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AS3429 Ch.6 Lecture notes (W2018)

Ch. 6 Premium Calculations

6.1 Summary 6.2 Preliminaries 6.3 Assumptions

6.4 Present Value of future loss random variable

6.5 Equivalence Principle

6.6 Gross Premium Calculation

6.7 Profit

6.8 Portfolio Percentile Premium Principle

6.9 Extra Risks

Note: additions//updates will be made to posted lecture notes as we work through the material

AS3429 Ch.6 Lecture notes (W2018)

6.1 Summary • Chapter focus is on how to determine premium calculations for

life insurance and life annuities, with a greater focus on life insurance

• Two different premium principles considered o Equivalence principle (most common for traditional policies)

o Portfolio percentile principle

6.2 Preliminaries

(i) Premium Types

(a) Net Premium

o considers benefit costs only, excludes expenses (and profit)

(b) Gross Premium

o explicitly allows for expenses

AS3429 Ch.6 Lecture notes (W2018)

6.2 Preliminaries(continued)

(ii) Premium Frequency Life Insurance o Typically periodic premium payments(eg annually, monthly, bi-weekly) o Periodic premiums are usually a level $ amount but they can be

varying (e.g. step rate)

Life Annuities o Whole/temporary life annuities usually single premium purchase o Deferred life annuities may be purchased with single premium or

periodic premiums during deferred period (pension funding applications)

(iii) Premium Timing and duration • Premiums are payable in advance with first premium payable when

policy is purchased

• Periodic premiums can be paid for the life or term of an Insurance policy or shorter(e.g. Limited Pay Whole Life), and cease to be payable on death of the insured

AS3429 Ch.6 Lecture notes (W2018)

6.3 Assumptions and Table sources

• The Standard Select Survival Model(SSSM)’ is default model for text examples and exercises. The model assumes i=5% and;

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)

• as noted previously, copy of Appendix D(SSSM) is posted - it also includes several values(e.g. select Insurance factors, pure

endowment factors, second moments for whole life insurance) - SOA LTAM tables now used are ultimate part of SSSM (copy posted)

• Standard Select Survival model(SSSM) Excel Worksheets - required for some text questions(as well as to understand some text examples) - recommend you set up own SSSM worksheet for your own use and/or use the

worksheet posted on OWL - author spreadsheet(SSSM) referenced earlier (Life Cons review notes) includes

several more factors than Table D

• ILT (copy posted)& other tables may be used for some class examples

AS3429 Ch.6 Lecture notes (W2018)

6.4 P.V. of the Future Loss Random Variable

•••• Cash flows associated with life insurance/annuity policies are generally life contingent

o This includes expenses and premiums as well as benefits •••• Can model (future outgo-future income) with the random variable

that represents the present value(P.V.) of future loss o Net Future loss- excludes expenses o Gross Future loss- includes expenses o Future loss random variable depends on time of death

Future Loss Random Variable Notation:

nL = PV benefit outgo – PV Net premium income

gL = PV benefit outgo + PV expenses – PV Gross premium income

AS3429 Ch.6 Lecture notes (W2018)

Example 1(6.4) A whole life insurance policy is issued to [45]. The sum insured is $25,000 payable immediately on death. Interest is at i =5%. A premium of $500 is paid at the beginning of each year for the policy. The policyholder dies at the end 25.8 years. Calculate the future loss at issue.

AS3429 Ch.6 Lecture notes (W2018)

Example 2 A whole life insurance policy is issued to [45]. The sum insured is $200,000 payable immediately on death. Premiums of amount P are payable annually in advance, ‘ceasing at age 65 or on earlier of death’ (20 Pay Whole Life Policy).What is the future loss random variable for this policy? •••• Distribution of future loss random variable used to determine

Premiums for given benefit/benefit for given premiums, using a Premium principle o Equivalence Principle (“benchmark principle”) o Portfolio Percentile Principle o Cash flow Profit Criterion Method (Ch.12)

AS3429 Ch.6 Lecture notes (W2018)

6.5 Equivalence Principle

•••• Under the equivalence principle, premiums are determined such that Expected value of the future loss is zero at contract issue

•••• Equivalence principle most common method for traditional insurances and is text default method for these policies (e.g Ch.6,7)

(i) Net Premiums

•••• Ignore expenses

•••• Benefits include either and/or both of death benefit/survivor benefits

•••• Equivalence principle means that E[L n

0] =0 , and since

L n

0 = PV benefit outgo – PV Net premium income, then under

equivalence principle EPVbenefit outgo= EPVnet premium income or PVFB = PVFP (at policy issue)

AS3429 Ch.6 Lecture notes (W2018)

6.5 Example 1 (SSSM is text default model) Consider a 20 pay whole life insurance with annual premiums issued to [50]. The death benefit(sum insured) is $500,000 payable at end of year of death.

(a) What is an expression for net future loss random variable?

