1. finance
2. personal finance
3. financial planning
4. retirement and pension

Actuarial Mathematics II Lecture Notes
DAW, expanded and revised from FC’s notes
April 27, 2015
1
Compound interest
1.1
Effective interest rates
• To specify an interest rate, one needs to define the basic time unit such as ‘year’ (e.g. annual
interest rate of 6%) and the conversion period at the end of which interest is paid (credited,
compounded).
• Effective Rate: this matches the time unit to the conversion period. So an interest rate of
10% per month over a conversion period of six months is not the effective rate.
• If the conversion period is equal to the basic time unit, then the interest rate is called effective,
e.g. 6% annual interest rate credited at the end of each year. Such rates are called Annual
Effective Rates: AER and denoted by i (e.g. i = 0.06).
• The point of an effective rate is to give us a baseline.
• We assume throughout that i > 0.
• Initial (time 0) amount F0 with AER i leads to Fn = (1 + i)n F0 after n years
• For example, 100 invested at AER 10% leads to F1 = (1.1)(100), F2 = (1.1)F1 = (1.1)2 (100),
and so forth.
• If you invest a further rk ≥ 0 at the end of year k = 1, . . . , n, then
Fk = (1 + i)Fk−1 + rk , k = 1, . . . , n
which leads to
n
X
Fn = (1 + i) F0 +
(1 + i)n−k rk .
n
(1)
k=1
• Some of this is best visualized as a timeline. (Inject rk at year k).
• (1 + i) is called the accumulation factor and (1 + i)k is called the accumulation value after
k years. That is, it is what you get back from initial investment F0 and no additional
investments.
1
• The equation may be rewritten as
Fn − F0 = (i)(
n
X
k=1
Fk−1 ) +
n
X
rk
k=1
which shows that the total increment in capital reflects an amount added because of interest
plus total deposits.
• The discount factor v =
amounts, as follows.
1
1+i
is used to determine the present value (value at time 0) of future
• Suppose that the future amount is C after k time units with effective rate i. Suppose that
A is its value at time 0. We must have C = (1 + i)k A, from which A = C/(1 + i)k = v k C.
v k C in cash is no better and no worse than investing v k C into the account.
• So, 121 in two years time at AER 10% is worth 121/(1.1)2 = 100 now.
• If we divide both sides of equation 1 by (1 + i)n , we have
v n Fn = F0 + Σnk=1 v k rk .
Note what this equation is telling us: the present value (v n Fn ) of the account is equal to
the present value of the initial capital plus the sum of the present values of every subsequent
addition.
1.2
Nominal interest rates
• If the basic time unit and the conversion period differ, the interest rate is called nominal.
• The relationship between nominal and effective rates is as follows. Divide the basic time unit
used for the effective rate by the number of conversion periods in the time units used for the
nominal rate. Let this be m. Use the notation i(m) for the nominal interest rate. Then the
effective rate is defined to be
i(m) m
i = (1 +
) − 1.
m
• For example, convert a nominal annual interest rate 0.06 with a conversion period of three
months into an annual effective rate.
• Basic effective time unit is one year. Nominal time unit is one year. m = 4 conversion
= 0.015 (i.e. 1.5%) is credited at the end of each
periods in the year. i(4) = 0.6. Thus, 0.06
4
quarter. The AER is (1 + 0.015)4 − 1 = 0.06136, or 6.136%. So initial capital F0 becomes
1.06136F0 after one year. So a nominal interest rate of 6% with a quarterly conversion period
is equivalent to AER of 6.136%.
• Note: the word ‘nominal’ is often deleted if it is clear what the main time unit is - usually
year. For the same reason ‘annual’ is often deleted.
• The relation can be reversed to report the nominal rate corresponding to an effective rate:
i(m) m
) =1+i
m
= m[(1 + i)1/m − 1]
(1 +
i(m)
2
• Alternative expressions for this are
1
i
(m)
= m(v
−1/m
− 1) = m
(1 − v m )
1
vm
.
• So, an AER of 10% and conversion every two months corresponds to m = 6 and the nominal
rate should be reported as = m[(1 + i)1/m − 1] = 9.61%. For AER 6% and m = 4 we have
nominal rate 5.87% similarly.
• See what happens as the number of conversion periods within the main time unit increases.
For i = 0.06,
m
1
2
3
4
6
12
52
365
∞
i(m)
0.06
0.05913
0.05884
0.05870
0.05855
0.05841
0.05830
0.05829
0.05827
• With i(m) = m[(1 + i)1/m − 1], what happens if m → ∞? (The limit exists because i(m) is
decreasing and non-negative.) Let
δ = lim i(m) .
m→∞
Now,
δ = lim i(m)
m→∞
1
= lim
m→∞
(1 + i) m − 1
1
m
x
(1 + i) − 1
x→0
x
f (x)
, with f (x) = (1 + i)x − 1, and g(x) = x
= lim
x→0 g(x)
f ′ (x)
= lim ′
by L’Hopital’s rule
x→0 g (x)
ln(1 + i)(1 + i)x
= lim
x→0
1
= ln(1 + i).
= lim
In using L’Hopital’s rule, we need to check that
lim f (x) = 0, lim g(x) = 0, g(x) 6= 0 for x 6= 0,
x→0
x→0
and that the limit exists. All these checks are satisfied.
• Note that there is a different proof for this limit in Gerber.
3
• δ = ln(1 + i) is called the force of interest equivalent to i.
• Note that the value of i corresponding to an infinite number of conversion periods, i.e.
(m)
continuous compounding, is ei − 1.
• Note that 1 + i = eδ and (1 + i)k = ekδ .
• This means that the accumulation factor and discount factors for k years with continuous
compounding (the limit case for m → ∞) are eδk = (1 + i)k and e−δk = v k , respectively.
1.3
Interest at start of period
• Interest can also be credited at the start of each conversion period; this is sometimes called
discount. Because the crediting of interest is immediate, there is also interest earned on the
interest, and so on. We denote by d the discount rate, i.e. the rate of interest-in-advance.
• The initial deposit F0 immediately becomes
2
F0 + dF0 + d F0 + . . . =
∞
X
F0 dj =
j=0
F0
,
1−d
provided that 0 < d < 1.
• Compare this to what happens to capital F0 invested with AER i. The return on capital is
(1 + i)F0 . Thus,
d
1
1
=i+1⇔i=
⇔d=
(i) = vi
1−d
1−d
1+i
so the discount rate is the discounted value of the interest rate.
• Note that d =
i
i+1
and i =
d
.
1−d
• The interpretation is that, per unit of capital invested, the interest payable at the end of the
year is the accumulated value of interest payable at the start.
• We may also consider a nominal rate d(m) where interest is applied at the start of each of m
conversion periods.
• As with interest paid in arrears, one can apply discount m times per year (or other unit of
time) and find the d(m) . We can find this as follows.
i
• As d = i+1
, we can match d(m) /m to the corresponding value of i(m) /m. These are the
interest rates paid per conversion period, paid at the start and in arrears respectively. Thus
d(m)
=
m
⇒
• Now it is easy to see that
d(m) =
i(m)
m
i(m)
+
m
(m)
1
i
i(m)
m
+1
.
lim d(m) = lim i(m) = δ.
m→∞
m→∞
This is logical because the difference between interest paid in advance and interest paid in
arrears disappears as the conversion period becomes very small.
