Actuarial Outpost TIA Exam 2 #5 Variance with a deductible
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#1
05-24-2018, 08:54 PM
 ericp Member Join Date: Aug 2007 Posts: 262
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.
#2
05-24-2018, 09:01 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 7,638

Quote:
 Originally Posted by ericp 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.
#3
05-25-2018, 10:00 AM
 Jim Daniel Member SOA Join Date: Jan 2002 Location: Davis, CA College: Wabash College B.A. 1962, Stanford Ph.D. 1965 Posts: 2,617

Quote:
 Originally Posted by Academic Actuary 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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