(b) Calculate the net annual premium using the SSSM

AS3429 Ch.6 Lecture notes (W2018)

Example 2 (a few working pages are attached) Consider an endowment insurance with term n years and sum insured S payable at the earlier of the end of the year of death or at maturity, issued to a select life aged x. Premiums of amount P are payable annually throughout the term of the insurance (same as saying P’s are paid for the life of the policy).

(A) Derive expressions in terms of S, P and standard actuarial functions for;

(i) The net future loss, Ln 0 (ii) The mean of Ln 0, (iii) The variance of Ln 0

(B) Assume S=150,000; n=20, [x]=50 and SSSM

(iv) Calculate the annual premium (equivalence principle ) (v) Suppose now the premium is payable quarterly. What is the quarterly premium? Assume UDD within each year of age and you are provided factors for α(4) and β(4).

AS3429 Ch.6 Lecture notes (W2018)

Example 2(working page)

AS3429 Ch.6 Lecture notes (W2018)

Example 2(working page)

AS3429 Ch.6 Lecture notes (W2018)

Example 2–annuity factor calculation for B (v)

)4(

|20:]50[ a&&

= )4(

]50[a&& )4(

70]50[20 aE &&−−−−

)m(

|n:]x[ a&&

= )m(

]x[a&& )m(

n]x[]x[n aE +− &&

Under UDD : )m(Ba)m(a ]x[ )m(

]x[ −α≈ &&&&

or use: ]x[)m(

)m(

]x[ A i

i A ≈

in identity:

)m(

]x[)m(

]x[ d

)A1( a

−

=&&

Also, can show under UDD;

≈ )m(

|n:]x[ a&& )E1)(m(Ba)m( ]x[n|n:]x[ −−α &&

)4(

|20:]50[ a&& 598403.12)E1)(4(Ba)4( ]50[20|20:]50[ =−−α≈ &&

AS3429 Ch.6 Lecture notes (W2018)

Example 3 (Text 6.4)

An insurer issues a “regular premium deferred annuity contract” to a select life aged x. Premiums are payable monthly throughout the deferred period. The annuity benefit of X per year is payable monthly in advance from age x+n for the remainder of the life of (x).

(a) Write down the net future loss random variable in terms of lifetime random variables for [x]. (b) Derive an expression for the monthly net premium.

(c) Assume now that, in addition, the contract offers a death benefit of S payable immediately on death during the deferred period. Write the net future loss random variable for the contract, and derive an expression for the monthly net premium

AS3429 Ch.6 Lecture notes (W2018)

Example 3 (Text 6.4)

An insurer issues a regular premium deferred annuity contract to a select life aged x. Premiums are payable monthly throughout the deferred period. The annuity benefit of X per year is payable monthly in advance from age x+n for the remainder of the life of x.

(a) Write down the net future loss random variable in terms of lifetime random variables for [x].

Ln0 = 0 – 12P |12/1K )12(

a +&& for T[x] ≤ n or K[x] (12)< n K ≡ K[x]

(12)

Ln0 = |n12/1K )12(naXv −+&& – 12P |n

)12( a&& for T[x] >n or K[x]

(12) ≥ n (b) Derive an expression for the monthly net premium.

AS3429 Ch.6 Lecture notes (W2018)

Example 3(working page)

AS3429 Ch.6 Lecture notes (W2018)

Example 4 You are given an extract(see below) from a life table with a four year select period. A three year term insurance with death benefit of $S is purchased by [41] with a net premium of $350 payable annually. Assume i=6% and that the death benefit is payable at end of year of death.

Calculate

(i) S(sum insured) assuming the equivalence principle (ii) standard deviation of L0 (iii) Pr [L0 >0]

[[[[x] ] ] ] l[[[[x]]]] l [[[[x]+]+]+]+1 l[[[[x]+2]+2]+2]+2 l[[[[x]+3]+3]+3]+3 lx++++4444 x++++4444 [40] 100,000 99,899 99,724 99,520 99,288 44 [41] 99,802 99,689 99,502 99,283 99,033 45 [42] 99,597 99,471 99,268 99,030 98,752 46

AS3429 Ch.6 Lecture notes (W2018)

Example 4(working page)

AS3429 Ch.6 Lecture notes (W2018)

Example 4(working page

(ii) standard deviation of L0 or √Var[L0] calculations : (using S=216,326.38 from (i))

k L0 Pr(K41=k) 0

49.731,203a350Sv 1

=− &&

q[41] = 0.0011322 =1— l[41]+1/ l[41]

52.849,191a350Sv 2

2 =− &&

1│q[41]= 0.0018737

11.640,180a350Sv 3

3 =− &&

2│q[41] = 0.0021943

≥3

69.991a3500 3

−=− && 3p[41] = 0.9947997

Var[L0]=E[L0 2]–[E[L0]]

2=E[L0 2], when equivalence principle used (to determine P)

can show Var[L0] =188,537,738 � √Var[L0] = 13,731

Table above provides Lo values for given values of k. General expression for Lo is;

