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					                             MATH20962 Tutorial 1



Life expectancy

1. Find values for the following using your table books (and write down the correct
actuarial notation):

(a) The expected future lifetime of a newborn (also known as the complete future
lifetime) assuming ELT15(Female) mortality.

(b) The expected complete future lifetime of a person aged exactly 55 assuming
ELT15(Female) mortality.

(c) The expected age at death (including fractional years) of a person aged exactly
55 assuming ELT15(Female) mortality.

Bonus question: explain why this is higher than the expected age at death of a
newborn.

(d) The expected curtate future lifetime of a person aged exactly 55 assuming AM92
mortality.


2. Is a person aged 60 assumed to live longer if he follows PMA92C20 or she follows
                      0
PFA92C20? Lookup e x for both tables and state a common conclusion for male
versus female pensioner life expectancy.


Looking up

3. Lookup:

(a) The expected present value of a 25 year endowment assurance of a life aged
exactly 40, using an interest rate of 6% and AM92 ultimate mortality. Write down the
correct actuarial notation.

(b) The expected present value of a whole life assurance payable at the end of year
of death of a life aged exactly 40, using an interest rate of 6% and AM92 mortality.

(c) The expected present value of a whole life assurance payable at the end of year
of death of a life aged exactly 40, using an interest rate of 12.36% and AM92 select
mortality. [Hint: 1:06^2 = 1:1236]


Tx, Kx

4. A person dies aged 82 yrs and 142 days. What are the values of T25 and K25?
Expected value and variance

5. April 2011 (subject CT5) Institute exam Q10:

Calculate the expected present value and variance of the present value of an
endowment assurance of 1 payable at the end of the year of death for a life aged 40
exact, with a term of 15 years.

Basis:
Mortality       AM92 Select
Rate of interest 4% per annum

                                      l55    9557.8179
[You may use:        15   p[ 40]                      0.969913 ]
                                     l[ 40] 9854.3036


General relationships between expected present value

6. State with reason whether the following statements are true or false. Where
necessary assume 5%pa interest.
(i)     A25:3|  A35:3|
(ii)     A35:10|  0.6
(iii)    A1         0.6
          35:10|




7. Evaluate A70:2| using AM92 ultimate mortality and 4%pa interest.
(You may assume             2   p 70 = l 72 / l 70  7637 .6208 / 8054 .0544  0.94830 )


8. Correct the following formulas
        n| Ax  v Ax  n
                 n
(i)
(ii)     Ax:n|  Ax:t|  v t t p x . Axt:nt|


9. State which has the highest value:
                                                     2
A[55 ]             A55           A25:5|                  A55
              2
The suffix indicates a modified interest rate of i’ where 1+i’ = (1+i)2
 Derivation of key formula

 10. April 2005 (subject CT5) Institute exam Q12:

 (i) By considering a term assurance policy as a series of one year deferred term
 assurance policies, show that:

              i
  A1             A1
   x:n|           x:n|


 (ii) Calculate the expected present value and variance of the present value of a term
 assurance of 1 payable immediately on death for a life aged 40 exact, if death occurs
 within 30 years.

 Basis:
 Interest 4% per annum
 Mortality AM92 Select
                                            l70    8054.0544
 You may assume            30   p[ 40]                      0.8173134
                                           l[ 40] 9854.3036


 Derivation of formula for the variance of the present value of a deferred
 assurance

 11. April 2007 (subject CT5) Institute exam Q10:

 Let X be a random variable representing the present value of the benefits of a whole
 of life assurance, and Y be a random variable representing the present value of the
 benefits of a temporary assurance with a term of n years. Both assurances have a
 sum assured of 1 payable at the end of the year of death and were issued to the
 same life aged x.

 (i) Describe the benefits provided by the contract which has a present value
 represented by the random variable X - Y.                                  [1 mark]

 (ii) Show that

              cov( X , Y ) 2 A1  Ax . A1
                                 x:n|         x:n|

 and hence or otherwise that

              var (X – Y) = 2 Ax ( n| Ax ) 2  2 A1
                                                     x:n|



 where the functions A are determined using an interest rate of i, and functions 2 A
 are determined using an interest rate of i 2  2i .




January 2012

				
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