• August 12th 2011, 04:03 PM
arcketer
Here's the problem I'm bugged about: An insurance company has 5000 policies and assumes these policies are all independent. Each policy is governed by the same distribution with a mean of $495 and a variance of$30,000. What is the probability that the total claims for the year will be less than $2,500,000? So I calculate the expected value of the 5000 policies as E(5000X) = 5000*E(X) = 5000*495 = 2,475,000. Then, the next step is to calculate the variance V(5000X) = 5000^2 * V(X) = 25,000,000 * 30,000 = 750,000,000,000. However, I am told that this is not the correct way to calculate the variance in this case. The property of V(aX) = a^2 * V(X) is one I've used for awhile and gotten comfortable with. Why does it not apply in this case? • August 12th 2011, 09:50 PM CaptainBlack Re: Question about calculating Variance Quote: Originally Posted by arcketer Here's the problem I'm bugged about: An insurance company has 5000 policies and assumes these policies are all independent. Each policy is governed by the same distribution with a mean of$495 and a variance of $30,000. What is the probability that the total claims for the year will be less than$2,500,000?

So I calculate the expected value of the 5000 policies as E(5000X) = 5000*E(X) = 5000*495 = 2,475,000.

Then, the next step is to calculate the variance V(5000X) = 5000^2 * V(X) = 25,000,000 * 30,000 = 750,000,000,000. However, I am told that this is not the correct way to calculate the variance in this case. The property of V(aX) = a^2 * V(X) is one I've used for awhile and gotten comfortable with. Why does it not apply in this case?

The mean is 5000*495

The variance is 5000*30000

CB
• August 13th 2011, 06:09 PM
arcketer
I know, the question however, is why does the variance identity V(aX) = a^2 * V(X) apply?
• August 13th 2011, 07:02 PM
CaptainBlack
$V(aX)=\int_D (ax-\overline{ax})^2f(x)\; dx=\int_D (ax-a\overline{x})^2f(x)\; dx$
........... $=\int_D a^2(x-\overline{x})^2f(x)\; dx=a^2\int_D (x-\overline{x})^2f(x)\; dx=a^2V(X)$