Q3.24 in the Q&A Bank

Discussion in 'CT5' started by Cathy, Jan 27, 2007.

  1. Cathy

    Cathy Member

    Hi

    Can anyone explain the solution to Q3.24 to me? When I attempted the question I zeroised the profit vector and got a value for the loss at the end of year 1. This is the same value as the expression in the solutions evaluates to (-46.48). But I don't understand the reasoning behind the solution given.

    Can anyone explain it to me (in simple terms)?

    Thanks
     
  2. Louisa

    Louisa Member

    I think they're saying that the loss at the end of year 1 needs to be set up so that (calculating at 6% interest) it can be split into 3 parts,
    the loss at the end of year 1
    a sum which when accumulated gives the loss at the end of year 2
    a sum which when accumulated gives the loss at the end of year 3.

    The p's come in because we need to do all calculations per policy in force at the start of year 1, whereas the year 2 and year 3 losses are given per policy in force at the start of yr 2 and yr 3 respectively.

    Don't know if that qualifies as "simple terms"!
    If you've got the right answer, you can surely write out the calculation you did as one formula and you'll have the right formula too. Don't think this is one to worry about, as I expect your method is more standard than theirs.
     
  3. Julie Lewis

    Julie Lewis Member

    The point that the solution is trying to make is that you don't take into account the positive cashflows at times 4 and 5. The losses are 30 at time 1, 12 at time 2 and 6 at time 3. These are all per policy in force at the beginning of the respective policy year. We want to get the value at time 1 of these losses, so we discount the time 2 and time 3 figures taking into account the probabilities that the policy is still in force at the start of years 2 and 3.

    The zeroisation method is equivalent to this and is therefore also fine. If you write out the formulae for the reserve at time 2, the new cashflow at time 2, the reserve at time 1, and the new cashflow at time 1, and tidy up, you should be able to come to the same algebraic answer as the solution.
     

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