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&= \operatorname{Var}[ \operatorname{E}( X_1 + \cdots + X_N )] + E[\operatorname{Var}( X_1 + \cdots + X_N )] \\
&= \operatorname{Var}[ \operatorname{E}( X_1 + \cdots + X_N )] + E[\operatorname{Var}( X_1 + \cdots + X_N )] \\
&= \operatorname{Var}(1, 000 N ) + \operatorname{E}(1, 000, 000 N ) \\
&= \operatorname{Var}(1, 000 N ) + \operatorname{E}(1, 000, 000 N ) \\
&= 1, 0002 \operatorname{Var}( N ) + 1, 000, 000 \operatorname{E}( N ) \\
&= 1, 000 ^ 2\operatorname{Var}( N ) + 1, 000, 000 \operatorname{E}( N ) \\
&= 1, 000, 000(4) + 1, 000, 000(4) = 8, 000, 000.
&= 1, 000, 000(4) + 1, 000, 000(4) = 8, 000, 000.
\end{align*}
\end{align*}

Latest revision as of 22:53, 5 July 2024

Solution: C

Let [math]N[/math] denote the number of hurricanes, which is Poisson distributed with mean and variance 4.

Let [math]X_i[/math] denote the loss due to the [math]i^{th}[/math] hurricane, which is exponentially distributed with mean 1,000 and therefore variance 1,0002 = 1,000,000.

Let [math]X[/math] denote the total loss due to the [math]N[/math] hurricanes.

This problem can be solved using the conditional variance formula. Note that independence is used to write the variance of a sum as the sum of the variances.

[[math]] \begin{align*} \operatorname{Var}(X) &= \operatorname{Var}[ \operatorname{E}( X | N )] + E[\operatorname{Var}( X | N )] \\ &= \operatorname{Var}[ \operatorname{E}( X_1 + \cdots + X_N )] + E[\operatorname{Var}( X_1 + \cdots + X_N )] \\ &= \operatorname{Var}(1, 000 N ) + \operatorname{E}(1, 000, 000 N ) \\ &= 1, 000 ^ 2\operatorname{Var}( N ) + 1, 000, 000 \operatorname{E}( N ) \\ &= 1, 000, 000(4) + 1, 000, 000(4) = 8, 000, 000. \end{align*} [[/math]]

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