Actuarial Science exam

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AS3429/9429a Ch.7 Lecture notes (W2018)

(iii) Recursive Formulas- 1st revisit previous examples-working page

(a) Example 1 revisited:

• 20 year endowment policy, no expenses • premium basis = policy basis • P=15,114.33 • showed 10V=190,339 and 11V=214,757

• work with tV to generate a recursive expression

−=

−+ |:[50] 500000A

t20tt V

|:[50] aP

t20t −+ &&

AS3429/9429a Ch.7 Lecture notes (W2018)

Recursive Formulas- 1st revisit previous examples-working page

years, and P was given (P=5200), and there were expenses

• showed 5V=29,067.11 and 6V=35,324.45

AS3429/9429a Ch.7 Lecture notes (W2018)

(c ) Example 4 (Text 7.4) REVISITED

A man aged 50 purchases a deferred annuity policy. Annual payments will be payable for life, with the first payment on his 60th birthday. Each annuity payment will be $10,000. Level premiums of $11,900 are payable annually for at most 10 years. On death before age 60, all premiums paid will be returned, without interest, at the end of the year of death.

The following basis is used for policy value calculations:

Survival model: Standard Select Survival Model Interest : 5% per year Expenses : 10% of the first premium, 5% of subsequent premiums, $25 each time an annuity payment is paid, and $100 when a death claim is paid

Expression for 5V in terms of 6V? 5V expression(Ex.4) is below, use this to get expression for 6V

|:55|:5560|:55|:55 AAA

5555 1

55 aP95.100aE025,10)I(PP5V &&&& −−−−++++++++++++====

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 4 REVISITED (working page)

from example 4:

|:55|:5560|:55|:55 AAA

5555 1

55 aP95.100aE025,10)I(PP5V &&&& −−−−++++++++++++====

Similarly

|:56|:5660|:56|:56 AAA

4564 1

46 aP95.100aE025,10)I(PP6V &&&& −−−−++++++++++++====

AS3429/9429a Ch.7 Lecture notes (W2018)

(iii) Policy Value Recursive Formulas

•••• look at formulas that express tV in terms of t+1V

•••• revisited some previous examples to illustrate idea and work towards text general formula

o Ex.1 for recursive formula when no expenses

o Ex.3 for recursive formula (expenses)

o Ex.4 for recursive formula when Death Benefit varies(plus there are expenses)

•••• there are some cases where policy values are easier to calculate using a recursive formula(Example 7.7 text)

AS3429/9429a Ch.7 Lecture notes (W2018)

Recursive Formula-General Case

•••• Consider a policy issued to (x) where all cash flows can occur only at the start or end of a year. Suppose this policy has been in force for t years, where t is a non-negative integer. Consider (t+1)st year, and let

P t ≡ the premium payable at time t

et ≡ the premium-related expense payable at time t

St+1 ≡ sum insured payable at t+1 if insured dies in the year

E t+1 ≡ the expense of paying the sum insured at time t+1

tV ≡ gross premium policy value for policy in force at time t,

t+1V ≡ gross premium policy value for policy “ “ “ time t+1.

it ≡ the rate of interest assumed earned in the year

Note: et , Et+1 q [x]+t and it all as assumed in the policy value basis.

( tV ++++Pt −−−− et)(1 ++++ it) = = = = q[[[[x]+]+]+]+t (St++++1 ++++ Et++++1) + + + + p[[[[x]+]+]+]+t (t++++1V) (1) or ( tV ++++Pt −−−− et)(1 ++++ it) = = = = t++++1V + q[[[[x]+]+]+]+t (St++++1 ++++ Et++++1 ---- t++++1V) (2)

•••• shaded term called Net Amount at Risk(NAR), Death Strain at Risk(DSAR) or Sum at risk

AS3429/9429a Ch.7 Lecture notes (W2018)

Formal derivation of general case recursive formula;

Lt= (1+it) -1 (St+1+ Et+1) – Pt +et if K[x]+t=0 with probability q[x]+t

Lt= (1+it) -1 Lt+1 – Pt +et if K[x]+t ≥1 with probability p[x]+t

tV = E(Lt)

=q[[[[x]+]+]+]+t(1+it) -1(St+1+Et+1)–(q[[[[x]+]+]+]+t+p[[[[x]+]+]+]+t )(Pt –et )+p[[[[x]+]+]+]+t (1+it)

-1E[Lt+1]

(tV+ Pt –et ) = q[[[[x]+]+]+]+t(1+it) -1(St+1+Et+1)+ p[[[[x]+]+]+]+t (1+it)