Ln0 = Sv K+1 – 350 |1Ka ++++&& for K[41]< 3 K ≡ K[41]

Ln0 = 0 – 350 |3a&& for K[41] ≥ 3

AS3429 Ch.6 Lecture notes (W2018)

Example 5 A select life aged 45 buys a policy with a single premium(P). The policy provides an annuity of $30,000 per year payable annually in advance from age 65. In the event of death before age 65, the premium is returned at the end of year of death. Assume SSSM applies (used Appendix D for calculations)

(a) Give an expression for Ln0

(b) Calculate P assuming the equivalence principle

(c) Now assume annuity is guaranteed to be paid for at least 5 years if insured survives to 65. How much does P increase by?

AS3429 Ch.6 Lecture notes (W2018)

Example 5 continued

A select life aged 45 buys a policy with a single premium(P). The policy provides an annuity of $30,000 per year payable annually in advance from age 65. In the event of death before age 65, the premium is returned at the end of year of death. Assume SSSM applies ( Appendix D used)

(a) Ln0 = Pv K+1 – P for K[45] < 20 (K=K[45])

Ln0 = |201K 20 av000,30

−−−−++++ && – P for K[45] ≥ 20

(b) Calculate P assuming the equivalence principle

(c ) Now assume annuity is guaranteed to be paid for at least 5 years if insured survives to 65. How much does P increase by?

AS3429 Ch.6 Lecture notes (W2018)

Example 5(working page)

AS3429 Ch.6 Lecture notes (W2018)

Example 6

A 20-pay $250,000 whole life insurance policy is purchased on (45) where the premiums are payable monthly and the death benefit is payable at moment of death. Assume that i=6% and ILT mortality applies. Also assume UDD within each year of age. Calculate the monthly premium(P).

Answer: P = $384.34

AS3429 Ch.6 Lecture notes (W2018)

Example 6 (working page)

AS3429 Ch.6 Lecture notes (W2018)

Notation used for net annual premiums for fully discrete Insurances (IAN)

x x

A P

&& ====

|n:x

x xn

A P

&& ====

|n:x

|n:x a

A P

&& ====

, |m:x

|n:x

|n:x m

A P

&& ====

|n:x

a&&

|n:x1

|n:x

A P ====

|n:x a&&

|n:x

A P ====

|m:x

|n:x

|n:x m

P &&

====

Default assumption: Periodic Premiums are payable for the life of the policy

Terminology “Fully discrete” : both premiums at benefits payable at discrete time points “semi-continous” : DB paid at moment of death, but Premiums paid at discrete points “Fully continuous” : premiums paid continuosly, DB paid at moment of death

AS3429 Ch.6 Lecture notes (W2018)

6.6 Gross Premiums

•••• apply equivalence principle to determine gross premiums, gross premiums consider insurers expenses

•••• Types of insurer expenses that need to be considered (a) Initial expenses, which are incurred when policy is issued. Examples

include agents commission and underwriting expenses

(b) Renewal expenses are normally incurred each time premium is payable: can include cost of premium processing, commissions etc

death of a policyholder (or annuitant) or on the maturity date of a term

insurance policy. Termination expenses are relatively small

•••• Timing and format of expenses o Convenient to assume expenses incurred at exact time as a premium or

benefit payment is made(in reality usually incurred slightly before/after) o Expenses can be proportional to premiums, proportional to benefits, or a

‘per policy’ expense

AS3429 Ch.6 Lecture notes (W2018)

•••• While GP calculations assumes expenses are known, expense

allocation is a complicated process

Gross Premiums

•••• Equivalence principle means that E[L g

0] = 0 , and since

L g

0 = PV benefit outgo + PV expenses – PV Gross premium income,

then under the equivalence principle

EPVbenefit outgo + EPVexpenses = EPVgross premium income or PVFB + PVFE = PVFGP (at policy issue)

AS3429 Ch.6 Lecture notes (W2018)

Example 1 (text 6.6)

An insurer issues a 25-year annual premium endowment insurance with sum insured $100 000 to a select life aged 30. The insurer incurs initial expenses of $2,000 plus 50% of the first premium, and renewal expenses of 2.5% of each subsequent premium. The death benefit is payable immediately on death.

(a) Write down the gross future loss random variable.

