Chapter 7 - ValuationsIn this chapter we examine the insurance company’s financialposition with respect to an insurance policy or annuity not just at thepolicy’s initiation, but while it is in force.

Example 7-1: Consider a person age 35 who takes out a $100,000whole life insurance policy (2 year-select) with payments made (aslong as the insured is alive) over a 30-year period until age 65.Premium payments are made at the beginning of each year andbenefit payment is made at the end of the year of death. We assumethe Standard Select Survival Model in our textbook and 5% interestrate. Expenses are 50% of the first premium, 2% of all subsequentpremium payments and $250 at death benefit payment. Pricing thepolicy via the equivalence principle yields:

7-1

The gross loss function random variable at t = 0 in this example is:

L0 = (100,250) νK[35]+1 + 305.49− (623.70)

(1− νmin[K[35]+1,30](

.051.05

) ),

and P is chosen so that

E [L0] = 0.

Now fast forward t > 0 years and assume that the policyholder is stillalive and about to make the t th premium payment (t < 30).The gross loss function random variable of a policy still in force aftert years is:

7-2

Here PVt denotes the present value measured into the future using tas the vision point. In the above example,

Lgt =

(100,250) νK[35]+t+1

−(623.70)

(1−νmin[K[35]+t+1,30−t](

.051.05

) )if 0 < t < 30

(100,250) νK[35]+t+1 if t ≥ 30

If the policyholder dies before t years have passed then the policy isno longer in force and is no longer a liability to the company’sfinancial position. However, all policies in force do obligate thecompany financially. The company must be prepared to meet theseobligations. The expected impact of a policy with gross loss functionLg

t at t years in force is described as the policy value at t and isdenoted:

7-3

The value tV represents the expected assets needed to fulfill thecompany’s obligation to the policyholder. The company must holdreserves (assets) sufficient to cover the sum of all policy values ofpolicies currently in force.

Returning to example 7-1: At t (integer) years (t > 0),

tV = E [Lgt ]

Evaluating this expression under the assumptions used to set thepremium (premium basis) produces:

7-4

These computations of policy values were all made under the sameassumptions used at t = 0 to set the premium. More typically, oncein force, policies are valuated based on a different set ofassumptions. These assumptions are called the policy value basis.They often include an alternative interest rate assumption - often alower value than the one used under the premium basis to produce amore conservative assessment of the policy from the company’sperspective [hence larger reserves]. But the policy value may alsoinclude an altered life distribution (life table) as the company updatesits experience with this type of insurance policy.

7-5

Example 7-2: A person age x takes out a $250K 20-year endowmentpolicy. Assume µx+t = .04 and δx+t = .05 for all t > 0. The deathbenefit is paid immediately at death or at the end of 20 years.Assume premiums are paid continuously at P per year andexpenses are $500 at initiation plus 1% of premiums.(a) Find the annual premium rate P under these assumptions andthe equivalence principle.

7-6

The expected gross loss at time t > 0 is:

tV = (250,000)Ax+t ;20−t | − (.99)(14,656.23)ax+t ;20−t | for 0 < t ≤ 200 for t > 20.

(b) Valuate this policy at t = 5 and t = 15 under the policy valuebasis that µx+t = .05 and δx+t = .04 for all t > 0.

7-7

Example 7-3: A discrete 2-year term life insurance pays benefitsaccording to the table below at the end of the year of death. Levelpremiums are paid at the beginning of each year. If i = .06, find 1V .x bx qx0 150 .101 300 .15

7-8

Example 7-4: You are given Ax :n| = .25 and d = .06. Calculate n−1Vfor an n-year endowment policy paying 1 at the end of the year ofdeath or n-years, assuming level annual payments at the beginningof each year.

7-9

Example 7-5: Consider a continuous whole life insurance with adeath benefit of 2,000 for the first 20 years and 1,000 after thatissued age 40. Given: δ = .06, A50 = .33, A

150:10| = .19781 and

10E50 = .40657. Find 10V if premiums are 66 for the first 20 yearspaid continuously.

