excans:9bf1ce362a: Difference between revisions
From Stochiki
(Created page with "'''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...") |
mNo edit summary |
||
Line 14: | Line 14: | ||
&= \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, | &= 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]]