-1E[Lt+1]

(tV+ Pt –et )(1+ it) = q[[[[x]+]+]+]+t(St+1+Et+1)+ p[[[[x]+]+]+]+t (t+1 V) t+1V = E[Lt+1]

( tV ++++Pt −−−−et )(1 ++++ it ) ==== q[[[[x]+]+]+]+t (St++++1 ++++ Et++++1 ) ++++ p[[[[x]+]+]+]+t (t++++1V) (1)

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 5 (text 7.7) Consider a 20-year endowment policy purchased by a life aged 50. Level premiums of $23,500 per year are payable annually throughout the term of the policy. A sum insured of $700,000 is payable at the end of the term if the life survives to age 70. On death before age 70 a sum insured is payable at the end of the year of death equal to the policy value at the start of the year in which the policyholder dies. The policy value basis used by the insurance company is as follows:

Survival model: Standard Select Survival Model(SSSM) Interest : 3.5% per year Expenses : nil

Calculate 15V the policy value for a policy in force at the start of the 16th year.

Why use recursive approach?

SSSM values used q69 = .009294 q68 = .008297 q67 = .007409 q66 = .006619 q65 = .005915

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 5 working page

AS3429/9429a Ch.7 Lecture notes (W2018)

(iv) Annual Profit

• recursive formulas for policy values show that if all cashflows between t and t+1 are as assumed (i.e. policy value basis assumptions are met), then insurer will be in a break-even position at time t+1, given they were in a breakeven position at time t o unlikely that all assumptions(interest, mortality, expenses and other

assumptions) met in any one year o insurers assets could be more or less than what is needed to match

actual t+1V values

• Analysis of surplus-break down profit/loss into its component parts o this analysis is done as a part of any valuation exercise o analysis helps to determine if assumptions need to be changed

amongst other things

• Sources of profit depend on assumption category and product o actual expenses < assumed expenses is a source of profit o actual interest < assumed interest is a source of loss o Actual Mortality < assumed mortality is a source of profit for insurance,

but a source of loss for annuities

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 6 (text 7.8) An insurer issued a large number of policies identical to the policy in Example 3 to women aged 60. Five years after they were issued, a total of 100 of these policies were still in force. In the following year, • expenses of 6% of each premium paid were incurred, • interest was earned at 6.5% on all assets, • one policyholder(p/h) died, and • a $250 expenses was incurred on payment of sum insured for the 1 p/h who died.

(a) Calculate profit or loss on this group of policies for this year.

(b) Determine how much of this profit/loss is attributable to profit/loss from interest, mortality, and from expenses.

( tV+Pt −−−− et)(1 + it )=q[x]+t (St+1 + Et+1)+p[x]+t (t+1V) (breakeven)

Total Gain/Loss=(tV +Pt ― e’t)(1 +i’t)―[q’[x]+t (St+1 + E’t+1) +p’[x]+t (t+1V)]

where ’ denotes actual experience

Note: Use 5V= 29,067.51, 6V= 35,324.17 (vs. Ex. 3 value caln’s)

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 6 (working pages)

Total “Profit (loss)” ≡ total Gain/(Loss)

Total “Gain (loss)” = 100(5V+.94P)(1.065) – (1)100,250 – (99)(6V)

Total “Gain (loss)” = $18,918.99

Sources of Gain/Loss:

1. Interest: gain

2. Expenses: loss 3. Mortality: loss

How much gain/loss is attributable to each assumption?

• Look at each factor. At each step, assume factors not yet considered are as specified in policy value basis, whereas factors already considered are as actually occurred

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 6 (working pages) (b) sources of gain(loss) summary: (i) Interest : 51,011 (ii) Expenses : (5,568) (iii) Mortality : (26,524)

Total Gain(Profit) : 18,919 Note: if assumption order is different, #s change a bit by source. For example looking first at expenses, interest, then mortality Sources of gain(loss) summary: (i) expenses : (5,490) (ii) Interest : 50,933 (iii) Mortality : (26,524) Total Gain(Profit) : 18,919

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 6 (working pages) _-another approach

100(tV+P-e)(1+i)-100(q[x]+t(St+Et)+p[x]+tt+1V) (assumption notes)

Interest actual 51,011 (actual i, tV basis for E&q)

expected 0 (all policy basis(tV) assms)

Gain(A-E): 51,011

Expenses: Actual 45,443 (actual i,E, tV basis for q

Expected

51,011 (actual i, tV basis for E&q)

