_2-1} = \frac{3}{4}\left ( \frac{1}{4}\right )^{n_1-1}, [[/math]]

which is the probability function of a geometric random variable with parameter [math]p=e^{-n_1} [/math]. The mean of this distribution is [math] 1/p = 1/e^{-n_1} = e^{n_1}[/math] and becomes [math]e^2[/math] when [math]n_1 = 2 [/math].

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