4
• We can also show after some rearrangement that
d(m) = m[1 − (1 + i)−1/m ] = m(1 − v 1/m ).
1.4
Perpetuities and annuities
• Annuity: a sequence of regular payments from a fund which accrues interest. The payments
exceed the interest gained, so the fund decreases. Payments cease when the fund runs out.
• Perpetuity: Similar to an annuity, but the fund does not decrease (it may increase), so
payments continue indefinitely. Examples are Consols (consolidated stock), e.g. British
Government war bonds.
• The idea is this: we buy an annuity by investing a lump sum. Certain amounts are then paid
back to us at fixed intervals of time from the lump sum, which is also accruing interest. We
can decide (using the discount factor) the sum we must invest today to get a certain overall
amount back out.
• We are assuming equal payments. Formulae for unequal and irregular payments are more
complicated.
• The Present value of the annuity (or perpetuity) is the amount of money at time 0 which
has the same value (at time 0) as all future payments, so it would be the money needed in
the fund.
• We assume annual payments of one unit (unless stated differently).
• We may (have to) buy an annuity as part of a pension. This will pay out until death, which
means a random number of payments. We shall consider this later in the course.
• Although phrased in terms of an investment you make, with payments made to you, the
ideas are exactly the same for loans made to you, where you make equal repayments subject
to some level of interest.
• There are two principal types of annuity (perpetuity), varying in the moment of each payment:
– Paid at the start of a (unit) time period (‘Due’);
– Paid at the end of a time period (‘Immediate’).
Again, unit time (one period) is assumed to be year unless stated differently.
• We use the following notation, given in Gerber. A mnemonic is that d for dots matches d for
due.
Present value
a
¨n
an
a
¨∞
a∞
Contract
Payments Paid at
Annuity-due
n
Start of year
Immediate annuity
n
End of year
Perpetuity-due
∞
Start of year
Immediate perpetuity
∞
End of year
5
• Recall that the present value of a payment of size 1 paid out k years into the future is v k ,
where v = 1/(1 + i).
• For the Immediate Annuity, the first payment is at the end of the year and will be worth v.
The second payment is at the end of the second year and is worth v 2 now. For the AnnuityDue, the first payment is made at time zero, with value 1. The second is made at the start
of the next year and is worth v. Thus,
v(1 − v n )
1−v
1 − vn
=
1−v
an = v + v 2 + v 3 + . . . + v n =
a
¨n = 1 + v + v 2 + . . . + v n−1
and so a
¨n = van .
• As 0 < v < 1 we may easily derive expressions for the corresponding perpetuities:
1
v
=
1−v
i
1
1
= 1 + v + v2 + . . . =
=
1−v
d
a∞ = v + v 2 + v 3 + . . . =
a
¨∞
a∞ and a
¨ ∞ = 1 + a∞ .
• a∞ = v¨
• To correspond to equal present value, the n equal payments need to correspond to
unit invested.
1
an
per
n
)
• Example. Annuity of 6000, interest rate 4%, n = 10, v = 1/(1.04) = 0.9615. a10 = v(1−v
=
1−v
8.1109 is the present value of the 10 payments made. For each unit invested, the annual
payment needs to be a 1 = 1/8.1109 = 0.1233, and for 6000 invested 6000/8.1109 = 739.75
10
is the annual payment that corresponds to that value. Note that the present value of investing
this amount is (1 + .04)10 (6000) = 8881.47.
• Every payment includes a payment for interest and a part of the annuity invested.
• We may also express:
1 − vn
i
1+i
1 − vn
=
=
d
i
an = a ∞ − v n a ∞ =
a
¨n = a
¨∞ − v n a
¨∞
• An n year annuity is equal to the difference between a corresponding perpetuity starting now
and a perpetuity starting in n years time, so also the present values of these are equal.
1.5
Several conversion periods
• Suppose instead that m equal payments of 1/m are made during each year or time unit,
(m)
(m)
at equal intervals. Let an and a
¨n be the present values of immediate and due annuities
under these conditions.
6
(m)
(m)
• Exercise. Find a∞ and a
¨∞ . For the due perpetuity, the amounts paid are 1/m, 1/m, . . .,
with present values 1, v 1/m , v 2/m , . . . ,, so that present value is
(m)
a
¨∞ =
1
1
1
(1 + v 1/m + v 2/m + . . .) =
.
m
m (1 − v 1/m )
The present value for the immediate perpetuity is the same, minus 1/m (which is paid at
the start for the due perpetuity, but not the immediate perpetuity):
(m)
(m)
a∞ = a
¨∞ −
(m)
(m)
n
1
.
m
n
• Exercise. Show that an = 1−v
and a
¨n = 1−v
. Begin with the due annuity. This is paid
i(m)
d(m)
1
2
1
in nm installments of m beginning now, and with present value m1 , m1 v m , m1 v m , etc. Writing
1
r = v m we have
(m)
a
¨n
=
1 (1 − rnm )
1 (1 − v n )
(1 − v n )
1
(1 + r + r2 + . . . + rnm−1 ) =
=
,
=
m
m (1 − r)
m (1 − v m1 )
d(m)
1
(m)
using an earlier expression for d(m) , namely d(m) = m(1 − v m ). For an , note that
(m)
an
1
(m)
¨n
= vma
=
1
= vm
(1 − v n )
,
i(m)
1 (1 − v n )
m (1 − v m1 )
using an earlier definition for i(m) .
1.6
Final values
• The final value (accumulated value) of an annuity is also of interest. Notation is as for present
value but replacing a by s.
• Recall that the final value of investment F0 is Fn = (1 + i)n F0 after n time units. Thus final
values can be computed as
v(1 − v n )
(1 + i)n − 1
=
1−v
i
n
n
(1 + i) − 1
(1 − v )
=
,
¨n = (1 + i)n
s¨n = (1 + i)n a
1−v
d
sn = (1 + i)n an = (1 + i)n
after some rearrangement and by recalling that d = vi.
• These are the values of the annuities at time t = n.
• What about the final value of a perpetuity? This is infinity, of course.
7
1.7
Remaining annuity after a period
• Suppose that we want to calculate the value of an annuity after h < n years.
• The retrospective method calculates the value of the original annuity after h years, minus the
accumulated value of the h payments made. Suppose that the annuity is F and the annual
payment is x = F/an .
• The value of the annuity after h years is (1 + i)h F after h years.
• The present value (T0 ) of the payments is ah x =
years of the payments is
xv(1−v n )
.
1−v
sh x = (1 + i)h xah = (1 + i)h F
The accumulated value after h
ah
.
an
• Thus, the value of the annuity after h years is
(1 + i)h F − (1 + i)h xah = (1 + i)h F (1 −
ah
).
an
• There is an alternative simpler prospective method, which is to recognise that the balance
remaining in the annuity must equal the present value of the remaining payments. There are
n − h remaining payments of size x, with present value
a
an−h x = F n−h .
an
1.8
Extended examples
• Example: Consider an annuity with payments of 20,000 per year for ten years, with AER
4%.
