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  • Question(s) for Professor Broverman

    Professor Broverman, I had another question from your 2005 Exam M study manual. For question 2 of your Section 19 Exercises on LC-226, you find that A50 equals .24467 and the second moment is .1201. For Var(L), you obtain 104,400. However, using the same numbers, I obtain 105,581 and I believe this difference is large enough to question. Am I doing something wrong, or have I made a mistake? Thank you for any clarification.
    act justly. walk humbly. .

  • #2
    Godspeed,

    You have it right. That was my mistake.
    This one is in the errata list, which can be
    found at http://www.sambroverman.com
    (you said your are using the 2005 edition,
    so take that link).

    Sam Broverman
    Sam Broverman

    [email protected]
    www.sambroverman.com

    Comment


    • #3
      Professor Broverman, I have another question that I didn't see addressed in your manual's errata. On LC-251, you calculate the benefit premium as 166.56. Yet, in your L formulations, this number fluctuates from 166.53 to 166.60. I'm thinking this is just a typographical error, but I just wanted confirmation. Thanks.
      act justly. walk humbly. .

      Comment


      • #4
        Godspeed,

        The premium should be 166.56 throughout the example.
        Thanks for pointing that out.
        Sam Broverman

        [email protected]
        www.sambroverman.com

        Comment


        • #5
          Three more questions, Professor Broverman: 1) In question 14 on LC-286, you list a(doubledot)40:10 as 7.6, but in the solution you list it as 7.7. Is the 7.7 figure the correct one? Also, a question on semantics: this same question asks about a 20-payment, 35-year endowment insurance. Does this mean that premiums are paid for 20 years and the insurance lasts for 35, thereby being unaffected by the 20 payments? It just seems like the test designers try to word as trickily as possible, but maybe it's just me.

          2) For part b) in question 23 on LC-288, I attempted to use the k=20 line to find Px from the relationship kVx=Ax+k-Pxa(doubledot)x+k

          Solving for Px, and using the Ax=1-d*a(doubledot)x relationship and the values on the k=20 line, I obtain:
          Px=(Ax+k-kVx)/a(doubledot)x+k
          =(1-.06*1.06*11.145-.24782)/11.145
          =.01089
          , which is exactly a tenth of the size of the answer you give as .1089. I believe the correct premium should be .01089.

          3) In the solution of question 31 on LC-298, you calculate abar50:10 as 6.5787. Yet, using the same numbers, I get 6.594, changing the final answer to 95.96.

          Your continued help is greatly appreciated. Thank you.
          Last edited by .Godspeed.; February 9 2006, 09:23 PM.
          act justly. walk humbly. .

          Comment


          • #6
            Godspeed,

            In LC-286 #14, the annuity value of 7.60 should
            have been 7.70, and in LC-288 #23 you are also
            right that the premium should be .01089 .
            Thanks for poitning those errors out. I have
            addedn them to the online errata list at my
            website.

            REgarding the use pf the phrase "n-payment insurance
            policy". This means that the policy has up to n scheduled
            premiums (if the policyholde lives that long), but the premiums
            would stop if death occurred before time n.
            For instance, a 30-payment whole life insurance refers
            to a policy with premiums for up to 30 years, but if the
            policyholder lives longer than 30 years, the policy continues
            but no more premiums are paid. Of course, this means
            that the premiums would be larger than if they were payable
            for whole life on the policy, so the policy can be paid
            off earlier. There is standard actuarial notation to
            represent an "n-payment policy". If the premium symbol for
            a policy with premiums for the full lifetime of the policy is P,
            then the n-payment premium policy with the same death
            benefits is denoted nP (where the n one the left is a
            subscript). So for the 20-payment, 35-year endowment
            policy you mentioned, the premium could be denoted
            20 P x:35 . This can be confusing when there
            is the n-year survival probability n p x, but the p is lower
            case for the survival probability and capital P for
            the premium notation.