(b) Calculate the gross premium assuming – Standard Select Survival Model and i=5%(Appendix D) – UDD within each year of age

AS3429 Ch.6 Lecture notes (W2018)

Example 1(working page)

AS3429 Ch.6 Lecture notes (W2018)

Example 1(working page)

30.295,2$ 475.a975.0

2000A100000 P

25:]30[

25:]30[ ====

−−−−

++++

==== &&

Appendix D values used (i =5%) : 384.19a ]30[ ====&& , 060.16a55 ====&& 37256.0E ]30[20 ==== 77772.0E505 ====

AS3429 Ch.6 Lecture notes (W2018)

Example 2 (text 6.7) Calculate the monthly gross premium for a 10-year term insurance with sum insured $50 000 payable immediately on death, issued to a select life aged 55, using the following basis:

Survival model: Standard Select Survival Model(SSM)

I assumed UDD for fractional ages

Interest : 5% per year

Initial Expenses : $500+10% of each monthly premium in the 1st year

Renewal Expenses: 1% of each monthly premium in the second and subsequent policy years

Factors provided:

α(m) = i d/ i(m)d(m) � α(12) = 1.000197 B(m) = (i– i (m))/i(m)d(m) B(12) = 0.466502

024955.0=1 |10:[55]

A 97723.0a )12(

|1:]55[ =&&

83389.7a )12(

|10:]55[ =&&

AS3429 Ch.6 Lecture notes (W2018)

Example 2(working page)

AS3429 Ch.6 Lecture notes (W2018)

Example 2

•••• In this Example, Year one premium income is not enough to cover year 1 expenses (‘new business strain’)

o Annual Premium amount =12P= 12(18.99)= $ 227.88

o 1st year expenses= ($ 500+ .10(12P) >> $ 227.88 •••• New business strain is common

o Insurer needs to ensure enough funds to sell polices and may have to borrow to write new business

o Early expenses gradually paid off by expense

loadings(loading=gross less net premium) in future premiums and the part of the premium that funds initial expenses is called deferred acquisition costs

AS3429 Ch.6 Lecture notes (W2018)

Example 3 A Whole Life insurance policy is purchased on (35). The death benefit is paid at the end of year of death and is $ 300,000 if death occurs within the first 30 years, and 75,000 thereafter.

Gross Premiums are payable annually, where the initial annual premium is level for the first 30 years, and thereafter annual premiums are 1/3 of the initial annual premium.

Expenses are 40% of the first year premium and 5% of all subsequent premiums. All expenses are payable at the beginning of the year.

Calculate the initial gross annual premium, assuming mortality follows the Illustrative Life Table (ILT) and that i=0.06.

Solution:

Set P = “Initial Gross Annual Premium”

EPVpremiums= EPVBenefits +EPVexpenses (or E[L G

0]=0)

consider info provided in ILT to streamline calculations

AS3429 Ch.6 Lecture notes (W2018)

Example 3(working page)

AS3429 Ch.6 Lecture notes (W2018)

Example 4 (text 6.9)

Calculate the gross single premium(P) for a deferred annuity of $80,000 per year payable monthly in advance, issued to [50] with the first annuity payment on the life’s 65th birthday. Allow for initial expenses of $1,000 and renewal expenses on each anniversary of issue date, provided policyholder is alive. Assume the renewal expense will be $20 on the first anniversary of the issue date, and that expenses will increase with inflation from that date at a compound rate of 1% per year.

Assume Standard Select Survival Model(SSSM) and i=5%.

The following SSSM values are provided (i=5%):

15E[50] =0.4616267 )12(

65a&& =13.08696 (used W3 approxm) )

AS3429 Ch.6 Lecture notes (W2018)

Example 4(working page)

AS3429 Ch.6 Lecture notes (W2018)

Example 5 A $250,000 Whole Life insurance policy is purchased on (35) with annual gross premiums(P) payable for a maximum of 10 years. The death benefit is paid at the end of year of death. You are given:

(i) Expenses of $200 are payable at end of each year, including year of death

(ii) i = 6% and mortality follows the Illustrative Life table(ILT)

Calculate P.

Solution:

-different but relatively simple expense structure in that the only expenses are dollar amount payable at end of each year. Equation below does properly reflect fact that, in the year of death, there is only one payment of $200(at the end of that year).

)1a(200A200,250A200a200A250000aP 353535353510:35 −−−−++++====++++++++==== &&&&

Solving for P using ILT values should give P=$4,530.37

AS3429 Ch.6 Lecture notes (W2018)

6.7 Profit

•••• can use future loss random variable to determine probability of a profit

•••• Note that while equivalence principle doesn’t explicitly allow for profit, once can implicitly allow for profit by providing for margins in assumptions (e.g. mortality, interest)

•••• assuming no margins though, equivalence principle has expected outcome of no profit(expected profit=0)

•••• actual profit outcome for a given policy can be positive or negative

•••• for actual profit for a group of policies to be close to expected, must sell large # policies to a group of independent lives (same age & risk) o fundamental insurance pricing concept(diversification) o as # contracts increase, probability of profit may decrease but probability

of a large aggregate loss (relative to total premiums) decreases o see text example with 1 year term-1 life versus 100 independent lives

same age, same risk, example looks at Pr [portfolio profit ≥ 0]

AS3429 Ch.6 Lecture notes (W2018)

Working page-text example:

AS3429 Ch.6 Lecture notes (W2018)

Example 1 (text illustration)

