16
Simplifications under level benefit contracts
The calculation of variances and other distributional features simplifies considerably when we have level benefits and constant interest. In fact, we can write down formulas for exact distributions of the major random variables of interest. Throughout this chapter, we consider the following setup. We have a general failure time T. We will consider insurances paying a level amount upon failure, either at the end of the year of failure or at the moment of such, and we will consider annuities paid prior to the failure of T with either a level payment or continuous payments at a level rate. In addition, we assume a constant force of interest δ. By taking T to be T(x), this will apply to level benefit, whole-life insurances and to level benefit whole-life annuities. By taking T = min{T(x), n}, this will apply to level benefit n-year endowment policies and to level benefit n-year temporary life annuities. Our assumption does not apply to term insurance, even when there is a level benefit during the term, since the benefit drops to zero after the expiration of the contract. However, in Section 16.5 we do illustrate that the calculation of exact distributions is possible for term or deferred insurances with a level death benefit paid over the benefit period. It is convenient to begin with a continuous failure time T. Consider an insurance policy paying 1 at the moment of failure. The discount function is given by v(t) = vt = e− δt. Let Consider an annuity with continuous payments at the rate of 1 per year, made prior to the occurrence of failure. If For the case where T = min{T(x), n}, we have already seen the first part of (16.16). This was the continuous ‘endowment identity’ given at the end of Section 8.8. Consider a contract that pays 1 unit at the moment of failure and has continuous level premiums at an annual rate of π payable prior to failure. (As a practical matter, this means that we are dealing with a net premium model that ignores expenses, which are unlikely to be level.) The prospective loss at time t is
where This greatly simplifies the calculation of the variance of the prospective loss. Compare the following with (14.14).
Suppose that π is an equivalence principle premium. Then Taking t = 0 in (16.16),
showing that
This reflects the fact that there is more risk involved in selling an insurance contract where premiums are payable over the entire life of the contract as opposed to the single-premium case. For failure occurring early, the insurer not only loses interest but will have collected relatively small amounts in premiums. For a final formula, express If π is an equivalence principle premium, we can substitute from (16.7) and take expectations to give a simple formula for the reserve at time t.
(For T(x), we obtained the discrete version of this in (6.6).) We now consider the case where failure benefits are paid at the end of the year of failure, and annuity benefits and premiums are payable yearly. All of the formulas in Section 16.2 have discrete counterparts, which for the most part are obtained by replacing T by If Z denotes the present value of the benefits for an insurance paying 1 unit at the end of the year of failure,
If Y denotes the present value of an annuity paying 1 unit yearly provided that failure has not occurred,
Now consider a contract which pays 1 unit at the end of the year of failure and has level annual premiums of π payable prior to failure. Then for any positive integer k,
Suppose that π is an equivalence principle premium. Then
One could possibly consider expenses when using the above simplified prospective loss formulas if the difference is only in the first year, as the following example shows. Example 16.1 An insurance policy provides 1000 at the end of the year of death with level premiums payable for life. There are expenses in the first year of 60% of the premium plus 50, and in the subsequent years of 15% of the premium plus 20. In addition, there is a death benefit settlement expense of 50. You are given that E(Z) = 0.4 and Var(Z) = 0.10 where Z is the present value of 1 unit of death benefit. The rate of discount is a constant 0.06. Find Var (L) where L is calculated using expense-augmented premiums, and including all expenses. Solution. We first compute the expense-augmented premium G as
so that
We know that Now consider a policy where the total inflow was 48.64 in every year. The value of L from that contract would differ from that on the one in question only by a constant amount, namely the extra amount in the first year due to higher expenses. Therefore, the variance of L would be the same, and we can use formula (16.6′), with π = 48.64, and the 1 replaced by the death benefit of 1050. From (16.6′),
In this section, we calculate the exact distributions for Of course, when X is continuous, The distribution functions we want are all easily expressed in terms of the distribution of T. Let N denote the least upper bound of the values of T. In the case that N = ∞ (as for example when T is exponential), the term vN in the formulas below will be equal to 0. The distribution of First note that arguing as in (16.10) gives
Using (16.2)
Substituting from above,
The minimum value of L will be
occurring for failure at time N, and its maximum value will be 1, occurring for failure at time 0. Using (16.16),
For the more general case, In the particular case where T is exponential with constant hazard function μ, we know that It is of interest to observe that if μ = δ in the exponential case, the exponent in the formulas above equals 1 so that Example 16.2 A company decides to add 20% to its equivalence principle premiums as a protection against unfavourable experience. In each of the following cases, find the probability that premiums will cover claims. Suppose that T is exponential with μ = 0.04 and that δ = 0.06. Solution. (a) The equivalence principle premium is 1/(μ + δ) = 10. The actual premium charged will be 12. From (16.16),
(b) The equivalence principle annual premium rate is just μ = 0.04, so the actual premium rate charged is 0.048. The probability that premiums cover claims is just