– Find the present value assuming this is an immediate annuity
– Find the present value assuming this is an annuity-due
– Repeat 1 and 2, but for perpetuities rather than annuities
– Find the final value assuming this is an immediate annuity
– Find the final value assuming this is an annuity-due
– Explain the difference between these two final values
• We can work assuming an investment of one unit and then multiply up at the end. The
formulae we need are as follows, with results for 20000 units at the end of each row.
i = 0.04
v = 1/(1 + i)
d = i/(1 + i)
a
¨10 = (1 − v 10 )/(1 − v)
a10 = v ∗ a
¨10
s10 = ((1 + i)10 − 1)/i
s¨10 = ((1 + i)10 − 1)/d
a∞ = 1/i
a
¨∞ = 1/d
= 0.9615
= 0.0385
= 8.4353,
= 8.1109,
= 12.0061,
= 12.4864,
= 25.0
= 26.0
8
⇒ 168706.6
⇒ 162217.9
⇒ 240122.1
⇒ 249727.0
⇒ 500000.1
⇒ 520000.0
• For the last part of the exercise, note that on the unit scale,
s¨10 − s10 =
10
X
k=1
k=0
X
(1 + i) −
(1 + i)k = (1 + i)10 − 1 = 0.4802.
k
9
Therefore the difference between the final values is 20000 × (1.0410 − 1) =
$9 604.89. This is the difference in present values between the final immediate annuity
payment and the initial annuity-due payment.
• To illustrate calculations for annuity remaining, imagine an annuity of 8000 with 10 payments
and AER 5%. i = 0.05, v = 0.9524, and find the value of the annuity after 6 payments.
– The annual repayments are 8000/a10 .
– For the retrospective method, we compute a10 =
repayments are 8000/7.7217 = 1036.04.
v(1−v ( 10))
1−v
= 7.7217. So, the annual
– The present value of these payments after 6 payments is
a6 (1036.04) = (5.0757)(1036.4) = 5258.62.
The accumulated value of these 6 payments is
s6 (1036.04) = (1 + i)6 a6 (1036.04) = (1.05)6 (5258.62) = 7047.05.
– In contrast, the value of the annuity after 6 years is (1 + i)6 (8000) = 10720.77.
– The difference between these is 10720.77 − 7047.05 = 3673.72.
• For the prospective method, x = 1036.04 and a4 = 3.5460, so the final value must be
1036.04 × 3.5460 = 3673.75. Slight difference reflects rounding error.
1.9
Repayment of a debt
• There are many possible ways of repaying debt. One can pay off debt in equal installments,
or one may pay off interest on the loan only, and repay the loan in full at the end of the
period.
• Denote the value of a debt at time 0 by S. The debt will be repaid in installments r1 , . . . , rn
at the end of years 1, . . . , n.
n
X
S=
v j rj
j=1
This is the present value of the first repayment, plus the present value of the second, etc.
• Exercise: Suppose all payments are equal, so rk = r for all k; derive r. We have
S=r
n
X
j=1
v j = rv(1 − v n )/(1 − v) = ran ,
so that r = S/an .
9
• Let Sk be the debt remaining immediately after rk has been paid, it is known as the principal
outstanding.
Sk = (1 + i)Sk−1 − rk for k = 1, . . . , n
with S0 = S
rk = iSk−1 + (Sk−1 − Sk )
• So the k-th payment is equal to the interest on the principal outstanding minus the reduction
in the principal outstanding.
• Exercise: Check that the results for repayment of a debt are similar to those for compound
interest (Section 1.1), effectively by replacing Fk by −Sk .
10
2
Lifetime Models
2.1
Introduction
• Let random quantity X denote the age of a person at death (in years).
• A person of age x, also called ‘a life aged x’ and denoted by (x), has random future lifetime
T (x) (sometimes denoted by T when it is clear to which (x) we refer).
• So given that X ≥ x:
X = x + T (x)
• T (x) is a random quantity for which we assume a probability distribution with cumulative
distribution function (CDF)
Gx (t) = Px (T (x) ≤ t) = P (T (x) ≤ t|X ≥ x)
• Assuming that this distribution is absolutely continuous, its probability density function
(PDF) gx (t) = G′x (t) exists.
• This can be interpreted as (see 1H Probability)
gx (t) · δ ≈ Px (t ≤ T (x) ≤ t + δ)
for small δ.
• So the probability of the event that someone aged x will die at an age between x + t and
x + t + δ is approximately gx (t) · δ if δ is small.
• Actuarial notation (which is unpleasant, but we will use it a lot):
t qx
= Gx (t) = P (T ≤ t|X ≥ x)
t px = 1 − Gx (t) = P (T > t|X ≥ x)
for t > 0. These are respectively, the probability of being dead and being alive at time t,
given life at time x.
• t qx is the probability that someone aged x is dead by age x + t.
• t px is the probability that someone aged x is still alive at age x + t.
• Clearly t qx + t qx = 1.
• For future intervals, define for s ≥ 0 and t ≥ 0:
s|t qx
= Px (s < T ≤ s + t|X ≥ x) = Gx (s + t) − Gx (s) = s+t qx − s qx .
This is the probability that death occurs between ages x + s and age x + s + t, given being
alive at x.
• This can be written in terms of life probabilities:
s|t qx
= s px − s+t px = G(s + t) − G(s)
11
• Note that the vertical slash in the notation
(unfortunately).
s|t qx
does not mean the conditioning symbol
• Remember that (T ≤ t) = P (T < t) is true in the continuous case but not in the discrete
case: Be careful as lifetime can be considered as both a continuous or a discrete quantity.
• Now consider the conditional probability that a life (x) survives a further t > 0 years given
it survives another s years. This is
t px+s
=P (X > x + s + t|X > x + s, X > x)
=P (T > s + t|T > s)
P (T > s + t)
P (T > s + t ∩ T > s)
=
=
=
P (T > s)
P (T > s)
1 − G(s + t)
=
1 − G(s)
s+t px
s px
All these probabilities are conditional on X > x.
• Question: Is
10 p40+10
=
10 p50 ?
• For lifetime models that are defined for x = 0, notice that
s+t ps
= t ps × s p0 .
• The conditional probability that a life (x) does not survive more than t ≥ 0 more years,
given it survives another s ≥ 0 years is:
t qx+s
= 1 − t px+s
s+t px
=1−
s px
s px − s+t px
=
s px
s|t qx
=
s px
G(s + t) − G(s)
.
=
1 − G(s)
• 1 qx , 1 px , and s|1 qx are also written omitting the one, i.e. as qx , px , and s| qx . These are
respectively, the chance of death within one year, the chance of being alive at the end of the
first year, and the probability of surviving s years and then dying within one year.
• For integer k
P (T > k) =P (T > k|T > k − 1)P (T > k − 1)
=P (T > k|T > k − 1)P (T > k − 1|T > k − 2)
. . . P (T > 2|T > 1)P (T > 1)
We identify P (T > k|T > k − 1) with P (T (x + k − 1) > 1), from which we obtain
P (T > k) =
k−1
Y
P (T (x + j) > 1)
j=0
12
and thence
k px =
k−1
Y
1 px+j =
j=0
k−1
Y
px+j
j=0
• The expected value of the remaining lifetime T (x) is
Z
Z ∞
Z ∞
[1 − Gx (t)]dt =
tgx (t)dt =
˚
ex = E(T (x)) =
0
0
∞
t px dt.
