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Old 05-24-2018, 08:54 PM
ericp ericp is offline
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Default TIA Exam 2 #5 Variance with a deductible

I don't understand why the solution to this problem used a method other than using the formulas given for the exponential but no mention is made why they were not.

Given an exponential with a deductible and we are asked to find the variance of the payment, why can't the formula (Ex^2 - E[x^d]^2) - (same for Ex)^2 be used?

The solution to this problem uses a form of the double expectation/variance formula for the variance. Well, that formula never made any intuitive sense to me so I am wondering why that was chosen for the solution to this problem rather than what would seem to be the simpler approach using the formulas given and then just plug in values.

Anyone know?
If not, does anyone understand the method that was used in the solution?

Thanks.
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Old 05-24-2018, 09:01 PM
Academic Actuary Academic Actuary is offline
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Originally Posted by ericp View Post
I don't understand why the solution to this problem used a method other than using the formulas given for the exponential but no mention is made why they were not.

Given an exponential with a deductible and we are asked to find the variance of the payment, why can't the formula (Ex^2 - E[x^d]^2) - (same for Ex)^2 be used?

The solution to this problem uses a form of the double expectation/variance formula for the variance. Well, that formula never made any intuitive sense to me so I am wondering why that was chosen for the solution to this problem rather than what would seem to be the simpler approach using the formulas given and then just plug in values.

Anyone know?
If not, does anyone understand the method that was used in the solution?

Thanks.
I don't know the exact problem but I assume it is find the variance of YL not YP given X is exponential. This problem can be done in less than a minute. If X is less than d, then the mean and variance of YL are both 0. If X is greater than d, then the mean is theta and the variance is theta^2. The probabilities are F(d) and 1 - F(d) from the exponential.

Calculate the variance as the mean of the variances plus the variance of the means (which can be done using the bernoulli shortcut). The formula simplifies and might be worth memorizing.
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Old 05-25-2018, 10:00 AM
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Jim Daniel Jim Daniel is offline
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I don't know the exact problem but I assume it is find the variance of YL not YP given X is exponential. This problem can be done in less than a minute. If X is less than d, then the mean and variance of YL are both 0. If X is greater than d, then the mean is theta and the variance is theta^2. The probabilities are F(d) and 1 - F(d) from the exponential.

Calculate the variance as the mean of the variances plus the variance of the means (which can be done using the bernoulli shortcut). The formula simplifies and might be worth memorizing.
Another way to think of this calculation is through mixture distributions. That leads to the fact that E[YL^k] = E[YP^k] x Pr[ there is a positive payment] . If X is Exponential with mean theta and there is just an ordinary deductible, then of course YP is Exponential with the same mean theta.
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