Gain(A-E): -5,567.57

Mortality Actual 18,919 (actual i, E, and q

Expected 45,443 (actual i,E, tV basis for q

Gain(A-E) -26,524.47

Net G/L 18,919.22

AS3429/9429a Ch.7 Lecture notes (W2018)

(v) Asset Shares

• Asset share is the share of the insurer’s assets attributable to each policy in force at any given time

ASt ≡ Asset share per policy surviving at time t (does NOT include/reflect Ps and related expenses due at time t so AS0=0))

• Asset Share(ASt) versus Policy value(tV)

o ASt is the amount the insurer actually has with respect to each surviving insured

o tV is the amount the insurer needs to have with respect to each surviving insured

• calculate ASt using actual experience and by assuming the given

policy is one of a large number of N identical policies o accumulate Premiums received less claims and expenses paid

to time t and divide by number of surviving policies to get ASt

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 7(text 7.9) Consider a policy identical to the policy studied in Example 7.4 and suppose that this policy has now been in force for five years. Suppose that over the past five years the insurer’s experience in respect of similar policies has been as follows.

• Annual interest earned on investments has been as follows;

Year 1 2 3 4 5 Interest % 4.8 5.6 5.2 4.9 4.7

• Expenses at the start of the year in which a policy was issued were 15% of the premium

• Expenses at the start of each year after the year in which a policy was issued were 6% of the premium.

• The expense of paying a death claim was, on average, $120. • The mortality rate, q[50]+t, for t=0,1,…,4 has been 0.0015.

Calculate the asset share for the policy at the start of each of the first six

years.

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 7(working page)

AS3429/9429a Ch.7 Lecture notes (W2018)

Table 7.1. Asset share calculation for Example 7.9

Fund Death Fund Cashflow at eoy Claims Fund at

Yr at b.oy. at b.oy. before claims & expenses at eoy Survivors ASt

1 0 10115N 10601N 18N 10582N 0.9985 N 10,598

2 10582N 11169N 22970N 36N 22934N 0.99852N 23,003

3 22934N 11152N 35859N 54N 35805N 0.99853N 35,967

4 35805N 11136N 49241N 71N 49170N 0.99854N 49,466

5 49170N 11119N 63123N 89N 63034N 0.99855N 63,509

can extend class approach used for AS1 and AS2 to generate rest of AS values

Note: AS5=63,509 versus 5V= 65,470

AS3429/9429a Ch.7 Lecture notes (W2018)

Recall Policy Value Recursive formula

( tV ++++Pt −−−− et)(1 + + + + it) = = = = q[[[[x]+]+]+]+t (St++++1 ++++ Et++++1) + + + + p[[[[x]+]+]+]+t (t++++1V) (1)

Analagous recursive asset share formula

(ASt ++++Pt −−−− e”t)(1 ++++ i”t) ====q”[x]+]+]+]+t (St++++1 ++++ E”t++++1) ++++p” [[[[x]+]+]+]+t(ASt+1) (i)

where ” indicates actual experience

Rearranging (i) gives

ASt+1 ====(1/(1/(1/(1/p” [[[[x]+]+]+]+t))))[[[[((((ASt ++++Pt −−−−e”t)(1 ++++ i”t) − q”x]+]+]+]+t (St++++1 ++++ E”t++++1)]

Text used modification of (i)

l”[x]+]+]+]+t (ASt ++++Pt −−−− e”t)(1 ++++ i”t) ==== d”[x]++++t (St++++1 ++++ E”t++++1) ++++llll” [[[[x]+]+]+]+t+1(ASt+1) where l”[x]+]+]+]+t ==== N N N N((((ttttpppp”xxxx) e) e) e) etctctctc

AS3429/9429a Ch.7 Lecture notes (W2018)

7.4 Policy Values for other discrete cashflows

• Section looks at determining tV when discrete cashflows are other than annual. General definitions below still hold, calculations can be more complex

� Gross premium policy value for a policy in-force at duration t(≥0)years after it was purchased is the expected value at that time of the gross future loss random variable on specified basis. The premiums used in the calculation are actual premiums payable under the contract.

� Net premium policy value for a policy in force at duration t years after it was purchased is the expected value at that time of the net future loss random variable on a specified basis (which makes no allowance for expenses). Premiums used in the calculation are the net premiums calculated on the policy value basis using the equivalence principle, not the actual premiums payable

• by example will show tV in theory can be calculated for any t (and recursive formulas can be used), and linear interpolation is useful approximation when t between premium payment dates (7.4.2)

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 8(text 7.10++) A life aged 50 purchases a 10-year term insurance with sum insured $500,000 payable at the end of the month of death. Level quarterly premiums, each of amount P=$460 are payable for at most five years.