7-10

Recursive Calculation of tV

The discrete life table settings are the best reflection of what occursin practice. But they are more difficult settings in which to changethe basis because it involves recalculation of A and a values underthe new basis. To facilitate the computation of tV , we examine it as acash flow setting. Consider the interval from t to t + 1 when thepolicyholder is alive at time t . Note that tV is the gross premiumpolicy value for the policy in force at time t. It incorporates futurebenefits, expenses and premiums. It does not involve anything paidor received at time t itself. Let

Pt = ( Premium received at time t) and

et = (Expenses paid at time t).

Because the person is alive at time t, there is no insurance benefitpaid at t. Note that tV represents the assets (cash flow) that needsto be available at time t to handle expected future obligations to thispolicyholder, given that the person is alive at t.

7-11

Since nothing is received or paid between t and t + 1 ( we areassuming year end benefit payments and/or premium payments ),the cash flow at t + 1 for this policy is

where it is the interest rate for the interval from t to t + 1 under thepolicy value basis.

This cash flow must be sufficient to paySt+1 =(Death benefit paid at t + 1 if the person dies between t and t + 1)andEt+1 = (Expenses associated with paying the death benefit at t + 1)Thus we must pay

(St+1 + Et+1) with probability q[x ]+t at time t + 1.

7-12

And if the policyholder lives to time t + 1 we must have assets oft+1V to cover the expected future cash needs associated with thispolicy. That is, we need

t+1V with probability p[x ]+t at time t + 1 .

The source of the assets and the expected distribution of thoseassets should have expected values that match, that is, we have therecursion relationship :

The quantity (St+1 + Et+1 − t+1V ) is referred to as the sum at risk attime t + 1.

7-13

Example 7-6: A woman age [40] purchases a 4-year $100Kendowment policy with a single premium. The benefit is paid at theend of the year of death or at t = 4. Assume the Standard life tables,i = .05, $500 initial expense and $100 at benefit payment.(a) What was the price of the policy?

7-14

(b)Valuate the policy for t > 0 (t an integer) with i = .04, theStandard life table and $100 claim expense as the policy basis.

7-15

Example 7-7: A three year endowment policy pays 1000 at the endof the year of death or three years. Level premiums are paid at thebeginning of the first two years. Expenses are 50 at benefit paymentand 25 at premium payment. Use i = .06 and the life table below.(a) Find the premium via the equivalence principle.(b) Find 1V recursively, when i = .05.x qx0 .11 .152 .22

7-16

Annual Profit

At the beginning of year t we hold tV in reserve for a particular policyunder the policy basis. Now suppose we have 100 policies identicalto this one. Then the assets carried to t + 1 for these policies is

100(tV + Pt − et )(1 + it )

and the expected assets needed at time t + 1 is

100[

t+1V + q[x ]+t (St+1 + Et+1 − t+1V )]

So if reality perfectly reflects what was assumed in the policy basis,then we will have (according to the recursion formula) exactly theproper amount to cover all claims plus provide the reserves neededat the start of the year t + 1, namely t+1V . But this may not hold true.

Interest - The actual asset growth over the year may not be it whichwas assumed in the policy basis.Expenses - The actual expenses et and Et may not match the policybasis assumptions.

7-17

Mortality - The actual number of deaths may not be equal to100q[x ]+t .

When these conditions do not match up it results in a profit or a lossincurred by the policies in question for the designated year.

Example - Continuation of Example 7-6: Suppose we have 100identical policies as described in force at t = 2. Consider thecompany’s position relative to these policies over the year t = 2 tot = 3. During that year one of the 100 policyholders died, thecompany’s assets grew by 4.5% and actual claims paymentexpenses were $250. How did the company fare relative to these100 policies?

7-18

Next we seek to break down the components(interest/expenses/mortality) to understand the marginal contributionof each to the profit (loss) incurred. We use the continuing examplefor illustration.