0
This result needs a bit of work to obtain. For a continuous nonnegative random variable,
Z ∞
[1 − G(x)]dx,
E(X) =
0
where G(x) is the cdf. The statement is not generally true for random variables which can
take negative values.
2.2
Force of mortality
• The pdf gx (t) can be difficult to interpret. It is the rate of death at x + t of lives aged x, but
not conditioned on being alive up until x + t!
• A more natural concept is the Force of mortality of (x) at age x + t is defined to be
µx+t =
gx (t)
d
d
= − ln[1 − Gx (t)] = − ln(t px )
1 − Gx (t)
dt
dt
Here we use that for function f (t) we have
d
ln[1 − f (t)] = −f ′ (t)/(1 − f (t)),
dt
or equivalently that
f ′ (t)
f (t)
=
d
dt
ln(f (t)).
• The Force of mortality is also known as the hazard rate or the failure rate.
• We can interpret this as the instantaneous rate of mortality conditional on having reached
that age. To see this, note that for small δ > 0
δ · µx+t =
P (t ≤ T < t + δ)
δgx (t)
≈
= P (T < t + δ|T ≥ t)
1 − Gx (t)
P (T ≥ t)
(remembering that g(t) × δ ≈ P (t < T < t + δ)).
• We can write many of our relations in terms of the force of mortality.
gx (t) = (1 − Gx (t))µx+t = t px µx+t
• The expected remaining lifetime may be written
Z ∞
˚
ex =
t · t px µx+t dt
0
13
• We showed above that
µx+t = −
d
ln(t px ),
dt
from which it follows that
t px
= e−
Rt
0
µx+s ds
.
This needs the fundamental theorem of calculus, summarised as F (x) =
Rx
0
f (x)dx.
• If µx+t is reasonably smooth, we have δ qx+t ≈ δµx+t for small δ > 0. (Note that the δ qx+t
expression has δ in the prefix.)
• This follows because the probability of death in a small interval is expressible as
δg(t) ≈ P (t ≤ T ≤ t + δ) = G(t + δ) − G(t)
for small δ, allowing us to express
g(t) ≈=
Recall next that
δ qx+t
=
G(t + δ) − G(t)
.
δ
G(t + δ) − G(t)
,
1 − G(t)
and we obtain the result after some manipulation and using the basic definition
µx+t =
2.3
g(t)
.
1 − G(t)
Central rate of death
• Recall that 1 qx is the probability a life now aged x dies within the next year, sometimes
called the initial rate of death at exact age x.
• The central death rate is defined to be
mx = R 1
1 qx
p dt
0 1 x
.
• This can be interpreted as the probability that a life aged anywhere between x and x + 1
dies before attaining age x + 1.
• Historically it was easier to actuaries to estimate mx from data than to estimate qx or µx+t .
• See Exercises 5 and 16 in section D.2.1 and refer back to page 22 of Gerber.
2.4
Basic parametric models
• Everything depends on our choice of probability distribution for lifetime remaining, Gx (t).
Yet this is unknowable. There have been many suggestions which are intuitive or mathematically attractive.
14
• De Moivre (1724):
gx (t) =
(
1
ω−x
0
0≤t≤ω−x
otherwise
for some choice ω. Note that the range can be rewritten as x ≤ x + t ≤ ω, which makes
plain that ω is the maximum possible age. Thus, the remaining lifespan of (x) is uniformly
distributed on [0, ω − x]. The force of mortality (check as an exercise) is
µx+t =
1
, 0≤t≤ω−x
ω−x−t
so µx+t → ∞ as (x + t) → ω.
• To prove this, we use µx+t = g(t)/(1 − G(t)). Here,
Z t
t
1
dt =
, 0 ≤ t < ω − x,
G(t) =
ω−x
o ω−x
after which the result follows.
• Gompertz (1824):
µx+t = Bcx+t
Two parameters (B, c), no maximum age assumed. The force of mortality grows exponentially
if c > 1 which reflects ageing.
• Makeham (1860):
µx+t = A + Bcx+t
Parameter A represents a constant risk of death. Bcx+t represents the effect of aging.
• Problems class question. Consider the Makeham lifetime model.
1. Find t px .
2. For which choices of parameters does this model correspond to memorylessness?
Solution. We can show that t px = exp(−At − mcx (ct − 1)), where m =
1. We have
t px
= e−
Rt
0
µx+s ds
B
.
ln c
,
so we need to evaluate
Z
0
t
(A + Bc
x+s
t
cs
)
)ds = As + Bc (
ln s 0
Bcx t
= At +
(c − 1).
ln c
x
For this, remember that the derivative of cf (x) is cf (x) f ′ (x) ln c. Here f ′ (x) = 1 is a
constant, so we obtain the integral just by dividing through by ln (c). The result now
follows.
15
2. Recall the memoryless property of exponential distributions from 1H Probability. Memorylessness is the property that P (X > n|X > m) = P (X > n), i.e. does not depend on
m. The exponential distribution is the only memoryless continuous probability distribution, and memorylessness characterizes it. (The only memoryless discrete distributions
are the geometric distributions.)
We obtain an exponential distribution if B = 0 or c = 1. When B = 0 the result is
t
) = t by L’Hopital’s rule, and the result then
obvious. For c = 1, note that limc→1 ( c −1
c
follows.
3. In these cases µx+t is constant and
t px
= e−(A+B)t = P (T > t),
so that P (T ≤ t) = 1 − e−(A+B)t , i.e. T (x) is exponentially distributed and does not
depend on current age x. That is, the probability of dying during a given period of time
is independent of the current age.
• Weibull (1939):
µx+t = k(x + t)n
Two parameters (k, n). This model is very popular in reliability theory.
k
• Exercise: Prove
exp(− n+1
[(x + t)n+1 − xn+1 ]). This just requires simple integration,
R t px =
noting that (x + h)n dx = (x + h)n+1 /n.
• Statistical methods can be used to fit and check models. For example, the parameters may
be estimated using historical data and using least-squares or maximum likelihood methods.
This is not discussed further in this module.
• Note that such parametric models do not easily allow for covariates such as medical history
and sex of the person.
2.5
Curtate future lifetime
• In insurance, lifetimes are often considered in years rather than smaller fractions.
• The Curtate future lifetime of (x) is the integer part of T (x), denoted by K(x) = [T (x)],
usually dropping notatiopn to just K. For k = 0, 1, . . .:
P (K = k) = P (k ≤ T < k + 1)
= G(k + 1) − G(k)
= k|1 qx
= k px · 1 qx+k
= k px · qx+k .
Here we are recalling one of our previous formulae, namely s|t qx = t qx+s × s px , and remembering that we drop the 1 preceding q and p by convention. We also remember here that G
is continuous, otherwise the result would not hold.
16
• The random quantity K has expectation
ex = E(K) =
∞
X
kP (K = k)
k=0
This can be written in different ways, for example as
∞
X
k=1
k k px · qx+k
• Perhaps the easiest is to notice that we need
(1)P (K = 1) + (2)P (K = 2) + (3)P (K = 3) + . . .
=P (K = 1) + P (K = 2) + P (K = 3) + . . .
+P (K = 2) + P (K = 3) + . . .
+P (K = 3) + . . .
=P (K ≥ 1) + P (K ≥ 2) + P (K ≥ 3) + . . .