(a) Calculate the (gross premium) policy values at durations 2.75, 3 and 6.5 years using the following basis.

Mortality : Standard Select Survival Model Interest : 5% per year Expenses : 10% of each gross premium

will show 2.75V = 3,090.30 and 3V=3,358.15

(b) How would you use recursive formulas to get 2.75V from 3V?

(c) Determine the policy value at each of (i) the end of 2 years 10 months and (ii) 2 years and 9.5 months, assuming the policy is still in-force at each of these times(will look at this exactly and by approximation(linear interpolation)

any required annuity/Insurance factors will be provided

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 8(working page)

AS3429/9429a Ch.7 Lecture notes (W2018)

Policy values between premium date approximation(working page attached)

• interpolate between policy value just after previous premium and policy value just before the next premium

o text formula: t+k+sV ≈ ( t+kV + Pt+k− Et+k )(1-s/k)+( t+2kV)(s/k)

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 8(working page)

AS3429/9429a Ch.7 Lecture notes (W2018)

• interpolate between policy value just after previous premium and policy value just before the next premium

Using t+k+sV ≈ (t+kV +Pt+k− Et+k )(1-s/k) + (t+2kV)(s/k)

and, from (a): 2.75V=3,090.30, 3V=3,358.15, (P2.75 - E2.75) =(.9)(460)

(ii)policy value at 2 years, 9.5 months (or t = 2.79167) k=3/12, s=0.5/12 so s/k=.5/3=.1667

2.79167V ≈ (2.75V+P2.75−E2.75 )(1-.1667)+(3V)(.1667) 2.79167V ≈3,480.55 versus Exact: $3,481

(i)policy value at 2 years, 10 months (or t = 2.8333) k=3/12, s=1/12 so (s/k)=1/3

2.833V ≈(2.75V+P2.75−E2.75 )(1-1/3) +(3V)(1/3)

2.833V ≈3,456.06 versus Exact: $3,456.73

AS3429/9429a Ch.7 Lecture notes (W2018)

7.5 Policies with continuos cashflows

• Recall, for policies with discrete cashflows:

( tV +Pt − et)(1 + it ) = q[x]+t (St+1 + Et+1) + p[x]+t (t+1V) (1) or ( tV ++++Pt −−−− et)(1 ++++ it) = = = = t++++1V + q[[[[x]+]+]+]+t (St++++1 ++++ Et++++1 ---- t++++1V) (2)

• Can rewrite (2) as

(t+1V −−−− tV) ==== tV it + (Pt −−−− et)(1 ++++ it ) −−−− q[[[[x]+]+]+]+t (St++++1 ++++ Et++++1 −−−− t++++1V) (1) change in Increase due increase due to decrease due to policy value to interest (Prems-expenses) mortality

• there’s a continuos version of the above formula called Thiele’s differential equation formula

AS3429/9429a Ch.7 Lecture notes (W2018)

• discrete version

(t+1V − tV) =tV it + (Pt − et)(1 + it ) − q[x]+t (St+1 +Et+1 − t+1V) (1)

• continuous version or Thiele’s differential equation

(d/dt)(tV) ==== tV δt + (Pt −−−− et) −−−− µ[[[[x]+]+]+]+ t (St ++++ Et −−−− tV), (2) where

Pt ≡ annual rate of premium payable at time t

et ≡ annual rate premium-related expense payable at time t

St ≡ sum insured payable at exact time t if insured dies at exact time t

Et ≡ expenses of paying the sum insured at exact time t

δt ≡ force of interest per year assumed earned a time t

tV ≡ policy value a policy in force at time t

all are continuous functions of t and as assumed in policy value basis

AS3429/9429a Ch.7 Lecture notes (W2018)

(t+1V −−−− tV) ====tV it + (Pt −−−− et)(1 + + + + it ) −−−− q[[[[x]+]+]+]+t (St++++1 ++++Et++++1 −−−− t++++1V) (1)

• Thiele’s differential equation

(d/dt)(tV) = tV δt + (Pt −−−− et) −−−− µ[x]+ t (St + Et −−−− tV) (2)