It is also possible to obtain exact distributions in some simple cases involving non-level benefits such as term or deferred insurance. Consider a contract that pays 1 at failure, provided failure occurs within n years. In order to handle failure times that are not continuous, we adopt the convention that a benefit is paid for failure at exact time n. We will compute the distribution of Example 16.3 Redo part (b) of Example 16.2, but now assuming n-year term insurance for n = 15 and 10. Level premiums are payable continuously for n years. Solution. The equivalence principle premium rate is still μ = 0.04 so the actual premium rate charged is still 0.048 and π/δ = 0.8. For the contract in Example 16.2, the value of L when T = n is 1.8e− 0.06n − 0.8. For n = 15, this is negative. This means that L becomes negative at some point prior to expiration of the term contract, so the probability that premiums cover claims is 0.58, exactly the same as it was in constant benefit case. When n = 10, the value of L for the contract in Example 16.1 is positive. Therefore, in order that L in this example be negative, it is necessary that T ⩾ 10, so that no benefits are paid. The probability of this is e− 0.6 = 0.67. A similar example is provided by deferred insurance. Consider a contract that pays 1 at failure, provided failure occurs after time n. We now adopt the convention that nothing is paid for failure at exact time n. In view of this convention, there need not be a maximum value of Refer again to Figure 16.1. Since nothing is paid for death at or before time n, which occurs with probability FT(n), the probability mass to the right of e− δn is picked up and set down as a point mass at 0. Utilizing (16.16), we can write
We next investigate a more complicated case where we compute the exact distribution of L in an annual premium policy. Consider a failure time T that is unbounded (i.e. N = ∞). We consider insurances which have premiums payable continuously at a level annual rate π for the duration of the contract. We will compare the distribution of L for a contract that pays 1 unit on failure or at time n if earlier, and a contract that pays 1 unit at failure provided this occurs within n years. These then correspond respectively to endowment and term insurance. To simplify the notation, let
In both cases, the value of L is given by (16.5) (with t = 0), provided that T takes a value less than n, which corresponds to L taking a value greater than b0. So for any interval (a, b) with a > b0, we have that P(a < L ≤ b) is the same in both cases and this value can be calculated directly from (16.16). Consider the remaining probability of sT(n). For the endowment contract, this will all be concentrated at the single point b0. For the term contract, it will all be concentrated at the single point c0, which lies in the interval ( − π/δ, b0). As n increases to ∞ both b0 and c0 approach − π/δ and the distribution approaches the whole-life case as given in (16.16). The various density functions are compared in Figure 16.2. The probability mass to left of the point b0 on the whole-life graph is picked up and set down as a point mass at b0 for the endowment policy, or at c0 for the term policy. Figure 16.1 Graph of Figure 16.2 Graph of fL(u) for various types of insurance 16.1 For a certain failure time T, an insurance contract pays 1 at the moment of failure, and has level premiums payable continuously prior to failure at the annual rate of 0.06. You are given that
where 16.6 An insurance contract, based on the failure time T, pays 1 unit at the moment of failure provided this occurs within 5 years. Nothing is paid for failure after that time. The force of interest is a constant 0.1. If T has the hazard function find the probability that the present value of the benefits is strictly positive, but less than or equal to e− 0.3.16.1 Introduction
16.2 Variance calculations in the continuous case
16.2.1 Insurances
be the present value of the benefits. In this case there is little simplification, and we know from the previous chapter that
16.2.2 Annuities
is the present value of the benefits, then
16.2.3 Prospective losses
and 
. This leads to
16.2.4 Using equivalence principle premiums
, so that from (16.16),
in terms of 
in (16.16). Then
16.3 Variance calculations in the discrete case
, 
by A, a by 
and δ by d. We will leave the formal derivations to the reader, but will list the formulas with the corresponding equation numbers as in Section 16.2, only with a prime to denote the discrete case.
, and substituting in the above, G = 80.75. This means that the total inflow after the first year is 0.85(80.75) − 20 = 48.64.
16.4 Exact distributions
and L. In each case, we will derive distribution functions. These will be given for values between the greatest lower bound and least upper bound of the values. (We know that F takes the value 0 for arguments less than the greatest lower bound, and 1 for arguments greater than the least upper bound.) It is convenient here to introduce some new notation. For any random variable X, let
and 
.16.4.1 The distribution of
is given by
16.4.2 The distribution of
16.4.3 The distribution of L
is given by the same formula, but with 
replacing 
and vN − t replacing vN.16.4.4 The case where T is exponentially distributed
and N = ∞. The above formulas simplify to
and L are all uniform random variables.
16.5 Some non-level benefit examples
16.5.1 Term insurance
, the present value of the benefits. The minimum positive value of 
is e− δn occurring for death at time n. Since nothing is paid for death strictly after time n, which occurs with probability sT(n), we can easily obtain the term distribution from the whole life distribution. Pick up the probability mass to the left of z = e− δn and set it down as a point mass at the point 0, as illustrated in Figure 16.1. From this, utilizing formula (16.16), we can read off this distribution of 
16.5.2 Deferred insurance
but e− δn is certainly an upper bound, since any benefits will be paid at a time later than n.
16.5.3 An annual premium policy
for various types of insurance
Exercises
Type A exercises
is the present value of the failure benefits, and 
is the present value of an annuity contract with continuous payments at the annual rate of 1, payable prior to failure. Find Var(L).
and Var(L) = 2, find Var(
).Type B exercises
; (b) 
; (c) the net annual rate of premium payment when premiums are payable continuously prior to failure, for an insurance paying 1 at failure and (d) the reserve at time k for the contract in (c).
is the present value of the benefits, find the 80th percentile of 
. That is, find the point z such that 
.
when the contract is a term insurance policy that pays 1 unit if failure occurs in the first 5 years.
is the present value of the benefits, calculate 
for all nonnegative values of z.
is the present value of the benefits, find the probability that: (a) 
(b) 
and (c) 
.