∞
X
=
P (K ≥ k)
=
k=1
∞
X
k px .
k=1
• ex is often easier to calculate than ˚
ex .
• With calculations mostly done on curtate future lifetimes, so discretized in years, sometimes
interpolation between consecutive discrete time points may be needed.
• One possibility is to use the function S(x) = T (x) − K(x) and to assume that S(x) is
uniformally distributed over (0, 1). This gives
˚
ex ≈ ex +
1
2
and it can also be used to approximate other quantities of interest (e.g. if month of death is
considered).
• The assumption of uniformity implies that S does not depend on the value of k, i.e. that
S(x) and K(x) are independent. This is often unreasonable.
2.6
Life tables
• A life table defines a probability model based on a large data set of lifetimes (usually without
using a full statistical framework to reflect the uncertainties involved; empirical proportions
are interpreted as probabilities). Creating life tables is not straightforward, this is not addressed in this course. Commonly, life tables are given for specific genders, risk groups,
etc.
17
• Start with a large group (cohort of people), with size l0 : d0 of these people die at age 0. In
general: lx people reach age x, dx die at age x, so
lx+1 = lx − dx
Probabilities for many events of interest can be defined as the corresponding proportions in
a life table.
• Proportion llx0 of the initial cohort reach age x. Proportion 1 −
died by age x + k.
lx+k
lx
were alive at age x, but
• We can use, for example
qx = 1 −
k px
=
lx+1
dx
=
lx
lx
lx+k
lx
• In this module, we use life tables from the book by Gerber (handed out; also available from
module webpage). We begin by using table E0 .
• This imagines that the initial cohort is 10,000,000 and shows the number of people still alive
at each age, lx , and the number who die before the next age, dx . It also shows qx , the
proportion expected to die that year.
• Observe that q0 is large, but drops away sharply. This is infant mortality. It starts climbing
again at age x = 12. Can make other comments, such as the fact that about 75% of people
reach age 65. This table has no probability for living beyond age 99, emphasizing that these
are approximations.
• Life tables for the UK are published via the Population page at http://www.ons.gov.uk/
• Example. Find, for the cohort aged 30, and using the life tables provided:
1. The probability to be alive at age 60
2. The probability to die before age 60
3. The probability to be alive at age 80
4. The probability to be alive at age 80 given alive at age 60
5. The probability to die next year
• Solutions:
= P (T > 30|X ≥ 30) =
l60
l30
=
= P (T > 50|X ≥ 30) =
l80
l30
= 0.412.
1.
30 p30
2.
30 q30
3.
50 p30
4.
20 p30+30
= 1 −30 p30 = 0.138;
5. q30 =
=
d30
l30
l80
l60
= 0.478(=
8118113
9501382
= 0.862;
0.412
)
0.862
= 0.00153. This is also reported directly in the table.
18
2.7
Life tables: fractions of a year
• Sometimes one wants to take the fraction of year lived in the year of death, S(x) = T (x) −
K(x), into account even when using a life table, for example by considering P (S ≤ u|X ≥
x) = u qx with 0 < u < 1.
• There are two common methods. These are both approximations as real information is not
available in the life table.
1. Assume that the death time is distributed uniformly over the year: u qx = (u)1 qx for
0 < u < 1.
qx
Exercise: Show that this gives µx+u = 1−uq
for 0 < u < 1. Proof of exercise. u px =
x
1 − u q x = 1 − u1 q x .
µx+u = −
d
1 d
ln (u px ) = −
(1 − u1 qx )
du
u px du
and the result follows.
2. Assume that µx+u is constant over 0 < u < 1.
(Some exercises on handout 2, and discussed P21 in Gerber.)
2.8
Practical problems with life tables
• We can consider life tables as similar to models such as Gompertz’s or Weibull’s, but the
model is incompletely specified (only discrete values given) and has far more parameters (one
for each year!).
• How to create a life table such that it is relevant for current generation? Life expectancy has
changed considerably in recent years.
• How to deal with covariates (e.g. gender, smoking habits, social class)?
• Life insurance is often only sold to people with good health (a medical test may be required).
So-called ‘select life tables’ take into account the effect of having joined the scheme recently,
and hence probably being healthier than the average person of same age in the scheme, for
a number of years (e.g. 5 years) after which the effect is assumed to have disappeared.
3
Life insurance
3.1
Introduction
• Life insurance contracts define future payment(s) of agreed amount(s) in return for a single
payment now or a defined sequence of payments. Payments are dependent on the insured
being alive at the time of payment:
– Whole life policies pay money when the policy holder dies
– Term policies pay money when the policy holder dies if this happens within a specified
period
– Deferred policies pay money when the policy holder dies if this happens after a specified
period
19
– Endowment policies pay money after a specified period if the policy holder is still alive
• Alternative criteria to death may also be used, e.g. disease or serious accident.
• The contracts typically relate to a single event with some uncertainty.
• Non-life insurance deals with multiple events, this is not discussed in this module as it
requires more knowledge of probability and stochastic processes. Risk to the insurer is far
more complicated in this field.
• For contracts where the benefit to the holder is a single payment, this payment is called the
sum insured.
• Denote by Z the present value of the contract, calculated using the technical rate of interest
i (used for calculation for the contract; it is usually assumed to be constant). Note that Z is
typically a random quantity if payment(s) depend on the policy holder being alive.
• E(Z) is called the net single premium (NSP) of the contract. This is the ‘fair price’ of the
contract. Normally an insurance company charges more, to cover expenses and to stay in
business given the random nature of future payments.
3.2
Whole Life Insurance
• Consider a contract which pays 1 unit at the end of year of death, so at time K(x) + 1. If
K(x) = k, the present value of the payment is v k+1 . Recall that this is the curtate future
lifetime.
• As usual, we drop the (x) notation where convenient, but remember that all results are
conditional on lifetime having reached age X ≥ x.
• To simplify some of the calculations, we use the indicator function 1[K=k] which equals 1 if
K = k, and 0 otherwise. Then,
K(x)
1[K=0]
1[K=1]
..
.
0
1
0
..
.
1
0
1
..
.
2
0
0
..
.
3
0
0
..
.
4
0
0
..
.
5
0
0
..
.
6
0
0
..
.
...
...
...
..
.
1[K=6]
..
.
0
..
.
0
..
.
0
..
.
0
..
.
0
..
.
0
..
.
1
..
.
...
..
.
6
7 ...
Year end 1 2 3 4 5
2
3
4
5
6
Z
v v v v v
v
v7 . . .
2
4
6
8
10
12
2
v v v v v
v
v 14 . . .
Z
• From this it is clear that we can write
Z=
∞
X
1[K=k] v k+1 = v K+1
k=0
and
Ex (1[K=k] ) = P (K = k).
20
• Note that only one term in the infinite sum is positive, all other terms are zero. The net
single premium is
E(Z) = E(v K+1 ) =
∞
X
v k+1 P (K = k) =
k=0
∞
X
v k+1 k px qx+k .
k=0
This is easy to calculate, something we consider later. For this result, we need to remember
the formula for P (K = k) which we dealt with when we looked at curtate future lifetimes.
• If we don’t discretize the time of death, we can imagine that the unit payment takes place
immediately after the death of the policy holder, i.e. with lifetime T . Then, the present
value of the payment is Z = v T and
Z ∞
Z ∞
T
t
E(Z) = E(v ) =
v g(t)dt =
v t t px µx+t dt.