• to apply (2) write (d/dt)(tV) ≈ ( t+hV −−−− tV)/h and assume small h (t+hV −−−− tV) ≈ tV δt h + ( Pt −−−− et)h −−−− hµ[[[[x]+]+]+]+ t (St ++++ Et −−−− tV) , or (tV)(1+ δt h) + ( Pt − et)(h ) ≈ t+hV+ hµ[x]+ t (St +Et − tV)

• this method for numerical solution of a differential equation is known as Euler’s method (which is a continuous time version of discrete recursive method) o the smaller the value of h, the better the approximation o start with the policy value at end of policy term(or just before it-

endowment insurance and work backwards, will show by example)

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 10(Text Example 7.12) Consider a 20-year endowment insurance issued to a life aged 30.The sum insured, $100,000, is payable immediately on death, or on survival to the end of the term, whichever occurs sooner. Premiums are payable continuously at a constant rate of $2,500 per year throughout the term of the policy. The policy value basis uses a constant force of interest, δ, and makes no allowance for expenses.

(a) Evaluate 10V.

(b) Use Euler’s method with h = 0.05 years to calculate 10V.

Perform the calculations on the following basis: Survival model: Standard Select Survival Model Interest: δ=0.04 per year

Use W3 formula where applicable (annual factors used provided in class)

(a) |:40|:4010 AV

1010 a2500100000 −=

AS3429/9429a Ch.7 Lecture notes (W2018)

(a) 10 V calculations

Assumptions: δ =0.04, and ‘Standard Select survival Model’: µx =A+Bc

x, A=0.00022, B=2.7x10-6, C=1.124

µ [x]+s = 0.9 2-s µx+s for 0 ≤ s ≤ 2 , l20=100,000(radix)

W3: )m(

xa&& ≈ xa&& – (m–1)/2m – (m2–1)/12m2[δ+µx ]

|:40|:4010 AV

1010 a2500100000 −=

21671.8aEaa 401010 =−= 5040|:40

,61285.20a =40 )3W(6358.18a =50

Used SSSM excel WS: i = 0.040810774 = eδ — 1 to calculate:

,11623.21a40 =&& ,13923.19a50 =&& 66518.0E4010 =

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 10 (working page)

AS3429/9429a Ch.7 Lecture notes (W2018)

(b) Use Euler’s method with h ====0.05 years to calculate 10V.

• rearrange t+hV — tV and then can generate 10V recursively

• will do first iteration by hand, rest was done in excel

• result is 10V= 46,635 using Eulers method and h=.05 (versus answer of 46,591 in (a))

• if reduce size of h, answer will be even closer to that in (a))

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 7.12 (h=.05)

x+t t ux+t tV 50 20.00 100,000.00

49.95 19.95 0.001147 99,675.67 49.9 19.90 0.001142 99,352

49.85 19.85 0.001136 99,029 49.8 19.80 0.001131 98,707 . . . 40.1 10.10 0.000513 47,069

40.05 10.05 0.000511 46,852

40.00 10.00 0.000510 46,635.12 (b)

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 11

You are given the following information on a fully continuous term insurance policy that was issued to (25):

The death benefit is $100,000 payable at moment of death

Premiums are payable continously at a rate of $700 per year

The force of interest is δ = 6%

The force of mortality at age 40 is 0.008

There are no expenses

(d/dt)(tV) at t = 15 is equal to -4.80 Calculate the policy reserve at time 15.

AS3429/9429a Ch.7 Lecture notes (W2018)

7.6 Policy Alterations

• In practice, not uncommon, after the policy has been in force for a period of time, for the policyholder to request a change in terms of the policy(e.g. a policy alteration)

• Alteration examples include; o policy cancellation: lapse or surrender o changing benefit amount or reducing premiums o converting from one type of coverage to a different type of coverage

(e.g whole life to term conversion)

• Policy Surrenders or lapse o when a policy is cancelled at request of policy holder before end of

policy term, it is said to lapse or be surrendered o a cash surrender value(CSV or Ct) is what policyholder would be

entitled to if policy lapses(CSV generally 0 for term insurance products) o there is some regulation on minimum surrender values (non-forfeiture

laws), which requires insurers to offer surrender values on certain contract types once they’ve been in-force a minimum number of years

AS3429/9429a Ch.7 Lecture notes (W2018)

• Reasons the CSV(or Ct) is often less than either of tV or ASt ; o anti-selection o associated expenses o liquidity risk

• Ct is often used towards the costs of alterations(other than surrenders), and, when it is, you would have following E.O.V for altered benefits;

Ct+(EPV future Ps, altered contract)t = EPV(future benefits+expenses, altered contact)t

• Paid-up Sum insured (use CSV as single P to purchase this) o any policy where no premiums are payable is said to be paid up o a policyholders that doesn’t want to pay more premiums but doesn’t

want to cancel policy can look at a policy alteration that results in a reduced sum insured

o equation highlighted above would be used to determined reduced paid up amount (but there would be no future P’s in this case)

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 12 Consider the policy in Examples 7.4 and 7.9(Ex. 4& 7 in lecture notes). You’re given that the insurer’s experience in the 5 years following the issue of this policy is as in Example 7.9. At the start of the 6th year, before paying the premium then due, the policyholder requests the policy be altered in one of the following three ways.