(a) (interest/expenses/mortality) = (actual vs policy basis/policy basis/policy basis)

7-19

(b) (interest/expenses/mortality) = (actual/actual vspolicy basis/policy basis)

(c) (interest/expenses/mortality) = (actual/actual/actual vs policybasis)

7-20

Of course,

46.275− .009− 3.757 = 42.509 total profit found earlier

It should be clear from this description that the breakdown of profit(loss) into its components of (interest/expenses/mortality) providesthe marginal contributions of each component given the changesmade earlier. The values obtained in this breakdown do depend onthe order in which they are introduced into the breakdown.

7-21

Example - Continuation of Example 7-7: Use 1V and 2V values fromexample 7-7. Now suppose the actual experience from t = 1 to t = 2shows qA

1 = .1, iA = .05, premium expense at t = 1 of 20 and benefitexpense at t = 2 of 40. Find the annual profit over that year andbreak it down by (mortality/interest/expense).

7-22

Asset Shares

The idea behind the policy value tV is to designate the amount ofassets that the insurance company needs to have available inreserves at time t in order to break even (in an expected valuesense) in the future transactions on a particular policy in force.

The concept of the asset share ASt of a policy is the amount ofassets that the company actually has available (in an expected valuesense) to cover the future transactions on a policy in force.

The asset share accumulates net premiums received before time tand deletes benefits and expenses paid on or before time t, savingthese assets to meet future needs for the policy.

7-23

Let

ANPt = (Accumulated value at time t of all net premiums

paid prior to time t)

and

ABEt = (Accumulated value at time t of all benefits and

expenses paid prior to or on time t)

More formally then, the asset share at time t is:

7-24

Because it is based on the past (looking back on transactions thatoccurred before time t), an asset share is typically computed underassumptions which represent the actual experience of the companybetween initiation of the policy and time t. So interest rates usedreflect the company’s actual asset growth rates and mortality mayalso reflect actual experience with this or similar products.

We next describe the recursive computation of an asset share.Imagine N identical policies all taken out at time 0 to persons age [x].At time t there are (on average) N (tp[x ]) of these policies still inforce. They each have an asset share ASt , so the total assetsbrought into the time interval from t to t+1 is

N (tp[x ]) ASt .

7-25

Each of these policies pays a premium of Pt at the beginning of thistime interval and all these assets grow by interest rate it between tand t+1 to reach the amount

at time t+1, where et is the expense associated with the premiumpayment at time t.

But, of course, some claims of amount St+1 must be paid at the endof the year to account for the

N (tp[x ]) (q[x ]+t )

expected deaths occurring between t and t+1. So the total remainingin the asset pool at t+1 is

where Et+1 is the expense attached to a benefit payment of St+1.7-26

This asset pool is shared among the N (t+1p[x ]) policies that remainin force at t+1. So the asset share of each individual policy is

N (tp[x ])[(ASt + Pt − et )(1 + it )− (St+1 + Et+1) q[x ]+t

]N (t+1p[x ])

.

Note that N cancels out here and thus plays no role in thiscomputation. Also since

t+1p[x ] = (tp[x ])(p[x ]+t ),

we have the recursive formula for ASt+t based on ASt and otheringredients

Note that asset shares by their nature provide a retrospectivecomputation. They begin with AS0 which is used to find AS1, thenAS2, etc. until reaching ASt at the desired time t.

7-27

The asset share , as we have described it, is typically computed onan actual experience basis. But this same computation can be madeunder the premium basis as well.- - - - - - - - - -Theorem: If(a) the premium is determined by the equivalence principle, and(b) both tV and ASt are computed under the premium basis, then

tV = ASt

- - - - - - - - - -This result is formally proved in the fully continuous case in theAppendix to this chapter. When conditions of the theorem hold, thecomputation of the policy value in the manner of tV is described asthe prospective computation of policy value, while computing it in themanner of ASt is described as the retrospective computation of thatsame policy value.