Consider a whole life Insurance policy issued to [30] where the 100,000 death benefit is payable at end of year of death. There’s an initial expense of $1,000 and ‘renewal expenses’ of $50(including at time of 1st premium). What is the probability the insurer makes a profit? Assume i=5% and the Standard Select Survival Model

Figure 6.1: graphs profit should death occur in a given year in terms of values at the end of a given year

The longer an insured survives, the greater the insurer’s profit, opposite is true for life annuities

AS3429 Ch.6 Lecture notes (W2018)

Example 1- working page

E[LG0 ] = 0 � P=498.45 (equivalence principle)

LG0 = 100,000 v K+1+ 1000 + 50 |1Ka +&& - P |1Ka +&& , K=K[30] (1)

Pr[ LG0 <0]=Pr[ (100,000–(50-P)/d)v K+1+1000+(50-P)/d < 0] (rewrote(1))

= Pr[ (100,000 – (50-P)/d)vK+1 < -1000 -(50-P)/d]

= Pr[ (109,417.45)vK+1 < 8,417.45]

= Pr[ vK+1 < 0.076929685]

= Pr[ (1.05)K+1 > 12.99888327]

= Pr[ (K+1)log(1.05) > log(12.99888327)]

= Pr[ (K+1) > log(12.99888327)/log(1.05)]

Pr[ LG0 < 0]= Pr[ (K +1) > 52.57] = Pr[K > 51.57] K=K[30]

= Pr[K[30] ≥ 52]=52p[30] =l82/l[30] =0.70704 (using SSSM)

There’s a profit if [30] survives 52 years & probability of this is 0.70704

AS3429 Ch.6 Lecture notes (W2018)

General model for Example1: Given: [x] = select issue age

S = sum insured for whole life policy payable at end of year of death

I = initial expenses

e = renewal expenses associated with each premium payment (including the first) Can show that

Pr [L g

0 <0] = Pr [ K[x] + 1>(1/δ) log((P- e + Sd)/(P – e – Id)) ] (6.4)

** not a useful formula to memorize(specific to one expense structure), but you are now able to derive it (general case of Ex.1)

AS3429 Ch.6 Lecture notes (W2018)

Example 2 ** An insurance company issues a deferred whole life annuity to [55]. Monthly payments of $1,000 will start 10 years from issue date. This annuity policy is being purchased with annual premium payments (P) during the deferred period. Assume there are no expenses. What is the probability that the insurance company makes a profit from this policy? Assume i =5%, the SSSM and UDD within each year of age.

You are given the following factors from the SSSM:

|10:]55[ a&&

= 8.021871 and 10E[55] )12(

65a&& = 7.768577

You can show(using equivalence principle & equation below) P = $11,621.09

]55[10|10:]55[ E12000aP ====&& )12(

65a&&

** Note: See Text Ex. 6.11, for another good life annuity profit question (deferred indexed life annuity purchased with a NSP)

AS3429 Ch.6 Lecture notes (W2018)

Example 2 An insurance company issues a deferred whole life annuity to [55]. Monthly payments of $1,000 will start 10 years from issue date. This annuity policy is being purchased with annual premium payments (P) during the deferred period. Assume there are no expenses. What is the probability that the insurance company makes a profit from this policy? Assume i=5%,

SSSM and UDD within each year of age. P =11,621.09

L0 = 0 – |1K aP

++++ && for K[55] ≤ 9 K=K[55]

−−−−==== −−−−++++ |1012/1K )12(10

0 )12( ]55[

av000,12L && |10aP && 10K )12(

]55[ ≥≥≥≥ **

Working with ** :

Pr[ L0 < 0] = Pr[(12,000v 10)(1- vK+1/12-10 ) < |10aP && ],

)12(

]55[KK ≡≡≡≡ d(12)

Pr[ L0 <0 ] = Pr[(12,000v 10)(1- vK+1/12-10 ) < 94,221.73]

d(12)

AS3429 Ch.6 Lecture notes (W2018)

Example 2 –working page

Pr[ L0 <0]= Pr[(12,000v 10)(1- vK+1/12-10) < 94,221.73] ,

)12( ]55[KK ≡≡≡≡

d(12)

=Pr [1- vK+1/12-10 < 0.62274824]

=Pr [ vK+1/12-10 > 0.37725176]

=Pr[(1.05)K+1/12-10 < 2.650749728] =Pr [ K+1/12-10 < log(2.6507497)/log(1.05)]

=Pr [ K+1/12-10 <19.98], )12( ]55[KK ≡≡≡≡

Pr[ L0 <0] =Pr [ )12( ]55[K +1/12 < 29.98]

AS3429 Ch.6 Lecture notes (W2018)

Example 2 –working page

Pr[ L0 <0] =Pr[ )12(

]55[K +1/12 < 29.98] = Pr [ )12(

]55[K < 29.897]

Pr[ L0 <0] = 29 11/12q[55] = 1 ― 29 11/12 p[55]=1―(29 p[55])(11/12 p84)