0
0
This needs remembering that we can write g(t) in terms of the force of mortality.
• Using life tables, this calculation is complex due to continuous time being used.
• To summarize, and introduce actuarial notation:
Ax = E(Z) = E(v
K+1
)=
∞
X
k=0
=
=
∞
X
k=0
∞
X
k=0
v k+1 k px × (1 −
v k+1 k px × 1 qx+k
k+1 px
k px
)
v k+1 (k px − k+1 px )
T
Ax = E(v ) =
Z
∞
v t g(t)dt.
0
• Clearly we must have Ax ≤ Ax , as K + 1 ≥ T .
• An approximate relation between these is
v 1/2 × Ax ≈ Ax ⇔ Ax ≈ (1 + i)1/2 Ax
based on the idea that if payment is deferred until the end of year of death, the insurance
company has possession of the money approximately for an extra half year.
• While Ex (Z) is the most important summary of the probability distribution of Z, it is also
useful to consider the second moment Ex (Z 2 ) to get an idea about the variation. This enables
calculation of the variance and is useful for assessing risk to the insurer as we will discuss
later.
• For a whole life insurance which pays one unit at the end of the year of death:
!2
∞
∞
X
X
k+1
2
1[K=k] v
= Ex (
1[K=k] v 2(k+1) )
Ex (Z ) = Ex
k=0
k=0
because only one term in this sum is positive. So Ex (Z 2 ) is equal to the net single premium
with discount factor v 2 instead of v.
21
• Higher order moments can be derived similarly, we do not consider these further.
• Let us consider in more detail approximations for whole life insurance with payment immediately after death, based on a life table.
• Assume that K(x) and S(x) = T (x) − K(x) are independent, so the moment of death within
the year of death is independent of the person’s age (usually this is not an unreasonable
assumption).
• Again we will drop (x) from the notation where convenient.
• Under this assumption, E(v T ) = E(v K+1 ) × E(v S−1 ) and so
Z 1
T
K+1
S−1
Ax = E(v ) = E(v
)E(v ) = Ax
(1 + i)1−s f (s)ds,
0
where f (s) is the distribution for the fractional lifespan S of (x) which ends in the interval
[k, k + 1) after x. Therefore, under this assumption, the correction factor depends only on
the model for S.
• Suppose that, with K and S independent, S is uniformly distributed on [0, 1). So f (s) = 1,
0 ≤ s < 1. Then
Z 1
Ax = Ax
(1 + i)1−s ds
0
1
(1 + i)1−s (−1)
= Ax
ln (1 + i)
0
i
= Ax ×
δ
where δ = ln(1 + i) is the force of interest. This value is close to that given for assumed
death in mid-year:
i
0.02
0.04
0.06
p
(1 + i)
1.00995
1.01980
1.02956
i
δ
1.00997
1.01987
1.02971
• There are other assumptions about S that can lead to useful approximations.
• Exercise: Find Ax and Ax if T (x) is exponentially distributed with force of mortality equal
to 0.04 and force of interest δ = 0.1.
– Recall that the exponential distribution with parameter λ has cumulative density function
P (T ≤ t) = G(t) = 1 − e−λt . The corresponding probability density function is g(t) =
λe−λt .
– We have t px = 1 − G(t) = e−λt .
– The force of mortality is
µx+t = −
d
ln( t px ) = λ.
dt
Therefore, we require λ = 0.04.
22
– Recall the relationship between the force of interest, δ, and i and v. From eδ = 1 + i, we
1
= e−δ . Hence
have that v = 1+i
Z ∞
Z ∞
λ
t
= 0.2857.
Ax =
v gx (t)dt = λ
e−(λ+δ)t dt =
λ+δ
0
0
– For the end-of-year payment we have
Ax =
∞
X
k=0
=
∞
X
k=0
v k+1 (k px − k+1 px )
e−(k+1)δ (e−λk − e−λ(k+1) )
−δ
= e (1 − e
−λ
)
∞
X
k=0
e−(δ+λ)k = e−δ (1 − e−λ )
1
= 0.2716.
(1 − e−(δ+λ) )
These calculations emphasize that for an exponential distribution of T (x) the net single
premium does not depend upon (x).
– Check performance of approximations:
√
√
√
1. Ax ≈ 1 + iAx . We have 1 + i = eδ/2 = e0.05 , and so 1 + iAx = 0.2855, which is
very close to 0.2857.
2. Ax ≈ δi Ax (Note we already know that S is not uniformly distributed). δi Ax =
(eδ −1)
Ax = 0.2856, which again is very close to 0.2857.
δ
3.3
Term Insurance
• A term insurance of duration n (years or other time units) pays out after the death of
the holder if this occurs within n years.
• This is useful in situations where the financial problems caused by death are significantly
higher over a given period of time e.g. mortgages (before they’re paid), self-employed people
with young children (before those children grow up and become self-sufficient).
• In case of payment of 1 unit at the end of year of death, the present value and net single
premium are
K+1
v
for K = 0, 1, . . . , n − 1
Term insurance: Z =
0
for K = n, n + 1, . . .
That is,
Z=
n−1
X
k=0
v k+1 1[K=k] + 1[K≥n] × 0 = 1[K<n] v K+1
• We introduce actuarial expectation for the expectation of this net single premium.
1
Ax:n
= E(
=
n−1
X
v k+1 1[K=k] ) =
k=0
n−1
X
k+1
v
k=0
n−1
X
k=0
(k px − k+1 px )
23
v k+1 k px qx+k
This is identical to Ax , except that the limit n − 1 replaces ∞.
• In case of pay out at the time of death the net single premium can be calculated, or if needed
approximated, similarly as presented for a whole life insurance. E(Z 2 ) is equal to the net
single premium at twice the force of interest. To see this, replace v by e−δ . (Check these
facts as exercise)
• Example. Find the value of A1x:n for a term insurance policy of duration n that is otherwise
identical to the policy described in the previous example.The force of mortality remains
λ = 0.04. For term insurance of duration n we have
1
Ax:n
−δ
−u
−δ
−λ
= e (1 − e
)
n−1
X
e−(δ+λ)k
k=0
e (1 − e )(1 − e−(δ+λ)n )
1 − e−(δ+λ)
= Ax (1 − e−(δ+λ)n ).
=
• Note in general that the ratio
unusual.
Ax
1
Ax:
n
will depend on x as well as n, but the exponential case is
• Putting in some values for n we find
n
1
Ax:n
1
0.0355
5
0.1367
10
0.2046
20
0.2551
1
Compare with Ax = 0.2716, and note that Ax:n
→ Ax as n → ∞).
• For more realistic lifetime models term insurance is generally considerably cheaper than whole
life insurance, since there’s a good chance that no claim will be made.
3.4
Commutation Functions
• Commutation is where a member of a pension scheme gives up part or their entire pension
in exchange for an immediate lump sum payment.
• Commutation functions are given in tables; we use those from the book by Gerber (in the
life tables handout).
• We therefore consider - for the sake of historical interest - how these calculations were once
performed. The methods considered are applied to life tables, so we start with some notation.
