I guess that kind of makes sense... On the test I'll just do E(X^2) - EX^2 to be safe instead of applying the variance...
what about the second question regarding the conditional probability? I'm a little weak on that subject so I'm actually not really sure what's going in that problem.
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Question on "Risk and Insurance"
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hey, it would not work that way...becuase it is mixed distribution....you could multiply just E(x) and E(x^2)...that fomula is not for Var(x)....i think you need to refer theory once agian...it might help you understand this concept better..hope it helps...
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Question on "Risk and Insurance"
Hi, new hereand I have two questions on the "Risk and Insurance" file found on the SOA website (http://www.soa.org/files/pdf/P-21-05.pdf)
On page 12 when they're calculating the variance of the insurance with 15% probability of actually paying a claim, while I understand how they got the variance, I don't know why it's not just Var(0.15X) = 0.15^2*Var(X), where X~exponential(0.1). So variance = 0.15^2 * 100 = 2.25 (which is vastly different from the 27.75 they calculated.
Also, on page 13, when they're calculating E(Y) = P(H=1)*P(X>5|H=1)*E(Y|X>5, H=1), where did the P(X>5|H=1) term go when they plugged in the variables? It looked like just P(H=1)*E(Y|X>5, H=1) when they did 0.15*integral.
Thanks so much and I look forward on being on the board!Tags: None
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