≈ 1 ― 29p[55](1―(11/12)q84) UDD assumption

Pr[ L0 <0] = 0.371603

AS3429 Ch.6 Lecture notes (W2018)

Example 3(Text 6.9) A 25-year endowment insurance policy is issued to [30] with basic sum insured of $250,000. Premiums are payable annually throughout the term of the policy. Initial expenses are $1200 plus 40% of the first premium. Renewal expenses are 1% of all subsequent premiums. The insurer allows for a compound reversionary bonus of 2.5% of the basic sum insured, vesting on each policy anniversary (including the last). The death benefit is payable at the end of the year of death. Assume the Standard Select Survival Model and i=5%

(a) Derive an expression for the future loss random variable (b) Calculate the annual premium (P) for this policy, (c ) calculate L0(k) for k=0,1,2…25 (d) Calculate the probability an Insurer makes a profit on this policy

AS3429 Ch.6 Lecture notes (W2018)

Example 3(Text 6.9) A 25-year endowment insurance policy is issued to [30] with basic sum insured of $250,000. Premiums are payable annually throughout the term of the policy.

Initial expenses are $1200 plus 40% of the first premium. Renewal expenses are 1% of all subsequent premiums.

The insurer allows for a compound reversionary bonus of 2.5% of the basic sum insured, vesting on each policy anniversary (including the last). The death benefit is payable at end of the year of death.

Assume the Standard Select Survival Model and i=5%

(a) Derive an expression for the future loss random variable set K≡ K[30] in my equation below

Lg0=250,000(1.025) min(K,25)(v)min(K +1,25)+1200 + .39P - .99P a&& |)25,1Kmin( +

AS3429 Ch.6 Lecture notes (W2018)

Example 3(Text 6.9) A 25-year endowment insurance policy is issued to [30] with basic sum insured of $250,000. Premiums are payable annually throughout the term of the policy. Initial expenses are $1200 plus 40% of the first premium. Renewal expenses are 1% of all subsequent premiums. The insurer allows for a compound reversionary bonus of 2.5% of the basic sum insured, vesting on each policy anniversary (including the last). Death benefit is payable at end of the year of death. Assume Standard Select Survival Model and i=5%.

(b) Calculate the annual premium (P) for this policy, given the following;

|25:]30[ a&&

= 14.73114, 25E[30] =0.289750771 (used 25E[30]= v 25

25p[30] )

|25:[30] A

i* = 0.0127074 ,where i*=(1.05/1.025) -1= 0.024390224

EPV(Premiums) = EPV(Benefits) + EPV(expenses)

P |25:]30[

a&& =EPV(Benefits) + 1200 + 0.39P + .01P |25:]30[

a&&

P(.99 |25:]30[

a&& - 0.39) = EPV(Benefits)+1200 =137,394.80 +1200

P = 9,764.44

AS3429 Ch.6 Lecture notes (W2018)

Example 3(Text 6.9)-working page

AS3429 Ch.6 Lecture notes (W2018)

Example 3(Text 6.10) A 25-year endowment insurance policy is issued to [30] with basic sum insured of $250,000. Premiums are payable annually throughout the term of the policy. Initial expenses are $1200 plus 40% of the first premium. Renewal expenses are 1% of all subsequent premiums. The insurer allows for a compound reversionary bonus of 2.5% of the basic sum insured, vesting on each policy anniversary (including the last). Death benefit is payable at end of the year of death. Assume Standard Select Survival Model and i=5%

(c) Let L0(k) denote the present value of the loss on the policy given that K[30] = k for k ≤ 24. Let L0(25) denote the present value of the loss on the policy given [30] survives to age 55. Using Excel or some other software, calculate L0(k) for k=0,1,2…25 From (a):

Lg0=250,000(1.025) min(K,25)(v)min(K +1,25)+1200+.39P ―.99P a&& |)25,1Kmin( +

(d) Calculate the probability the insurer makes a profit on this policy.

AS3429 Ch.6 Lecture notes (W2018)

Example 3(Text 6.9)- part c-generated values(excel)

k L0(k) 0 $233,436.57 1 $218,561.17 2 $204,259.14 3 $190,506.40 4 $177,279.93 5 $164,557.73 6 $152,318.77 7 $140,542.97 8 $129,211.12 9 $118,304.86

10 $107,806.63 11 $97,699.66 12 $87,967.91 13 $78,596.02 14 $69,569.34 15 $60,873.82 16 $52,496.05 17 $44,423.20 18 $36,642.97 19 $29,143.62 20 $21,913.91 21 $14,943.08

22 $8,220.84 23 $1,737.34 24 -$4,516.87 25 -$1,178.61

AS3429 Ch.6 Lecture notes (W2018)

6.8 The Portfolio Percentile Premium Principle •••• look at large portfolio of n identical and independent(i.i.d.) policies

and set premium (P) so there’s specified probability(α) that total future loss is negative

