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### Transcript of Chapter 4 - Insurance...

• Chapter 4 - Insurance BenefitsSection 4.4 - Valuation of Life Insurance Benefits

(Subsection 4.4.1) Assume a life insurance policy pays \$1immediately upon the death of a policy holder who takes out thepolicy at age x .

The present value of this payment is:

where δ = ln(1 + i). This present value Z is a random variable,because the future life time of this person, Tx , is itself a randomvariable.

4-1

• The Expected Present Value (EPV) of this future payment is:

=

∫ ∞0

e−δ t fx (t) dt

=

∫ ∞0

e−δ t tpx µx+t dt

The second moment of this random present value is :

E[

Z 2]

= E[{e−δ Tx }2

]= E

[e−2δ Tx

]≡ 2Ax

4-2

• Here is superscript 2 on the left indicates that this computation of Axuses twice the force of interest (which is NOT the same thing astwice the interest rate). Consequently, the variance of the randompresent value Z is:

Of course, if the policy pays a benefit of s dollars, then the expectedpresent value is

EPV = E [s Z ] = s Ax ,

and the random benefit’s variance is

Var [s Z ] = s2( 2Ax − {Ax}2 ).

4-3

• Example 4-1: Suppose fx (t) = 2tω2 for 0 < t < ω, and zero elsewhere.Find the EPV and Var[ Z ] in this setting.

4-4

• Example 4-2: Assume T40 is a continuous random variable withconstant force of mortality .03. If a whole life policy pays 100immediately upon death and the force of interest is δ = .04, find theEPV of this insurance claim.

4-5

• (Subsection 4.4.2) Assume the life insurance policy pays \$1 at theend of the policy year in which the person dies.

The present value of the random benefit is now:

Z = νKx+1

where Kx = bTx c is the curtate future life time random variable.

The Expected Present Value is therefore,

= ν P[Kx = 0] + ν2 P[Kx = 1] + ν3 P[Kx = 2] + · · ·

= ν qx + ν2 1|qx + ν3 2|qx + · · ·

4-6

• Recall that when k ≥ 1,

The second moment of this random present value is:

2Ax = E [Z 2] =∞∑

k=0

(νk+1

)2k |qx

=∞∑

k=0

(ν2)k+1

k |qx

which is the same as the computation of Ax when the force ofinterest is doubled, i.e. the effective interest rate is j satisfying

4-7

• The variance of the random present value is

Var [ Z ] = 2Ax − (Ax )2.

Example 4-3: Suppose qx+0 = .1, qx+1 = .4, and qx+2 = 1.0 withi = .25 then find Ax and Var [ Z ].

4-8

• (Subsection 4.4.3) Assume the life insurance policy pays \$1 at theend of the ( 1m )

th fraction of the policy year in which the person dies.Typically, m = 1,2,4 or 12.

With K (m)x =(# of sub-periods suvived)

m , the EPV of the benefit is:

A(m)x = E[νK

(m)x +

1m]

=(ν

1m)

1m

qx +(ν

2m)

1m

∣∣∣ 1m

qx +(ν

3m)

2m

∣∣∣ 1m

qx + · · ·

The variance of the random present value is

Var [ Z ] = 2A(m)x − (A(m)x )

2,

where 2A(m)x is computed like A(m)x with the force of interest doubled,

i.e. with ν replaced with ν2.

4-9

• Example 4-4: A three year old red fox buys a whole life policy thatpays 1 at the end of the half-year in which death occurs. Let i = .05and assume UDD within each year. Use the life table to find the EPVof this insurance claim.

x lx dx3 2194 19314 263 2635 0

4-10

• (Subsection 4.4.4) Recursive Computation

In discrete cases with benefit payments of \$1 at the end of the policyyear of death the values of Ax are computed based on a life table.These tables have a maximum life length ω beyond which no personis assumed to live. That is, there is a value ω such that

qω−1 = P[ person dies before age ω | person lives to ω − 1 ] = 1.