(a) The policy is surrendered immediately.

(b) No more premiums are paid and a reduced annuity is payable from age 60.In this case, all premiums paid are refunded at the end of the year of death if the policyholder dies before age 60.

(c) Premiums continue to be paid, but the benefit is altered from an annuity to a lump sum (pure endowment) payable on reaching age 60. Expenses and benefits on death before age 60 follow the original policy terms. There is an expense of $100 associated with paying the sum insured at the new maturity date. Calculate the surrender value(a above), the reduced annuity (b above) and the sum insured (c above) using the assumptions below(see next page)

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 12(continued)

Calculate the surrender value(a), the reduced annuity (b) and the sum

insured (c) using the Ct assumptions below and life insurance and annuity factors as provided(from Example 4) (i) 90% of the asset share less a charge of $200, or (ii) 90% of the policy value less a charge of $200

together with the assumptions in the policy value basis when calculating revised benefits and premiums Notes:

– required life insurance/annuity factors provided(used previously) – previously calculated values: 5V=65,470 (notes Ex. 4), AS5=63,509 (Ex.7)

(a) Surrender Value (policy is surrendered -calculate CSV)

(i) C5 = (.9)(63,509) – 200 = $56,958

(ii) C5= (.9)(65,470) – 200 = $58,723

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 12(working page

Ex4(for reference): A man aged 50 purchases a deferred annuity policy. Annual payments will be payable for life, with the first payment on his 60th birthday. Each annuity payment will be $10,000. Level premiums of $11,900 are payable annually for at most 10 years. On death before age 60, all premiums paid will be returned, without interest, at the end of the year of death.

The following basis is used for policy value calculations:

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 13(text exercise 7.11) A10-year endowment insurance is issued to a life aged 40. The sum insured is payable at the end of the year of death or on survival to the maturity date. The sum insured is $20,000 on death, $10,000 on survival to age 50. Premiums are paid annually in advance.

(a) The premium basis is: Expenses : 5% of each gross premium including the first Interest : 5% Survival model: Standard Select Survival Model

Show that the gross premium is $807.71.

(b) Show that 4V=3,429.68 (policy value just before 5 th P is paid)

(c)Just before the fifth premium is due the policyholder requests that all future premiums, including the fifth, be reduced to one half their original amount. The insurer calculates the revised sum insured – the maturity benefit still being half of the death benefit – using the policy value in part (b) with no extra charge for making the change. Calculate the revised death benefit

AS3429/9429a Ch.7 Lecture notes (W2018)

(b) Example 13 (working page)

10E[40] = 0.60929 6E44 = 0.7422401

a&& [40]:10 = 8.087046 a&& 44:6 = 5.319477

A[40]:10 = 1― d a&& [40]:10 A44:6 = 1― d a&& 44:6

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 14 –Ch.7 Par Insurance example(text exercise 7.10)

A life aged 50 buys a 10,000 participating whole life insurance policy.The sum insured is payable at the end of the year of death. The premium is payable annually in advance. Profits are distributed through cash dividends paid at each year end to policies in force at that time. The premium basis is:

Initial expenses: 22% of the annual gross premium plus $100 Renewal expenses: 5% of the gross premium plus $10 Interest: 4.5% Survival model: Standard Select Survival Model

a) Show that the annual premium, calculated with no allowance for future bonuses, is $144.63 per year. b) Calculate the tV at each year end using the premium basis. c) Assume earns insurer earns i= 5.5% each year. Calculate DIV payable each year assuming (i) the policy is still in force at the end of the year, (ii) experience other than interest exactly follows the premium basis, and

(iii) that 90% of the profit is distributed as dividends to policyholders.

d) Calculate the expected present value of the profit to the insurer per policy issued, using the same assumptions as in (c). (e) What would be a reasonable surrender benefit for lives surrendering their contracts at the end of the first year