• Recall that using life tables, we have for that cohort: t px =
lx+t
lx
and qx =
dx
.
lx
• Recall (section on life tables) that the probabilities t px and qx can be approximated by lx+t
lx
P
p
and dlxx , respectively. Recall also (material on curtate lifetimes) that E( K) = ∞
k=1 k x =
P∞ lx+k
k=1 lx .
24
• To obtain Ax we have
Ax =
∞
X
k=0
=
∞
X
v k+1 (k px − k+1 px )
v k+1 (
k=0
lx+k lx+k+1
−
)
lx
lx
∞
1 X k+1
=
v dx+k ,
lx k=0
as dx = lx − lx+1 .
• Commutation tables, such as Table E1 in Gerber, are constructed for each AER i of interest.
1
Table E1 is for i = 5%, and correspondingly for v = 1+i
= 1/1.05. The following columns
are constructed.
• Define, for x = 1, 2, . . .,
Dx = v x l x
Cx = v x+1 dx
∞
∞
X
X
x+k+1
Mx =
v
dx+k =
Cx+k
k=0
∞
X
Nx =
k=0
Dx+k
k=0
then
Ax =
Mx − Mx+n
Mx
1
=
and Ax:n
.
Dx
Dx
• The calculation for a single net premium is reduced to a single division (followed by multiplication by the sum insured). For a policy payable at time of death we just correct to get
Ax as usual.
• Example. Find the NSP of a whole life insurance policy on x = (40), paying amount 1 at the
end of year of death when the interest rate is i=0.05. Repeat for x = (60).
– From the tables we have
A40 =
and
A60 =
M40
275145.23
= 0.2080
=
D40
1322891.9
186711.648
M60
= 0.4259.
=
D60
438355.9
– Note that A60 > A40 , as you would expect as the insurance company can expect more time
for the initial premium to grow to match the unit payout.
• Example Consider term periods of 1, 5, 10 and 20 for current age x = 40. We earlier calculated
A40 = 0.2080, so we expect the NSP for each term insurance to be rather less.
25
– From tables, D40 = 1, 322, 891.9, and we also require M40 , M41 , M45 , M50 and M60 . For
1
A40:1
we need
1
=
A40:1
M40 − M41
275145.23 − 271642.654
= 0.002648.
=
D40
1322891.9
– The remaining calculations give answers as follows.
n
1
5
10
20
∞
1
A40:n 0.002648 0.01377 0.02918 0.06685 0.2080
A160:n 0.01311 0.06869 0.1443 0.2983 0.4259
– As we would expect, a term contract of given duration is more expensive for current age
(60) than (40).
3.5
Other life insurance contracts
• Deferred whole life insurance. Payment at the end of year of death, if that occurs after n
years.
• A pure endowment of duration n pays 1 unit at the end of year n if the insured is still alive
at that moment (else no payment).
• For a standard endowment the sum insured is paid at the end of year of death if it occurs
within n years, but otherwise paid at the end of year n.
• In the following table, we have already dealt with the first two.
Whole life insurance: Z =
Term insurance:
Z=
Deferred whole life:
Pure endowment:
Z=
Z=
Standard endowment: Z =
3.5.1
v K+1
for K = 0, 1, . . .
v K+1
0
for K = 0, 1, . . . , n − 1
for K = n, n + 1, . . .
0
v K+1
for K = 0, 1, . . . , n − 1
for K = n, n + 1, . . .
0
vn
v K+1
vn
for K = 0, 1, . . . , n − 1
for K = n, n + 1, . . .
for K = 0, 1, . . . , n − 1
for K = n, n + 1, . . .
Deferred whole life insurance
• The present value for a unit sum insured is Z = 0 for K = 0, 1, . . . , n − 1 and Z = v K+1 for
K = n, n + 1, . . .. We can write this as
Z = 1[K≥n] v K+1
∞
X
=
v k+1 1[K=k]
=
k=n
∞
X
k=0
v
k+1
1[K=k] −
26
n−1
X
k=0
v k+1 1[K=k]
• Now take expectations to obtain the net single premium. Notice that the expectations of
these two halves has already been obtained separately; the first has expectation Ax and the
1
second has expectation Ax:n
. Hence, and using notation n| Ax , we find that
n| Ax
1
= E(Z) = Ax − Ax:n
Mx Mx − Mx+n
Mx+n
=
−
=
.
Dx
Dx
Dx
• So the NSP for a deferred whole life insurance is the difference between the NSP for a whole
life insurance and the NSP for a term insurance.
• For payment at the moment of death, computation (or approximation) is again similar as for
whole life or term policies.
3.5.2
Pure endowment
• Its present value and NSP are
Z = 1[K≥n] v n
E(Z) = v n P (K ≥ n)
= v n n px
lx+n
= vn
lx
Dx+n
=
Dx
= Ax:n1
• Note that the payment is not at a random time.
3.5.3
Standard endowment
• The present value and NSP are
Z = 1[K<n] v K+1 + 1[K≥n] v n
Mx − Mx+n + Dx+n
1
Ax:n = Ax:n
+ Ax:n1 =
Dx
• Example.
– Calculate standard endowments for a life aged 40. We have
M40 − M60 + D60
≈ 0.398,
D40
M40 − M70 + D70
≈ 0.279
=
D40
A40:20 =
A40:30
and the longer term endowment has lower NSP as the payment is likely to be paid rather
later.
27
– The same policies for a life aged 60 have net single premiums
M60 − M80 + D80
≈ 0.478,
D60
M60 − M90 + D90
≈ 0.431
=
D60
A60:20 =
A60:30
and these are rather more similar as there is a much greater risk of the payment being due
to the death of the insured.
• For payment at the time of death the term insurance part can be approximated as before;
the pure endowment part remains the same. So, we correct the term A1x:n to (i/δ)A1x:n and
add this to the term Ax:n1 .
3.5.4
General life insurance policies
• Variations and generalizations of standard contracts are straightforward.
• For example the defined benefit may change depending on year of death: cK+1 with present
value v K+1 cK+1 .
• Exercise: Check that, for increasing payments cK+1 , the contract can be regarded as a
sum of deferred whole life insurances; hence derive an expression for the NSP in terms of
commutation functions. (See Gerber p27).
3.6
Risk to the insurer
• Insurance companies must consider many contracts and ensure that they can meet their
obligations. They need to consider further aspects of the probability distributions of the
random payments to get an idea of the risk involved.
• Most of this risk results from the possibility of earlier than expected payments, where particularly for term insurance contracts the risk can be substantial (why?).
• The variance is useful, particularly for use in the Central Limit Theorem.
• Varx (Z) = Ex (Z 2 ) − (Ex (Z))2 ; we have calculated the NSP, Ex (Z), already for each type of
policy.
• For a whole life insurance for (x), with 1 unit payment at the end of the year of death,
2
E(Z ) =
∞
X
(v
k+1 2
)
k px qx+k
k=0
=
∞
X
v 2(k+1) k px qx+k
k=0
• This is equal to the NSP of a whole life insurance for (x) with AER equal to (1+i)2 −1 = i2 +2i,
or equivalently the NSP for force of interest 2δ, as v 2 = e−2δ where δ = ln(1 + i). To see this,
suppose that v 2 = vj = 1/(1 + j), whence j = 1/v 2 − 1 = i2 + 2i.
• For a whole life insurance with payment at time of death E(Z 2 ) is also equal to the corresponding NSP with force of interest 2δ.