If a policy was taken out by someone age ω − 1, then

Aω−1 = E[νKω−1+1

]= ν. (because Kω−1 ≡ 0)

If a policy was taken out by someone age ω − 2, then

4-11

• Similarly, a policy taken out at age x has expected present value

Ax = νqx + ν2pxqx+1 + ν3pxpx+1qx+2 + · · ·

= ν[qx + px

{νqx+1 + ν2px+1qx+2 + ν3px+1px+2qx+3 + · · ·

}]showing that

So Ax can easily be computed as long as Ax+1 is known. Since weknow the final value, Aω−1 = ν, we can use backward recursion tocompute Ax values for all the ages x in the life table.

4-12

• Example 4-5: Use i = .05 and complete the Ax column.x qx Ax0 .012461 .022452 .086193 .377454 1.00

4-13

• (Subsection 4.4.5) Term Insurance

Under a term insurance policy payment is only made if the insuredperson dies before a fixed term (n years) ends. After the n years thepolicy pays no benefits (has no value).

(a) Continuous Immediate Benefit Case: Here the benefit of \$1 ispaid immediately upon death of the policy holder. So the presentvalue of the benefit is:

Z ={

e−δTx if Tx ≤ n0 if Tx > n

The expected present value of the benefit is

4-14

• The second moment of the random present value is

2A1x :n| =

∫ n0

e−2δt tpxµx+tdt .

which results from doubling the force of interest. Its variance is then

Var [Z ] = 2A1x :n| −

(A

1x :n|)2.

(b) Discrete Annual (mthly ) Benefit Case:When the benefit is paid at the end of the year in which deathoccurs, the expected present value of the benefit is

and when it is paid at the end of the ( 1m )th portion of the year in

which death occurs it is

A(m)1x :n| =nm−1∑k=0

k+1m)

km

∣∣∣ 1m

qx .

4-15

• Example 4-6 : Suppose that beyond age x there is a constant forceof mortality µx and a constant force of interest δ. Find A

1x :n|.

4-16

• (Subsection 4.4.6) Pure Endowment

A pure endowment benefit pays \$1 at the end of the policy term of nyears if the insured person of age x survives the policy term of nyears.

The present value of the benefit is the random variable

Z ={νn ≡ e−δn if Tx ≥ n0 if Tx < n

The expected present value of the benefit is

There is no difference between continuous and discrete random lifelengths in this setting. Pure endowment is a tool for describing morecomplex insurance instruments and is not sold as a separate policy.

4-17

• The symbols

are used interchangeably. We see this as a discounted survivalfunction.

4-18

• Example 4-7 : Suppose Tx is de Moivre (0, ω − x) and the interestrate is constant, find A

1x :n|.

4-19

• (Subsection 4.4.7) Endowment Insurance

Endowment insurance pays a benefit of \$1 if the insured person ofage x dies before the end of the policy term of n years or it pays \$1at the end of the term if that person survives the term of n years.

The present value of the benefit is the random variable

Z ={νTx ≡ e−δTx if Tx < nνn ≡ e−δn if Tx ≥ n

= νmin(Tx ,n).

The expected present value of a continuous benefit paidimmediately at death is therefore

4-20

• Its variance is:

Var [Z ] = 2Ax :n| −(

Ax :n|)2.

where 2Ax :n| is computed in the same manner as Ax :n| only with adoubled force of interest, i.e. δ is replaced by 2δ and ν is replaced byν2.

The annual and mthly discrete cases are completely analogous,producing EPV’s of:

Ax :n| = A1x :n| + A

1x :n| annual

and

A(m)x :n| = A(m) 1

x :n| + A1

x :n|. mthly

4-21

• Note, for example that in the annual case,

Z ={νKx+1 if Tx < nνn if Tx ≥ n

= νmin(Kx+1,n)

and so

4-22

• Example 4-8 : You are given A1x :n| = .4275, δ = .055 and

µx+t = .045 for all t . Find Ax :n|.

4-23

• (Bonus Subsection 4.4.7.5) Special Continuous Models

(a) Constant Force of Mortality and Constant Force of Interest:Consider a policy that pays \$1 immediately at the death of theinsured. Assume also that there is a constant force of interest δ anda constant force of mortality µx that describes survival beyond agex . Now we recall that

tpx = e−µx t and thus fx (t) = µxe−µx t for t > 0

For whole life it follows that

Ax =∫ ∞

0e−δ t µxe−µx tdt = µx

∫ ∞0

e−(δ+µx ) tdt

=−µx

(δ + µx )e−(δ+µx ) t

∣∣∣∞0

which produces the formula

4-24

• It quickly follows that

E[

Z 2]

= 2Ax =µx

(2δ + µx )and

Var [Z ] =µx

(2δ + µx )−( µx

(δ + µx )

)2Example 4-9 Same Setting: For term insurance, find A

1x :n| and

Var [ Z ] .