            Sincerely,
            Sam Broverman

            [email protected]
            www.sambroverman.com

            Comment


            • #7
              Thank you very much; that cleared it up.

              It appears that you replied as I was editing my post, so here is the last question I had that your probably didn't see:

              3) In the solution of question 31 on LC-298, you calculate abar50:10 as 6.5787. Yet, using the same numbers, I get 6.594, changing the final answer to 95.96.

              Also, on LC-299, on the line reading "but it is not always true that", I believe the "-" should be a "=".

              Thanks again.
              Last edited by .Godspeed.; February 11 2006, 05:48 PM.
              act justly. walk humbly. .

              Comment


              • #8
                Thanks for the updated errata.
                Last edited by .Godspeed.; February 15 2006, 12:30 AM.
                act justly. walk humbly. .

                Comment


                • #9
                  Professor Broverman, I had a question regarding your August 1, 2005 Exam M Question of the Week. I spotted what look to be two transcription errors: 20E35 should be .286 and not .266; and 2^A35 should be .03488 and not .3488.

                  I now get E(S)=27420*1000 and Var(S)=656,890,715*1000.
                  Using these updated values, I now get a desired premium of 28,459 as opposed to 24,740. Could you please confirm this? Thank you very much.
                  act justly. walk humbly. .

                  Comment


                  • #10
                    Godspeed,

                    Thanks for pointing that out. I tried working the
                    arithmetic with the corrected values, but I didn't get
                    the same numbers as you. I would like to confirm
                    your calculations before I post a correction. Can you
                    point out how you got 27,420 a nd 656,890,715 ?
                    Thanks.

                    SB
                    Sam Broverman

                    [email protected]
                    www.sambroverman.com

                    Comment


                    • #11
                      .Godspeed,

                      I think your formulation for E[W^2] is incorrect.
                      W pays a death benefit of 200,000 for death prior to 65
                      and 100,00 for death after 60. Then W^2 wouldpay
                      200,000^2 prior to 65 and 100,000^2 after 65 (with a
                      doubled force of interest for present value calculation).

                      You have written
                      E(W^2)=200,000^2*2^A35-100,000^2*30p35*v^60*2^A65 .
                      This would pay 200,000^2 for whole life, but then 100,000^2
                      would be cancelled off for death after age 65. This results in
                      200,000^2 - 100,000^2 paid for death after age 65,
                      but the payment should be 100,000^2 (which is not equal
                      to 200,000^2 - 100,000^2). Thiis formulation worked for E(W)
                      since 200,000 - 100,000 = 100,000 .

                      Sincerely,

                      Sam Broverman
                      Sam Broverman

                      [email protected]
                      www.sambroverman.com

                      Comment


                      • #12
                        Ok, I deleted the previous post so others might not be confused. I think I have it now. I think I was trying to take a shortcut without fully checking my logic.

                        We agree that E(S)=1000*E(W)=19,620,000.

                        E(W^2)=200,000^2*(2^A35-v^60*30p35*2^A65)+100,000^2*v^60*30p35*2^A65
                        =200,000^2*(.03488-*1.06^-60*7533964/9420657*.23603)+100,000^2*1.06^-60*7533964/9420657*.23603
                        =1,223,536,132

                        Var(W)=E(W^2)-E(W)^2=1,223,536,132-19,620^2=838,591,732
                        Then, Var(S)=1000*Var(W)=838,591,732,000

                        Thus, (1000C-19,620,000)/sqrt(838,591,732,000)=1.282-->C=20794

                        Is this better? Thank you for the continued help.
                        Last edited by .Godspeed.; March 11 2006, 01:31 PM.
                        act justly. walk humbly. .

                        Comment


                        • #13
                          .,Godspeed,

                          I finally had a chance to recheck the arithmetic.

                          I agree with your answer of 20,794 .

                          Thanks for finding that. I'll post a correction.
                          Sam Broverman

                          [email protected]
                          www.sambroverman.com

                          Comment

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