7-28

Example - Continuation of Example 7-6: Previously we found:Single Premium: 82.86542 Premium Expense: .500Endowment Benefit: 100 Experience produces:

Year 1 2 3 4it .05 .045 .05 .055Ben.Exp .10 .20 .20 .25

Use the standard life tables to find the asset shares:

7-29

Example 7-8: A whole life policy is created which pays 200 at theend of the year of death. Expenses are not included, when the levelpremium paid at the beginning of each year is determined.

Year Premium Basis Actual Basisx qx i qx i0 .25 .05 .30 .061 .70 .05 .80 .0552 1.0 .05 1.0 .05

(a) Find the net premium value.(b) Compute the asset share values.

7-30

Section 7.4 - Alternate Valuation Times

Example 7-9: A $100K whole life policy is issued to a man age(40) with payments of P made monthly while the man is stillalive with the premium payments made for at most 20 years.The death benefit is paid at the end of the month of death andthe future life distribution is deMoivre (0, 50). Assume δ = .04.Expenses are not included. (a) Find P.

Benefit:

A(12)40 =

600∑k=1

e−.04(k/12)

50(12)=

e−.0412

600(1− e−

.0412 (600))

1− e−.0412

= .4316122

Premium:

A(12)140:20| =

240∑k=1

e−.04(k/12)

50(12)=

e−.0412

600(1− e−

.0412 (240))

1− e−.0412

= .27487688

7-31

7-32

(b) Find 4 512

V , assuming the policyholder is alive at this time andabout to make a premium payment.The future life distribution of the policyholder isdeMoivre (0, 50 - 4 5

12 ) = (0, 54712 ).

Future Benefit:

A(12)44 5

12=

547∑k=1

e−.04(k/12)

(54712 )(12)

=e−

.0412

547(1− e−

.0412 (547))

1− e−.0412

= .45911336

EPV of Future Benefit = 100,000A(12)44 5

12= $45,911.34.

Premium:

A(12)1

44 512 :15 7

12 |=

187∑k=1

e−.04(k/12)

547= .25397061

A(12)

44 512 :15 7

12 |= .25397061 + 15 7

12E44 5

12= .606832656

7-33

a(12)

44 512 :15 7

12 |=

1− .606832656.0399334

= 9.8455765

Approximating the valuation at dates that are not payment dates

So far we have only valuated policies at dates that are eitherpremium payment dates or potential benefit payment dates.Suppose k represents the interval between payment dates.(typically, k is 1 (annual payments) or 1

12 (monthly). Now suppose0 < s < k and we wish a valuation at time t + k + s.

7-34

An approximation can be found by interpolating between

the reserves available just after the premium payment of (Pt+k ) attime t + k , minus the expenses of receiving and processing thatpayment (et+k ) and

t+2kV ,

the reserves available at the next premium (benefit) payment time.

A linear interpolation produces as approximate valuation of

7-35

Example 7-10: Use the setting of example 7- 7 and find theapproximation for(a) 1.6V and (b) 2.2V .

7-36

Section 7.5 - Valuation for Continuous Payment Models

Example 7-11: Consider a whole life policy for someone age(x) with a benefit of $1000 paid at moment of death, premiumsof P paid annually smeared over time, constant force ofmortality µ = .02 and constant force of interest δ = .04. Weignore expenses.

(a) P is determined by

1000Ax = Pax

1000µ

µ+ δ= P

1µ+ δ

So P = µ (1000) = 20 .

(b) Find 10V .

10V = 1000Ax+10 − 20ax+10

7-37

Example 7-12: Consider the same setting as example 7-11, exceptfuture life length is Uniform (0, ω-x)

(a) P is determined by

1000Ax = Pax = P(1− Ax )

δor

P =1000(.04)Ax

(1− Ax )where here Ax =

1δ(ω − x)

(1− e−δ(ω−x)).

