Exercise


ABy Admin
May 04'23

Answer

Solution: C

Let X be the number of tornadoes and Y be the conditional distribution of X given that X is at least one. There are (at least) two ways to solve this problem. The first way is to begin with the probability function for Y and observe that starting the sums at zero adds nothing because that term is zero. Then note that the sums are the first and second moments of a regular Poisson distribution.

[[math]] p(y) = \operatorname{P}[Y = y] = \operatorname{P}[X = y | X \gt 0] = \frac{\operatorname{P}[X=y]}{\operatorname{P}[X\gt0]} = \frac{3^ye^{-3}/y!}{1-e^{-3}}, \, y = 1,2, \ldots [[/math]]

[[math]] \operatorname{E}(Y) = \frac{1}{1-e^{-3}} \sum_{y=1}^{\infty} y\frac{3^ye^{-3}}{y!} = \frac{1}{1-e^{-3}} \sum_{y=0}^{\infty} y \frac{3^ye^{-3}}{y!} = \frac{3}{1-e^{-3}} [[/math]]

[[math]] \operatorname{E}(Y^2) = \frac{1}{1-e^{-3}} \sum_{y=1}^{\infty}y^2 \frac{3^ye^{-3}}{y!} = \frac{1}{1-e^{-3}} \sum_{y=0}^{\infty} y^2 \frac{3^ye^{-3}}{y!} = \frac{3 + 3^2}{1-e^{-3}} [[/math]]

[[math]] \operatorname{Var}(Y) = \frac{12}{1-e^{-3}} - \left( \frac{3}{1-e^{-3}}\right)^2 = 2.6609. [[/math]]

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

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