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Microeconomics questions answer them well.the questions below are complete (ii) the profit signature per policy sold if Sterling reserves are zeroised. (c) The office now

Microeconomics questions answer them well.the questions below are complete

(ii) the profit signature per policy sold if Sterling reserves are zeroised.

(c) The office now wishes to make an allowance for surrenders. It assumes that, at the end

of the first and the second policy years, 3% of the surviving policyholders will surrender (just

before payment of the second and third annual premiums respectively.) Surrender values are

equal to the value of the policyholder's units (after deduction of fund management charges),

with a surrender penalty of 10. Calculate

(i) the revised profit signature per policy sold, ignoring any need to maintain Sterling Reserves

at the end of each year,

(ii) the revised profit signature per policy sold if the Sterling Reserves are zeroised, and

(iii) the net present value, at a risk discount rate of 15% per annum, of the revised profit

signature per policy sold, assuming that the Sterling Reserves are zeroised.

(a) If a profit test for a unit-linked policy reveals negative cash flows in the second or any

subsequent policy year, it is customary to eliminate these negative values by setting up sterling

reserves at the end of each year.

Describe briefly the technique ("zeroisation") by which these reserves are calculated.

(b) An office issues a 3-year unit-linked policy with a yearly premium of 500. The death

benefit, payable at the end of the year of death, is 1,000 or the bid value of units if greater.

The maturity value is the bid value of the units at maturity.

95% of each premium is invested in units at the offer price. The bid price of units is 95% of

the offer price. Management charges of 1

4% of the bid value of the units are deducted at the

end of each year (before payment of death and maturity claims).

The office expects to incur expenses of 75 at the start of the first year and 25 at the start

of each subsequent year.

Using a profit testing analysis, calculate for a life aged 60 at entry

(i) the expected profit in each of the 3 years per policy in force at the beginning of the year,

(ii) the net present value at the issue date of the expected profit from one policy assuming a

risk discount rate of 10% per annum.

Assume that the unit fund grows at 8% per annum (before deduction of management charges),

that sterling reserves need not be maintained at the end of each year, and that the possibility

of surrender may be ignored. The mortality of policyholders follows A1967-70 ultimate and

sterling reserves earn interest at 6% per annum during each policy year.

If a profit test for a unit-linked policy reveals negative cash flows in the second or any subsequent policy year, it is customary to eliminate these negative values by setting up sterling

reserves at the end of each year.

Calculate the sterling reserves required at the end of each policy year, per policy then in force,

for a 3- year policy for which the profit signature (with no allowance for sterling reserves at

the end of each year) is (250, - 100, - 50), given that the rate of mortality is 0.01 per annum

at each age and sterling reserves earn interest at 8% p.a.

An office issues a unit-linked endowment assurance with annual premium 400 and term five

years to a life aged 60 who is subject to A1967-70 ultimate mortality.

The sum assured, payable at the end of year of death or at the maturity date, is the bid value

of the units held, subject to a guaranteed minimum death benefit of 2,000. The allocation

proportion is 70% for the first annual premium and 98% for all subsequent annual premiums.

For units the bid/offer spread is 5% and the annual rate of management charge is 0.75%.

In determining the sterling reserves necessary for the policy the office makes the following

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and display the values of p, and q, for ages 0 to 140. As a check on your output, input the parameter values given in {3.7.1} and compare your a: values with those for :r = 13, 14. . . . in the Illustrative Life Table in Ap- pendix 2A. We will refer to this computing object as your Illustrative Life Table. When the Makeham parameter values are not stated, those of (3.7.1) are implied. IRemark: With a Makeham Table, 30:} > 0 for all x :> I}, so a: does not exist as dened in Section 3.3.1. For the parameter values of the Illustrative Life Table, qua is zero to eight decimal places; thus we choose [a = 140 for our Illustrative Life Table, that is, Table 2A.] 3. In your Illustrative Life Table use the forward recursion formula l\" 1 = (pxlxj and initial value 13 = 96,80?.83 to calculate the ix values of Table 2A. [Remark The Makeham law was not realistic for ages less than 13, so the Illustrative Life Table is a blend of some ad hoc values from 0 through 12 and the Makeham law table from age 13 up.] . Illustrate the result of Exercise 3.41 by doubling the A and B parameter values in your Illustrative Life Table. . Use the backward recursion formula of Table 3.5.1 to calculate values of eI in your Illustrative Life Table for ages 13 to 140. . Compare the values of ex at x = 20, 40, 6t}, 80, and 100 in your Illustrative Life Table with those found when the force of mortality is doubled. :. Use the backward recursion formula of Table 3.5.1 and the trapezoidal rule to calculate values of at in your Illustrative Life Table for ages 13 to 110. '. Verify the following backward recursion formula for the temporary curtate life expectancy to age 3;: Em=Px+PxEx+rm forx=0,1,...,y-1. Determine an appropriate starting value for use with this formula. For your Illustrative Life Table calculate the curtate temporary life expectancy up to age 45 for ages 13 to 44. .-. Verify the following backward recursion formula for the 11-year temporary eurtate life expectanCy: era 2 p1\" npr) + p1: ex+1=| forx = 0' 1r ' ' ' I m _ 1' Determine an appropriate starting value for use with this formula. For your Illustrative Life Table calculate the 10-year temporary curtate life expectancy for ages 13 to 139. '. \"Look up\" Email in your Illustrative Life Table. [Hint Since the C(x) term in the relation in Exercise 3.53 does not depend on n, it may be more efcient to view 815.3 as a curtate temporary life expectancy to age 40 for (15).]

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