4-25

• Under these assumptions, the pure endowment term is:

nEx = e−δn npx or

Of course, endowment insurance of \$1 paid immediately at death orthe end of the term (whichever comes first) has an EPV of

Ax :n| = A1x :n| + nEx .

4-26

• (b) Uniform Distribution of Deaths and Constant Force of Interest:Consider a policy that pays \$1 immediately at the death of theinsured. Assume also that there is a constant force of interest δ anda uniform (de Moivre) distribution of deaths beyond age x . Now werecall that

tpx =ω − x − tω − x

and fx (t) =1

(ω − x)both for 0 < t < ω − x .

For whole life it follows that

Ax = E[e−δ Tx

]=

∫ ∞0

e−δ t fx (t)dt

=

∫ ω−x0

1(ω − x)

e−δtdt =−1

δ(ω − x)(e−δt

)∣∣∣ω−x0

or

4-27

• Moreover, under these assumptions, the term insurance EPV forn < (ω − x) is

It is the same integral as above only with n as the upper bound.

Under these assumptions, the pure endowment term is:

nEx = e−δn npx or

4-28

• (Subsection 4.4.8) Deferred Insurance Benefits

Sometimes a policy does not begin to offer death benefits until theend of a deferral period of u > 0 years.

Consider a term life insurance policy that does not begin insurancecoverage until time x + u and ends insurance coverage at x + u + n.The present value of a \$1 benefit paid immediately at death is:

Z ={

0 if Tx < u or Tx > u + ne−δTx if u < Tx ≤ u + n

and the expected present value is:

Changing the variable of integration from t to s = t − u in thisintegral produces,

4-29

• u

∣∣∣A 1x :n| = ∫ n0

e−δ(s+u)s+upx µx+s+uds

= e−δu∫ n

0e−δsupx spx+u µx+s+uds

= e−δuupx∫ n

0e−δsspx+u µx+s+uds, that is,

The nature of the integral boundaries in the original description ofthis EPV yields

as an additional way to express the EPV of a deferred benefit interms of EPV’s of those that are not deferred.

4-30

• Deferred policies are useful in describing or decomposing otherpolicies. For example, a 10 year term policy has EPV

A1x :10| = A

1x :6| + 6

∣∣∣A 1x :4|.That is, it is equivalent to a 6 year term policy plus a 6-year deferred4-year term policy. A whole life policy is equivalent to a n-year termpolicy plus a n-year deferred policy. So its EPV will satisfy

Ax = A 1x :n| + n∣∣∣Ax or Ax = A 1x :n| + n∣∣∣Ax .

This implies, for example,

Therefore we can compute EPV’s of term insurance policies from theEPV’s of whole life policies.

4-31

• Example 4-10: A 3-year deferred whole life policy pays 1 at themoment of death. You are also given thatµt = .01 for 0 ≤ t ≤ 2 and = .02 for t > 2. We also haveδt = .05 for 0 ≤ t ≤ 2 and = .06 for t > 2. Find the actuarial presentvalue of this insurance.

4-32

• Section 4.5 - Relationships among Ax , Ax and A(m)x

As the number of potential benefit payment points per year, m,increases death benefits are paid sooner (or possibly at the sametime point). Since the payment amount is the same, discounting overa shorter period produces a larger expected present value.

4-33

• Therefore,

Ax < A(2)x < A

(4)x

and, in general, A(m)x is an increasing function of m.

It follows that when m > 1,

But the differences between these values are relatively small andcan be approximated.

4-34

• (a) Approximations assuming UDD (uniform distribution of deaths)within each year

Let K (m)x =( # of sub-periods survived)

m and examine

A(m)x = E[νK

(m)x +

1m

]=∞∑

k=0

E

[νK

(m)x +

1m

∣∣∣∣∣ death in[x + k , x + k + 1)]

P

[death in

[x + k , x + k + 1)

]

Before continuing, we examine the conditional expected value on theright.

