excans:Fd9e9305f8: Difference between revisions
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(Created page with "'''Solution: E''' Let S be the speed and X be the loss. Given S, X has an exponential distribution with mean 3X. Then, noting that the variance of an exponential random varia...") |
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<math display = "block"> | <math display = "block"> | ||
\begin{align*} | \begin{align*} | ||
\operatorname{Var}(X) &= \operatorname{Var}[E(X | S) ] + \operatorname{E}[ \operatorname{Var}(X | S) ] \\ | \operatorname{Var}(X) &= \operatorname{Var}[\operatorname{E}(X | S) ] + \operatorname{E}[ \operatorname{Var}(X | S) ] \\ | ||
&= \operatorname{Var}[3S] + \operatorname{E}[9S^2] \\ | &= \operatorname{Var}[3S] + \operatorname{E}[9S^2] \\ | ||
&= 9(20-5)^2/12 + 9[(20-5)^2/12 + 12.5^2] \\ | &= 9(20-5)^2/12 + 9[(20-5)^2/12 + 12.5^2] \\ | ||
Latest revision as of 02:04, 9 May 2023
Solution: E
Let S be the speed and X be the loss. Given S, X has an exponential distribution with mean 3X. Then, noting that the variance of an exponential random variable is the square of the mean, the variance of a uniform random variable is the square of the range divided by 12, and for any random variable the second moment is the variance plus the square of the mean:
[[math]]
\begin{align*}
\operatorname{Var}(X) &= \operatorname{Var}[\operatorname{E}(X | S) ] + \operatorname{E}[ \operatorname{Var}(X | S) ] \\
&= \operatorname{Var}[3S] + \operatorname{E}[9S^2] \\
&= 9(20-5)^2/12 + 9[(20-5)^2/12 + 12.5^2] \\
&= 1743.75
\end{align*}
[[/math]]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.