A life office sells 5-year term assurance policies to lives aged 60. Each policy has a sum assured of £10,000 payable at the end of the year of death. Premiums of £200 are payable annually in advance throughout the 5-year term or until earlier death. Let L denote the present value of the insurer’s loss on one of these policies, at policy outset, ignoring expenses. (i) Write down an expression for L . (ii) Assuming AM92 Ultimate mortality and 5½% pa interest, calculate the expected value and standard deviation of L . The approach in CMP seems to be convenient to calculate the variance but I found it difficult to see where the values of Loss L came from. I approached from calculating the expected value of benefits and premium but got a slightly different result. Could you help me understand the working in this question, especially the loss amount? I want to know the logic and understanding behind using the loss amounts with their associated probabilities. Thank you lots in advance!
Hi Sudhan, L is the future loss random variable. It is the present value of loss on 1 policy from the life office’s point of view i.e. PV of outgo (claims) minus PV of income (premium). If the life dies within 5 years (i.e. if K60 = 0,1,2,3,4) then a sum assured of 10,000 will be paid out at the end of year of death. The PV of this is 10,000v^K60+1 premiums will be paid up until the life dies i.e. an annuity (in advance) paid for K60+1 years. Therefore PV of loss (L) = 10,000v^K60+1 – 200adue:<K60+1> If the life survives the 5 years (i.e. K60 >= 5) then no sum assured will be paid premiums will be paid for the full 5 years i.e. an annuity (in advance) paid for 5 years. Therefore PV of loss (L) = -200adue:<5> For the EPV, it is not possible to use annuity/assurance functions from the tables because interest is not at 4 or 6%. We therefore need to determine the PV of loss for each value of K60 then multiple by the associated probabilities and sum to determine the EPV. If K60 = 0, L = 10,000v – 200adue:<1> = 10,000(1.055)^-1 – 200 x 1 = 9,278.67 Pr(K60 = 0) is the probability that the 60 year old dies in the first year i.e. q60 which can be taken directly from the AM92 If K60 = 1, L = 10,000v^2 – 200 adue:<2> =10,000(1.055)^-2 – 200 x [ (1 – (1.055)^-2) / ( 0.055/1.055) ] = 8,594.95 Pr(K60 = 1) is the probability that the 60 year old dies in the second year i.e. p60q61 which can be calculated by taking q61 directly from the AM92 and multiplying by (1-q60)… or using dx and lx suggested in the solution. etc… I’m not sure which approach you’ve taken when you said you approached by calculating the EPV of benefits and premiums but got a slightly different result. If the interest was 4% or 6% you could use term assurance / temporary annuity functions from the tables to determine the EPV of the benefits and premiums (and therefore the loss) i.e. 10,000TA:60:<5> - 200adue:60:<5>. However, this is not possible because interest is at 5.5% so you need to follow the approach outlined above. Hope helpful but feel free to come back to me for clarification Thanks, Michael