4-35

• E

[νK

(m)x +

1m

∣∣∣∣∣ death in[x + k , x + k + 1)]

=

(1m

)νk+

1m +

(1m

)νk+

2m + · · ·+

(1m

)νk+

mm

=

(1m

)νk(ν

1m + ν

2m + · · ·+ ν

mm

)

= νk+1

((1−ν)ν

m(ν−1m − 1)

)= νk+1

(i

i(m)

)where the last step follows because (1− ν)/ν = i andi(m) = m

((1 + i)

1m − 1

).

4-36

• Substituting this in the term at the bottom of the previous pageshows that under the UDD assumption

A(m)x =∞∑

k=0kpx qx+k νk+1

(i

i(m)

)

=

(i

i(m)

) ∞∑k=0

νk+1 kpx qx+k so

Taking the limit as m→∞ shows that under this UDD assumption

4-37

• So motivated by the UDD assumption, we have the followingapproximations;

A(m)x ≈(

ii(m)

)Ax

Ax ≈( iδ

)Ax and

Ax :n| ≈( iδ

)A1x :n| + nEx .

4-38

• (b) A claims acceleration approximation

This method is motivated by a UDD assumption, because under aUDD assumption a death is equally likely to occur in any one of them time segments of the year. In this case the average claimpayment time during a given year is

1m

( 1m

)+

1m

( 2m

)+ · · ·+ 1

m

(mm

)=

1m2

m∑j=1

j =m + 1

2m.

The approximation to A(m)x is made by making any payments duringthe year at this average time within each year, producing

A(m)x ≈ νm+12m qx + ν1+

m+12m 1

∣∣qx + ν2+m+12m 2∣∣qx + · · ·= ν

m+12m −1

∞∑k=0

νk+1 k∣∣qx

= (1 + i)m−12m

∞∑k=0

νk+1 k∣∣qx

4-39

• Typically, we expect claims to increase over time. So when using theaverage time within the year, we expect to move more claims back intime than forward, thus accelerating claims (making them earlierthan anticipated). The above computation motivates

Letting m→∞ produces

Likewise,

Ax :n| ≈ (1 + i)12 A1x :n| + nEx .

4-40

• Example 4-11: Consider the following: (1) Ax with CFM µ∗ and CFOIδ, (2) A

′x with CFM µ∗ + c and CFOI δ, and (3) A

′′x with CFM µ∗ and

CFOI δ + c. Assume c > 0. Find the ordering among these 3 EPV’s.

4-41

• Section 4.6 - Variable Benefits

When life insurance benefits are solely a function H(Tx ) of the futurelife length of the policyholder, then

and the variance of the present value is

Var(PV ) = E[ν2Tx H2(Tx )

]−(EPVH

)2.

Note that the second moment does not just require doubling theforce of interest, the benefit amount must also be squared.

4-42

• When the benefit function is linear, e.g. H(t) = a + b t with a and bconstants, then

EPVH = a E[νTx]

+ b E[Tx νTx

]In the continuous case with the benefit being paid immediately upon

death, the notation is

in a whole life insurance setting and

(̄IĀ)1

x :n| ≡∫ n

0t e−δ t tpx µx+tdt

in a term insurance setting.

4-43

• In a discrete setting with payments at the end of the year of death,the present value random variable is

Z = νKx+1 (Kx + 1)

where Kx = bTxc.

Thus the EPV of whole life insurance, for example, is

(IA)

x =∞∑

k=0

νk+1 (k + 1) k∣∣qx

and for term life insurance it is

with k∣∣qx = kpx qx+k = lx+k−lx+k+1lx coming from a life table.

4-44

• There are some problem settings in which the life insurance benefitsincrease exponentially over time, i.e. they are multiplied by (1 + j)Tx .The present value of the benefit is then

Z = νTx (1 + j)Tx =(1 + j

1 + i

)Tx≡ νTx∗

where

ν∗ =1

1 + i∗=(1 + j

1 + i

), that is

So the EPV’s

Ax :i∗ A1x :n|i∗ Ax :i∗ and A

1x :n|i∗

for the continuous and discrete cases are computed as before, onlywith a new interest rate i∗.

