Exercise


May 06'23

Answer

Solution: C

The four possible outcomes for which X + Y = 3 are given below, with their probabilities.

[[math]] \begin{align*} (0,3)&: e^{-1.7} \frac{2.3^3e^{-2.3}}{3!} = 2.0278e^{-4} \\ (1,2)&: 1.7 \frac{e^{-1.7}}{1!} \frac{2.3^2e^{-2.3}}{2!} = 4.4965e^{-4} \\ (2,1) &: \frac{1.7^2 e^{-1.7}}{2!} \frac{2.3e^{-2.3}}{1!} = 3.3235e^{-4} \\ (3,0) &: \frac{1.7^3 e^{-1.7}}{3!}e^{-2.3} = 0.8188e^{-4}. \end{align*} [[/math]]

The conditional probabilities are found by dividing the above probabilities by their sum. They are, 0.1901, 0.4215, 0.3116, 0.0768, respectively. These apply to the X – Y values of –3, –1, 1,and 3. The mean is

–3(0.1901) –1(0.4215) + 1(0.3116) + 3(0.0768) = –0.4498. 

The second moment is

9(0.1901) + 1(0.4215) + 1(0.3116) + 9(0.0768) = 3.1352. 

The variance is 2.9329.

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

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