The question said the premium $200 is payable in advance for a 5 years term assurance. Anyone can explain why there is a 6th premium payment at age 65 that result the -901.03 in the table? Thanks.
Its not 6th year. Its loss if life survives 5 years. Think it is as expressing random variable... \( X = V^{K_{60} + 1} K_x < 5 \\= 0 K_x \geq 5\) I haven't understood how the loss \(L_i\) is calculated at each duration. But i used 1st principle i.e. \( E\left( L_i\right)\) and got the correct answer And i am struggling with variance. My value of variance is approx. \(17000^2\)
Thanks for the help. I had no problem with variance. You need to square the Li at each duration and multiply by its respective probability. Li at each duration are: L1 = 10,000V^1 - 200(annuity in advance for 1 year) ......Keep doing it until L5, which has 2 possibility. L5 = 10,000V^5 - 200 (Annuity advance for 5) L5 = -200 (Annuity advance for 5) Then sum them all, you will get the variance. By the way how do you type annuity function nicely?
Thanks. But am not able to find how \(L_i\) calculated at each duration and hence not able to derive Variance using given table. Using 1st principal \( V\left(X\right) = E\left(X^2\right) - E^2\left(X\right)\) In this eqn am not able to calculate \(E\left(X^2\right)\) correctly. Annuity functions using LATEX