4-45

• Example 4-12: Consider a whole life policy payable immediately atdeath, with constant force of mortality of µx , constant force ofinterest δ, and variable payment of eα t where µx , δ and α are allpositive constants with α < µx + δ. Find the EPV.

4-46

• Example 4-13: Consider a whole life policy payable immediately atdeath, with constant force of mortality of µx = .03 for all t > 0.Assume δt = .01 for 0 ≤ t < 65 and = .02 for t ≥ 65. In addition thebenefit payment is bt = 200 for 0 ≤ t < 65 and = 100 for t ≥ 65.Find the EPV.

4-47

• Supplement to Chapter 4

Pure Endowment Revisited

The pure endowment term

assumes compound interest using a constant force of interest. Here

νn =( 1

1 + i

)n= e−δ n

with δ = ln(1 + i).

Note there are no assumptions made here about survival. It onlyassumes use of the survival function tpx . The term nEx representsdiscounting over a survived term of n years.

4-48

• Moving the Vision Point When Computing an EPV

Sometimes potential insurance payments (benefits) don’t begin rightaway. Pictured below is a 20-year term life insurance policy paying\$1 at the end of the year of death, but it is deferred 10 years, i.e.there are no benefits paid if death occurs during the first 10 years forthe policyholder who buys the policy at time t = 0 when this personis age x .

4-49

• Earlier we described the EPV of this deferred life insurance as:

But we now approach this computation in a different manner.Suppose we view the EPV of this policy from the vision point oft = 10.

4-50

• If the policy had been purchased at t = 10 the EPV would be

A 1x+10:20|.

But, of course, it was purchased at t = 0. We must discount the EPVback to t = 0 via

Here the discounting process must account for more than just theinterest rate. In order for any benefits to be paid, the policyholdermust first survive from age x to age x + 10. Thus we must discountover a survived period of time, using the survival discount factor of10Ex . Therefore, in general

4-51

• This process of discounting over a survived period, enables us tocompute EPV’s by breaking the time line (or potential benefitpayments) into disjoint insurance benefit segments and thendiscounting each segment to time t = 0, taking the necessarysurvival into account.

We refer to

In an expected life insurance benefit computation, if we were tomove the vision point into the future from, for example, t = 0 to t = uwhere u > 0, we would divide by the survival discount factor, in thiscase uEx+0, provided there were no potential benefit payments inthe interval [0,u] .

4-52

• Example 4-14: Consider a 8-year deferred 40-year term lifeinsurance policy that pays 100 during the first 20 years and 200during the second 20 years of insurance coverage with all paymentsat the end of the year of death. Find expressions for the purchaseprice (premium) of this policy at t = 0 and also the price if it waspurchased at t = 5.

4-53

• Putting the Pieces Together

Knowing the formulas for the exponential and de Moivre distributionsand knowing how to discount while taking survival into accountenables us to quickly solve complex problems by segmenting theminto simpler settings in which both can be applied.

Example 4-15: A policyholder age x = 40 begins an endowmentpolicy that pays 200 at the moment of death or at age 65, whichevercomes first. Assume a de Moivre (0,50) distribution for future lifelength and δ = .05. Find the EPV and Var[ Z ].

4-54

• Example 4-16: A whole life policy pays 100 immediately at death.We are given δt = .04 for 0 ≤ t ≤ 10 and = .05 for t > 10 plusµx+t = .06 for 0 ≤ t ≤ 10 and = .07 for t > 10. Calculate thepremium for this insurance.

4-55

• Example 4-17 (Example 4-10 revisited): A 3-year deferred whole lifepolicy pays 1 at the moment of death. You are also given thatµt = .01 for 0 ≤ t ≤ 2 and = .02 for t > 2. We also haveδt = .05 for 0 ≤ t ≤ 2 and = .06 for t > 2. Find the actuarial presentvalue of this insurance.

4-56

• Example 4-18: A whole life policy pays 1 immediately at death forthe first 10 years and .5 after that. Given a constant force ofmortality of µ∗ and a constant force of interest of µ∗, plusEPV ≡ E[ Z ] = .3324, find Var[ Z ].

4-57