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 14(working page)

t p[50]+t tV Profitt DIVt 0 0.9989670 0 1 0.9987360 3.061 0.128 $ 0.1154 2 0.9985310 123.847 1.305 $ 1.1756 3 0.9983770 248.226 2.512 $ 2.2645 4 0.9982030 376.909 3.756 $ 3.3861 5 0.9980072 509.948 5.043 $ 4.5469 6 0.9977876 647.389 6.373 $ 5.7476 7 0.9975408 789.276 7.748 $ 6.9886 8 0.9972635 935.634 9.167 $ 8.2704 9 0.9969519 1086.477 10.630 $ 9.5935

10 0.9966018 1241.804 12.139 $10.9583

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 14(working page)

AS3429/9429a Ch.7 Lecture notes (W2018)

7.7 Retrospective Policy Values

• tV = E[Lt] is considered a prospective policy value

• Retrospective policy value (looks at past) equals: accumulated Premiums less accumulated benefits for a large group of identical policies and dividing this amount equally amongst survivors o it is a “theoretical asset shared based on a different set of assumptions

(i.e. is based on policy value assumptions)”

• retrospective policy value is not used much anymore o seldom equals prospective value, due mostly to fact that experience

often differs from assumptions and policy basis assumptions often differ from premium basis assumptions – tV

Retro = tV Prospective only if (i) contract P determined using equivalence

principle and (ii) same basis is used for P and tV Retro , tV

Prospective

o retrospective policy values not used by countries that use gross premium policy values reserves,

AS3429/9429a Ch.7 Lecture notes (W2018)

When are retrospective Policy values useful?

• key retrospective measure is Asset share which measures the accumulated contributions of each surviving policy to the insurer’s funds

• however, when insurer uses net premium policy value, retrospective policy value may be useful

o recall (Ch. 7.2), net premium policy value uses premium based on valuation basis(policy basis) (regardless of true premium)

o IF P is calculated using the equivalence principle the retrospective and prospective net premium policy values will be the same

o retrospective policy value can be easier to calculate for complicated benefit or premium structures

o in U.S. some policies still valued using net premium policy values and in these cases retrospective formula is sometimes used

AS3429/9429a Ch.7 Lecture notes (W2018)

Defining Retrospective Net premium policy value

Define;

tV P = Prospective net premium policy value (tV

P = E[Lt])

tV R = Retrospective net premium policy value

(accumulated Prem less accumulated benefits for a large group of identical policies and divides this amount equally amongst survivors)

Also define;

L0,t = (PV, at issue, of future benefits payable up to time t − (PV,at issue of future net premiums payable up to time t)

tV R ≡ -E[L0,t](1+i)

t/tpx = -E[L0,t]/tEx

Assuming P is calculated using equivalence principle, same basis is used for

tV R and tV

P , can show;

0 = E[L0] 0 = E [L0,t] + v

t tpx E[Lt] , and rearranging gives

(0- E [L0,t])/( v t tpx ) = E[Lt]

tV R = tV

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 1: A whole life insurance with sum insured of $10,000 is issued to a life age 30. The Net annual Premium(P) is determined using the equivalency principle and the death benefit is paid at the end of the year of death. Express the net premium policy value at time t in terms of standard actuarial functions using (i) the prospective and (ii) the retrospective method. Verify that these expressions are equal.

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 2: A whole life insurance policy is issued to (40). The death benefit in the first five years of the contract is 5,000 and if death occurs after five years the death benefit is $100,000. Net Annual premiums are payable for a maximum of 20 years. Premiums are level for the first five years (P) then increase by 50% (1.5P). (a) Provide an expression for net policy values: 4V

R and 4V P

(b) Provide an expression for net policy values: 20V

R and 20V P

Note: should be able to show that, say 4V

R =4V P (same with t=20)

– as per Example 1 approach – start with equation for 0V=E[L0]=0 (equivalence principle) – alegbraically break up(1st 4 year versus the rest), and rearrange

AS3429/9429a Ch.7 Lecture notes (W2018)

7.8 Negative Policy Values

• can have negative policy values in early durations but in practice they would be set to 0

• negative policy values can reflect poor policy design

AS3429/9429a Ch.7 Lecture notes (W2018)

7.9 Deferred Acquisition Expenses & Modified

Premium reserves

• the most common application of policy value calculations are for reserve determination (i.e. actual capital held in respect of a policy)

• in reserve determination, a net premium approach with modification may be used to approximate a gross premium policy value (will look at impact of acquisition costs on policy values first)

