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Long-Term Actuarial Math Old Exam MLC Forum

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  #1  
Old 11-09-2019, 06:07 PM
erichuang1996 erichuang1996 is offline
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Default Covariance of Life Annuity and Life Insurance?

So this question:

Z is the present value random variable for a unit whole life insurance on (x) with benefits paid at the
end of the year of death.

Y is the present value random variable for a unit whole life annuity-due on (x).

With the given values, Ax = 0.3, 2Ax = 0.16, d = 0.04

and calculate: Cov(Z,Y)

I did it by doing

Cov(Y,Z) = E(YZ) - E(Y)E(Z)

= E(Y*(1-dY))-E(Y)E(Z)

= E(Y)-dE(Y^2) - E(Y)E(Z)
= (1-.3)/.04-.04*(1-.16)/.04^2 - (1-.3)/.04*.3
= -3.5 - 5.25
= -8.75


I used the fact that Z = 1-dY. However, this doesn't work out in the problem. The solution used Y = (1-Z)/d, which changes

E(YZ) to E(Z(1-Z)/d) which is positive 3.5 instead of negative 3.5

Am I missing something? This is driving me crazy

Last edited by erichuang1996; 11-09-2019 at 06:10 PM..
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Old 11-09-2019, 07:06 PM
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tkt tkt is offline
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Your E[Y^2] is incorrect. You can't calculate it that way because E[Y^2] is NOT equal to the EPV of annuity evaluated at twice of force of interest.

To get E[Y^2], the quickest way is to calculate Var[Y] first, which is (0.16-0.3^2)/(0.04^2) = 43.75. Then, E[Y^2] = Var[Y] + (E[Y])^2 = 43.75 + [(1-0.3)/0.04]^2 = 350.

Thus, E[YZ] = (1-0.3)/0.04 - 0.04*350 = +3.5.
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  #3  
Old 11-09-2019, 10:49 PM
erichuang1996 erichuang1996 is offline
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thank you. I thought there was something wrong with my arithmetic. Turns out it was because my method was wrong!
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  #4  
Old 11-10-2019, 01:56 AM
Academic Actuary Academic Actuary is offline
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There is a much easier way to calculate Cov (Y,Z). Substitute Y in terms of Z and use the fact that Cov(Z,Z) is the variance. The results is -Var(Z)/d.
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