For example, when x = 40 and ω = 100,

A40 =1

(.04)(60)(1− e−(.04)(60)) = .3788675 and

P = 24.40.

7-38

(b) Find 10V .

10V = 1000 Ax+10 − Pax+10

Again using x=40 and ω= 100, produces

A40+10 =1

(.04)(50)(1− e−(.04)(50)) = .4323324 and

10V = 1000(.4323324)− 24.40(1− .4323324)

.04= 86.06.

Here the tV is positive for all t > 0 and goes to 1000 as t → 60. Buta uniform death distribution is also unrealistic.

Next we examine a truly flexible continuous model.

7-39

Let

Pt = smeared rate of premium paid at time t,

(meaning Pt (∆t) is paid in a small window of width ∆t around timet.)

et = smeared expense rate at time t,

(Pt − et ) = smeared assets received rate,

St = Benefit paid at t when it is moment of death,

Et = Expenses attached to paying the benefit at time t

(St + Et ) = Sum insured plus expenses if death occurs at t,

µt = force of mortality of x at time t and

δt = force of interest at time t.

7-40

Recall that accumulation satisfies

A(t) = A(0)e∫ t

0 δr dr .

Thus when discounting, we use

νt ≡ e−∫ t

0 δr dr

and when discounting from time t+s back to time t, we use the factor

e−∫ t+s

t δr dr =e−

∫ t+s0 δr dr

e−∫ t

0 δr dr=νt+s

νt.

Likewise, when viewing survival from age x+t to age x+t+s, we notethat

t+spx = (tpx )(spx+t ) or

spx+t =t+spx

tpx.

7-41

For a policy issued to age x, we seek tV , the reserves needed tyears into the life of the policy, given that the insured is alive at time t.

tV =(

EPV Benefits and payment Expenses to be paid beyond t)

−(EPV Net Premiums received beyond t)

≡ EPVBEt − EPVNPt .

Here,

EPVBEt =

∫ ∞0

(νt+s

νt

)(St+s + Et+s)spx+tµx+t+sds

=

∫ ∞t

(νr

νt

)(Sr + Er )

r px

tpxµx+r dr . So

7-42

Similarly,

EPVNPt =

∫ ∞0

(νt+s

νt

)(Pt+s − et+s)spx+tds and

νt (tpx )EPVNPt =

∫ ∞t

νr (Pr − er )r pxdr .

Therefore, tV satisfies

(7.5.1)

If we can find the integrals on the right, we can solve for tV . This canbe done numerically for a specific value of t.

Next we take the derivative of both sides of equation (7.5.1). Notethat

7-43

ddtνt =

ddt

e−∫ t

0 δr dr = e−∫ t

0 δr dr(

ddt−∫ t

0δr dr

)= −δtνt .

Also recall that

ddt

(tpx ) =ddt

(1− Fx (t)) = −fx (t) = −tpxµx+t .

So the derivative of the left-hand side (LHS) of equation (7.5.1) is

−δt (LHS)− µx+t (LHS) + νt (tpx )( d

dt tV)

7-44

When taking the derivative of the right-hand side of equation (7.5.1)with respect to t, we note that t only occurs in the lower bound of theintegral. Thus the derivative of the right-hand side of equation (7.5.1)is

νt (tpx )[(Pt − et )− (St + Et )νx+t

].

Setting the two sides of the derivative of equation (7.5.1) equal toone another and cancelling νt (tpx ) off both sides produces

which is known as Theile’s differential equation.

7-45

This differential equation enables us to adjust tV for small changesin t. For small h,

t+hV − tVh

.=

ddt

(tV ),

That is,

t+hV − tV.

= tV (h δt ) + (Pt − et )h − (St + Et − tV ) h µx+t or

Accumulated New assets Sum at Prob. ofReserves Added Risk Death

Recall that tV is computed from the end of the policy in the directionof the beginning. While h in the above must be small, it can benegative, creating a scheme for finding an approximate amount ofreserves, tV , at any time t while the policy is in force.