•••• Notation: N is the # of identical and independent Policies

L0,i is the future loss random variable for policy i and mean/variance for each L0,i equals mean/variance of L0,1

L is the total future loss in the portfolio where L = ∑ L0,i

E[L] = ∑ E[L0,i] = N E[L0,1] and Var[L] =NVar[L0,1]

• set P so that Pr[L<0] = α , use CLT for distribution of L o Φ is cumulative distribution function of the standard normal distribution o for large n, distribution of L is approximately Normal with

mean =NE[L0,1] and Variance (V) = NV[L0,1] , can calculate P using

Pr[L < 0] =Pr[(L − E[L]) < 0− E[L] = α = Φ(-E[L]/√ V[L])

√V[L] √V[L]

AS3429 Ch.6 Lecture notes (W2018)

Example 1: An Insurance Company issues identical whole life insurance policies to n individuals aged 50 with independent future lifetimes. You are given the following:

(i) Each policy has sum insured of $150,000 payable at end of year of death

(ii) Issue expenses are 25% of the first year premium and renewal expenses are 5% of each subsequent annual premium

(iii) The assumed interest rate is i = 6% and mortality is based on the Illustrative Life table(ILT required factors provided)

(iv) The annual premium for these policies was calculated using the percentile premium principle and the normal approximation, such that the probability of a loss on the portfolio of n policies is 5%

What is the annual premium per policy for a portfolio of n=10,000 policies?

AS3429 Ch.6 Lecture notes (W2018)

Example 1-working page

AS3429 Ch.6 Lecture notes (W2018)

Example 1-2nd working page

Pr[L > 0 = 0.05] � Pr[L ≤ 0 = 0.95]

Pr[L < 0] =Pr[(L − E[L]) < 0− E[L] = α = Φ(-E[L]/√ V[L]) √V[L] √V[L]

Using N(0,1) Pr[ z < 1.645] = 0.95

1.645 = 0 − E[L] , or √V[L]

1.645√V[L] = − E[L]

1.645√n (150,000+16.7833P)√0.03273)= −n[37,357.5 —P(12.40346)]

1.645(150,000+16.7833P)√0.03273)= −√n[37,357.5 —P(12.40346)]

plug in n=10,000 and solving for P gives P= 3,060.17

AS3429 Ch.6 Lecture notes (W2018)

Example 2(Text 6.11) An insurer issues whole life insurance policies to select lives aged 30. The sum insured of each policy is $100,000 is paid at the end of the month of death. Level monthly premiums are payable throughout the term of the policy. Initial expenses, incurred at the issue of the policy, are 15% of the total of the first year’s premiums. Renewal expenses are 4% of every premium, including those in the first year. Assume the SSSM with interest at 5% per year.

Calculate the monthly premium using the portfolio percentile principle, such that probability that the future loss on the portfolio is negative is 95%. Assume a portfolio of 10,000 identical, independent policies.

Compare this to the Premium calculated using the equivalence principle. You are given the following values (using SSSM)

d(12) = 0.0486911 a&& (12)

[30] = 18.922102

A(12) [30] = 0.0786618, 2A(12) [30] = 0.011539249

First, can show P=36.39 using equivalence principal

AS3429 Ch.6 Lecture notes (W2018)

Example 2 (working page)

(ii) Monthly premium calculations: Portfolio percentile principle An insurer issues whole life insurance policies to select lives aged 30. The sum insured of each policy is $100,000 is paid at the end of the month of death. Level monthly premiums are payable throughout the term of the policy. Initial expenses, incurred at the issue of the policy, are 15% of the total of the first year’s premiums. Renewal expenses are 4% of every premium, including those in the first year. Assume SSSM(i=5%) & given:

d(12) = 0.0486911 a&& (12)

[30] = 18.922102

A(12) [30] = 0.0786618, 2A(12) [30] = 0.011539249

K=K(12)[30]

|12/1K )12(12/1K

i,0 a)P12(96.)P12)(15(.v100000L ++++ ++++

−−−−++++==== && ,

30 )12()12(

]30[i,0 a)P12(96.)P12)(15(.A100000]L[E &&−−−−++++==== ,

E[L0,i] = $7,866.18 - 216.18P (used given values)

and can show V[L0,i] =(100,000+236.59P) 2(0.0053515)

then √V[L0,i] = (100,000+236.59P) (0.073154)

AS3429 Ch.6 Lecture notes (W2018)

Example 2- working page

AS3429 Ch.6 Lecture notes (W2018)

Table 6.4

Premiums according to portfolio size (Example 14)

n P

1,000 38.31 2,000 37.74

5,000 37.24 10,000 36.99 20,000 36.81

• P decreases as n increases. • insurer diversifies mortality risk as n � ∞

AS3429 Ch.6 Lecture notes (W2018)

Example 3(determine percentile of L given P)