• Denote tV n and tV

g as gross and net premium policy values using the equivalence principle and using original premium interest &mortality basis

tV n = EPVfuture benefits – EPV future net premiums

tV g = EPVfuture benefits + EPV future expenses – EPV future gross premiums

0V n = 0V

g =0

• Rewrite tV g as

tV g =EPVfuture benefits+EPVfuture expenses – (EPVfuture net premiums+EPVfuture expense loadings)

= tV n + ( EPVfuture expenses – EPVfuture expense loadings)

tV g = tV

n + tV e where tV

e = (EPVfuture expenses – EPVfuture expense loadings)

AS3429/9429a Ch.7 Lecture notes (W2018)

tV g = tV

n + tV e where tV

e = (EPVfuture expenses – EPVfuture expense loadings)

• Note that tV e is generally negative � tV

n > tV g (same assumptions for

both), and this reflects the impact of the initial(acquisition) expenses;

o Pg – Pn = Pe where Pe is called the expense loading(or expense premium)

• note that if only had renewal expenses then Pe would equal the expenses, but when have acquisition/initial expenses Pe is greater as it has to fund both renewal and initial expenses

Example 1: Calculate the expense loading and 10V

n, 10V g and 10V

e for a $100,000 whole life insurance policy issued to a life aged 50. Premiums are payable annually in advance throughout the term of the contract. The following assumptions are used for both premium and policy value calculations:

Initial expenses: 50% of gross premium plus $250 Renewal expenses: 3% of the gross premium plus $25 Interest: 4% Survival model: Standard Select Survival Model (required factors below)

a&& [50] =19.35185, A[50] =0.255698, a&& 60 =16.562066, A60 =0.362997 (i=4%)

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 1 (working page):

aP n &&

[50] A100000=

[50] 31.321,1P n

======>==>==>==>

aP g &&

[50] A100000=

[50] )P03.25(225P47.0 gg ++++ a&&

[50]

89.435,1P g

======>==>==>==>

aP e &&

[50] )P03.25(225P47.0 gg +++= a&&

[50]

58.114Pe ===>

58.114PPP Nge

====−−−−====

So, the expense loading portion(of gross premium) is $114.58

Note: renewal expenses = 68.08 = 0.03*PG+25 acquisition expenses = 967.95 =(0.5Pg )+250

so (114.58-68.08)=46.50 of the expense loading is used to amortize the acquisition expenses

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 1 (working page):

10V calculations using P n =1,321.31, Pg =1,435.89, Pe = 114.58

60 n

10 A100000V = 60 naP &&− 08.416,14=

60 G

10 A100000V = 60a25&&+ 60 gaP97.0 &&− 88.645,13=

= e

10V 60a25 && 60 GaP03.0 &&+ 60

eaP &&− 19.770−=

10V G = 10V

n + 10V e , as expected

Notes: negative expense reserve referred to as DAC

tV n is conservative as doesn’t allow for DAC reimbursement

AS3429/9429a Ch.7 Lecture notes (W2018)

7.9Deferred Acquisition Expenses & Modified Premium reserves(continued)

• negative expense reserve is referred to deferred acquisition cost(DAC)

• net premium reserve can be viewed as being overly conservative but can look at a modification of net premium reserve to approximate gross premium reserves without having to directly account for expenses o Full Preliminary Term(FPT) is a modified premium reserve

method and the one that is most commonly used

• FPT method (intended to approximate tV G esp. in early years)

o FPT uses net premium policy value approach, allows for a lower initial premium to implicitly consider DAC

o FPT considers policy as two policies-a 1 year term and a separate contract issued to same life 1 year later, if living

o Notation used 1P[x] ≡ single premium to fund year 1 benefits (cost of insurance (COI))

Pn[x]+1 ≡ net premium in all subsequent years

o under FPT 1V = 0

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 1 revisited:

(a) Calculate the modified premiums for this policy

(b) Compare net premium policy value, gross premium policy value, and FPT reserve at durations 0,1,2, and 10

Net Policy value (tV

n) Gross Policy Value (tV g) FPT (tV

FPT) time

0 0 0 0 1 1,272.15 383.67 0 2 2,573.97 1,697.21 1,318.50

10 14,416.08 * 13,645.88 * 13,313.21 * previously calculated

additional factors provided (to those in Example 1) q[50]= 0.0010333

a&& [50]+1 =19.105668, A[50]+1 =0.2651666, a&& 52 =18.853734, A52 =0.274856

AS3429/9429a Ch.7 Lecture notes (W2018)

Example 1 revisited(working page)