7-46

Example 7-13: Consider a 4-year endowment policy for $100K onselect age [40], which pays immediately at death or at the end ofyears (whichever comes first). Premiums are paid continuously at$22K per year and δt = .03 + t(.005). Use the text’s select life tablesand Assume UDD within each year. Also assume a $100 expense atbenefit payment.

- - - - - - - - - -We start at the end with

4V = 100.1K

Let h = - .01 and approximate

= 99.82995K . Then continuing back:

3.98V = (99.82995)[1− (.01)(.4995)] + 22(−.01)

−(100.1− 99.82995)(.01)µ43.99 (7.5.2)

What do we use for µ43.99?7-47

Suppose

t = j +k

100for k = 1,2, · · · ,100.

When t is an integer j = t-1, otherwise j = btc. The force of mortalityis:

7-48

So when j = 43 and k = 99, we get

f (t)S(t)

=

(65.08

99169.41

)(

1− (.99)(

65.0899169.41

)) .= .00065668

Substituting into expression (7.5.2) produces

7-49

Section 7.6 - Policy Alterations

A policy that is cancelled at the request of the policyholderbefore the term of the policy, is said to lapse or to besurrendered. The value of the policy at this point in time iscalled the surrender value or the cash value.

Typically, the customer wants to use this cash value towhatever advantage the policyholder can get from it.

Example 7-14: Consider a 4-year 100K endowment policy witha single premium. One that uses the standard life table withi = .05 issued to a woman age [40]. See the earlierdiscussions of this setting on pages 7-14 and 7-29.

Premium is: P = 82.86542K

2V = 92.55033 under the policy value

AS2 = 90.36376 under actual experience

7-50

Suppose at time t = 2, this woman wants to surrender theendowment policy in favor of a paid-up whole life insurance policy.What would be the sum insured for this whole life policy?

The typical reason for altering a policy is that the customer no longerwants to pay premiums. This customer desires a paid-up policyinstead. Since the process is stopping short on the stream ofpremium payments the sum insured is not going to be the same asunder the original agreement, but is now a paid-up sum insured.

Non-forfeiture laws specify that no changes can be made in thepolicy during the first few years it is in force. This is so that theinsurance company can recover its initial expenses in establishingthe policy.

7-51

Recall that

tV is the reserves which the insurance company should haveavailable to cover expected future liabilities associatedwith the policy as computed under a policy value basis.

ASt is the assets actually available to cover expected futureliabilities attached to a policy as computed under anactual financial experience basis.

If experience used in ASt is close to the policy value basisassumptions, then the values of tV and ASt should be closetogether. The cash value attached to a policy change willcorrespond to one or both of these values, but at a reduced rate. Itmay explicitly include expenses attached to altering the policy.

7-52

Example 7-15: Consider a setting in which T0 ∼ Uniform (0,100) inwhich a $1,000 whole life policy is written to someone age (40). Letδ = .04 and assume continuous payments of P per year.

On page 7-38 we found that the premium should be P = 24.40.Suppose after 10 years, this policyholder wants to cease payingpremiums and surrender this policy. What would be the sum insuredfor the paid-up policy?

On page 7-39 we found

10V = 86.06.

Now suppose the actual capital growth experienced by the companyduring the first 10 years of this policy was δ = .03. Thus the assetshare of this policy is difference between the premium accumulation:

7-53

and the payout experience accumulation

∫ 10

01000e.03(10−t) 1

60dt = e(.3)

(10001.8

)(1− e−.3) = 194.37,

divided by the probability of surviving those 10 years

10p40 =(50

60

). Therefore

AS10 =262.02− 194.37(50

60

) = 81.18.

The cash value of the policy is 95% of the smaller of 10V and AS10,that is,

7-54

This cash amount is used to purchase a whole life policy for theindividual who is now age 50. Let S be the sum insured. Find S .