A 20-year endowment policy with sum insured of 250,000 payable at the end of year of death is issued to an individual aged 40. You are given that the annual premium payable for the life of this policy using the equivalence principal is P=$7,333.84. You are also given that standard deviation of the net future loss random variable using this premium is $14,485 for this policy. Assume now that 10,000 of these identical independent policies are sold. Estimate the 99th percentile of the corresponding net future loss random variable, using the premium given above. Know that

485,14]L[V i,0 =

AS3429 Ch.6 Lecture notes (W2018)

Example 3(working page):

AS3429 Ch.6 Lecture notes (W2018)

6.9 Extra Risks

•••• looks at three different methods used to determine policy premiums policy when there is extra mortality risks insured is a substandard risk

1. Age rating Method

• may be used when an insured suffers from a medical condition (referred to as an impaired life)

• can reflect extra risk by assuming insured is older (e.g. insured is 50, impairment is such that their mortality risk is that of a 60 year old, then determine premium assuming insured is age 60)

2. Constant addition to force of mortality (µ x )

• this method used when extra risk is essentially independent of age (e.g. insured participates in risky hobbies)

• can be computational shortcuts using this method

3. Constant multiple of mortality rates

• reflect extra mortality as a multiple of standard mortality, • e.g. q' [x]+t = 1.2 q[x]+t (the ' refers to the impaired life)

AS3429 Ch.6 Lecture notes (W2018)

Example 1(Text 6.12)

Calculate the annual premium for a 20-year endowment insurance with sum insured $200,000 issued to a life aged 30 whose force of mortality at age 30+s is given by µ[30]+s+ .01

Allow for initial expenses of $2,000 plus 40% of the first premium, and renewal expenses of 2% of the second and subsequent premiums. Use the Standard Select Survival Model with interest at 5% per year.

You are given that |20:]30[

a&& j = 12.0717, where j =0.065523

AS3429 Ch.6 Lecture notes (W2018)

Example 1(working page)

Calculate the annual premium for a 20-year endowment insurance with sum insured $200,000 issued to a life aged 30 whose force of mortality at age 30+s is given by µ[30]+s+ .01

You are given that |20:]30[

a&& j = 12.0717, where j =0.065523

Using equivalence principle gives

P a&& ” [30]:20 = 200,000A ”

[30]:20 + 2000 + 0.38P+0.02P a&& ”

[30]:20

re-arranging gives,

P = (200,000A”[30]:20 + 2000) ___________________________________

(0.98 a&& ”[30]:20 ― 0.38) & will show P=$7600.82

AS3429 Ch.6 Lecture notes (W2018)

Example 2(Text 6.13)

Calculate the monthly premium for a 10-year term insurance with sum insured $100,000 payable immediately on death, issued to a life aged 50. Assume that each year throughout the 10-year term the life is subject to mortality rates that are 10% higher than for a standard life of the same age. Allow for initial expenses of $1000 plus 50% of the first monthly premium and renewal expenses of 3% of the second and subsequent monthly premiums. Use UDD assumption where appropriate, and use Standard Select Survival Model with i = 5%.

You are given α(12) = 1.0002 and β(12) = 0.4665 and the following mortality rates from the Standard Select Table

q[50] =.001033 q[50]+1 =.001264 q52 =.001469 q53 =.001623

q54 =.001797 q55 =.001993 q56 =.002212 q57 =.002459

q58 =.002736 q59 =.003048

Note: q'[x]+s=1.10 q[x]+s

AS3429 Ch.6 Lecture notes (W2018)

age q (SSSM) q' t tp'[50] tǀq'[50] v t v

t+1 vttp'[50] vt+1tǀq'[50]

[50] 0.001033 0.00114 0 1.0000 0.0011 1 0.95238 1.0000 0.0011

[50]+1 0.001264 0.00139 1 0.9989 0.0014 0.9524 0.90703 0.9513 0.0013

52 0.001469 0.00162 2 0.9975 0.0016 0.9070 0.86384 0.9047 0.0014

53 0.001623 0.00179 3 0.9959 0.0018 0.8638 0.82270 0.8603 0.0015

54 0.001797 0.00198 4 0.9941 0.0020 0.8227 0.78353 0.8178 0.0015

55 0.001993 0.00219 5 0.9921 0.0022 0.7835 0.74622 0.7774 0.0016

56 0.002212 0.00243 6 0.9899 0.0024 0.7462 0.71068 0.7387 0.0017

57 0.002459 0.00270 7 0.9875 0.0027 0.7107 0.67684 0.7018 0.0018

58 0.002736 0.00301 8 0.9849 0.0030 0.6768 0.64461 0.6666 0.0019

59 0.003048 0.00335 9 0.9819 0.0033 0.6446 0.61391 0.6329 0.0020

10 0.9786

q'[x]+s=1.10 q[x]+s 8.0516 0.0158

ǁ ǁ

Rounded #s for presn only (ä [50]10)’

(A'[50]10)’

AS3429 Ch.6 Lecture notes (W2018)

Example 2(working page)