7-55

In general, the cash value, Ct , is used to create an altered policywhich satisfies:

Care must be taken to ensure there is no anti-selection taking placewhich will disadvantage the insurance company. For example, aperson could be buying life insurance because the individual knowshe/she will soon die.

7-56

Example 7-16: Consider the setting of example 7-8, page 7-30.(a) Find the tV values for t = 0,1,2,3 under the premium basis.(b) At the end of year 1, the policyholder surrenders the policywithout paying the second premium, deciding in favor of a paid-upwhole life policy. If the cash value is 85% of the smaller value amongasset share and valuation, what is the benefit of this paid-up policy?

7-57

Section 7.8 - Negative Policy Values

Recall that

tV = EPV(future benefits and expenses)

- EPV(future premiums)

If initial (start-up) expenses are high, the premium amount willbe larger. Note that 0V = 0. When t is small but positive, thestart-up expenses are no longer in the future, but the largerpremiums typically will remain in the future. Thus for smallvalues of t, it is possible that tV is negative. The insurancecompany will not include a negative tV in reservescomputations, because the policyholder might lapse the policyat any time. The company will replace any negative tV valueswith zero to ensure sufficient reserves are available.

7-58

Appendix to Chapter 7 - Proof of Theorem on page 7-28 in theFully Continuous Case

Let h(Tx) denote the present value random variable (at t=0) offuture premiums.

Let k(Tx) denote the present value random variable (at t=0) offuture benefits and expenses.

By the equivalence principle, we have

E[h(Tx)

]= E

[k(Tx)

]. (7.A.1)

For an arbitrary t∗ > 0, examine

E[h(Tx)

]=

∫h(t)fx(t)dt

=

∫t≤t∗

h(t)fx(t)dt +∫

t>t∗h(t)fx(t)dt

7-59

Therefore,

E[h(Tx )

]ν t∗ t∗px

=

∫t≤t∗ h(t)fx (t)dt

ν t∗ t∗px+

∫t>t∗

ν−t∗h(t)(

fx (t)t∗px

)dt

=

∫t≤t∗ h(t)fx (t)dt

t∗Ex+ E

[ν−t∗h(Tx ) |Tx > t∗

]The second term on the right is the expected present value at t=t∗ offuture premiums beyond t∗ given that the policyholder is alive at t=t∗.

The first term is the expected present value at t = 0 of all premiumsprior to t = t∗, accumulated forward to t = t∗ by dividing by t∗Ex .

Break downE[

h(Tx )]

ν t∗ t∗pxin the same manner. Equation (7.A.1) then

shows

7-60

∫t≤t∗ h(t)fx (t)dt

t∗Ex+ E

[ν−t∗h(Tx ) |Tx > t∗

]=

∫t≤t∗ k(t)fx (t)dt

t∗Ex+ E

[ν−t∗k(Tx ) |Tx > t∗

]This can be rewritten as

E[ {ν−t∗k(Tx ) − ν−t∗h(Tx )

} ∣∣Tx > t∗]

=

∫t≤t∗ [h(t)− k(t)]fx (t)dt

t∗Ex. (7.A.2)

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The left-hand side of equation (7.A.2) is tV computed prospectivelyinto the future beyond time t = t∗. The function k(t), for example,discounts future (beyond t∗) benefits and expenses back to t = 0 andthe factor ν−t∗ moves the vision point from time 0 to time t∗. Thiscomputation involves future benefits and expenses minus futurepremiums in an effort to assess the expected future liabilitiesattached to the policy.

The right-hand side of equation (7.A.2) is tV computedretrospectively in the fashion of ASt , examining the EPV of premiumsminus benefits and expenses in an effort to determine how muchhas been put aside up to time t∗ for future liabilities on this policy.

To be equal the equivalence principle must be used to determine thepremium and both of these computations must be computed underthe PREMIUM BASIS ASSUMPTIONS.

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