• The same applies to higher moments: E(Z m ) is the NSP with force of interest mδ.
28
• Example. For the exponential lifetime model with λ = 0.04 and δ = 0.1, find the variance
and standard deviation of Z.
– We have previously demonstrated that Ax =
Ex (Z 2 ) =
– This gives us Varx (Z) =
3.6.1
1
6
− ( 72 )2 =
25
294
λ
λ+δ
=
2
7
= 0.2852. Therefore
λ
4
1
=
= .
λ + 2δ
24
6
= 0.08503 and hence SD = 0.292.
Combinations of policies
• Suppose the insurer offers n unit value whole life contracts (payable at time of death) to
independent lives (xi ). Let Zi = v Ti for each i (so xi dies at time Ti ), then
E(
n
X
Zi ) =
i=1
Var(
n
X
n
X
E(Zi )
i=1
Zi ) =
i=1
n
X
Var(Zi )
i=1
(For dependent Zi covariances must be taken into account for the variance.)
• The Central
Limit Theorem can be used to assess risks for total of the n payments for large
Pn
n, as i=1 Zi is approximately normally distributed.
• Example. Find the mean, variance and standard deviation of the sum of 30 contracts of the
type described in the previous example, assuming all lives are independent.
– We have
E(
30
X
Zi ) = 30Ax = 8.571
i=1
and
Var(
30
X
Zi ) = 30Var(Z1 ) = 2.551.
i=1
which results in a standard deviation of 1.597.
– Early deaths (small values of Ti ) mean greater payouts. We use the CLT To assess the
probability of a specific total payout. Approximately,
Y =
30
X
i=1
Zi ∼ N (8.571, 1.5972 ).
– This can be used to make calculations about the probability for hitting certain targets.
For example, a 95% probability interval for the total return Y is 8.571 ± (1.96)(1.597) =
(5.44, 11.70). There is a 0.025 chance that policies with a total income of 8.571 incur costs
exceeding 11.70.
– This can be dealt with by the insurance company building in a safety margin by setting
the premium at 11.70
= 0.390, rather than 0.2852.
30
29
3.7
Summary of net single premiums
Whole
Term
Ax =
1
=
Ax:n
∞
X
k=0
n−1
X
v k+1 P (K = k)
Mx
Dx
v k+1 P (K = k)
Mx − Mx+n
Dx
k=0
1
= Ax − Ax:n
Deferred
n| Ax
Pure
Ax:n1 = v n n px
1
+ Ax:n1
Standard Ax:n = Ax:n
30
Mx+n
Dx
Dx+n
Dx
Mx − Mx+n + Dx+n
Dx
4
Life annuities
4.1
Whole life annuity-due
• A whole life annuity provides a sequence of payments, usually monthly or annually, until
death. We mostly assume a unit payment each year.
• Remember that we work with the present value of all contracts, i.e. we deal with everything
in terms of today’s money.
• The present value of a life annuity is denoted by Y .
• A whole life annuity-due pays a unit amount at the start of each year until death.
• The policy holder dies at time T ∈ [K, K + 1) so payments are at times 0, 1, . . . , K.
1 + v + . . . + v n−1 , (standard due annuity) so if K = k the present value of the
• Recall a
¨n = P
payments is kj=0 v j = a
¨k+1 .
• Further,
P (Y = a
¨k+1 ) = P (K = k) =
k px qx+k .
• We may write Y as
Y =
∞
X
1[K=k]
k
X
∞
X
j
v =
j=0
k=0
j=0
v
j
∞
X
1[K=k] =
∞
X
v j 1[K≥j] .
j=0
k=j
(The order of summation can be exchanged because all terms are positive - try writing out
the series.)
• The net single premium is (new notation)
a
¨x = E(Y ) = E(
∞
X
1[K=k]
j=0
k=0
=
∞
X
Px (K = k)
k=0
k
X
k
X
vj =
j=0
∞
X
k=0
vj ) =
∞
X
Ex (1[K=k]
k px qx+k
vj )
j=0
k=0
k
X
k
X
vj =
j=0
∞
X
¨k+1
k px qx+k a
k=0
• It can also be written as
a
¨x = E(Y ) =
∞
X
j=0
j
v Px (K ≥ j) =
∞
X
v j j px
j=0
• Written in that form you can see that a whole life annuity-due is equal to a collection of pure
endowments.
• Recall that a pure endowment of duration j pays 1 at end of year j (hence the start of year
j + 1), as long as the policy holder is still alive. So, we can consider a whole life annuity-due
as simply a collection of pure endowments; one with period 0, one with period 1, and so on.
And, indeed, the net single premium of a j-year pure endowment is v j j px , as needed.
31
1−v k+1
,
d
• We saw before (lectures on annuities) that a
¨k+1 =
rate.
• Therefore
Y =
∞
X
1[K=k] a
¨k+1 =
k=0
where K is random.
∞
X
1[K=k]
k=0
where d = i/(1 + i) is the discount
1 − v K+1
1 − v k+1
=
d
d
• This allows us to make a link to whole life insurance.
• This is because a whole life insurance has present value Z = v K+1 and net single premium
Ax = E(Z) = E(v K+1 ). Now we can see that Y = (1 − Z)/d.
• It follows that
• For the variance,
a
¨x = E(Y ) =
1 − Ax
1 − E(Z)
=
d
d
Var(Z)
Var(1 − Z)
=
,
2
d
d2
which can be used to assess risk to the insurer as before.
V ar(Y ) =
• Using commutation functions and Ax = Mx /Dx :
a
¨x =
Dx − M x
dDx
• Alternatively, recall that Dx = v x lx so that we may write
P∞ x+k
∞
∞
X
X
v lx+k
v x+k lx+k
k
= k=0
a
¨x =
v k px =
x
v lx
Dx
k=0
k=0
• Define a further commutation function
Nx =
∞
X
v x+k lx+k =
k=0
Dx+k .
k=0
(Note this was misdefined in an earlier lecture as
• Then
∞
X
a
¨x =
P∞
k=0
dx+k .)
Nx
.
Dx
• Example: Calculate the NSP for a whole life annuity-due for the exponential lifetime model
for (x) with constant force of mortality λ = 0.04 and force of interest δ = 0.1. The notation
(x) is an abbreviation for a person currently aged x.
– This is an example we considered previously. We found that Ax = 0.2716.
– Since i = eδ − 1, we have that d =
i
i+1
= 0.09516.
– Therefore the NSP for the whole life annuity due is given by
a
¨x =
0.7284
1 − Ax
=
= 7.6544.
d
0.09516
32
• Example: Calculate the NSP for a whole life annuity-due using the provided commutation
columns, so with i = 0.05, for (30) and (50), and briefly discuss these.
– We have that
a
¨x =
Nx
.
Dx
– Using our tables, we have
N30
39698231
= 18.058
=
D30
2198405.5
N50
=
= 14.636.
D50
a
¨30 =
a
¨50
– As a check we can use e.g. a
¨30 =
which also gives us 18.058.
1−A30
.
d
Here A30 = M30 /D30 = 0.1401 and d = 0.05/1.05,
– Note that it is entirely reasonable that a
¨30 > a
¨50 , since there will be more payments made
to a younger policy holder.
4.2
Temporary life annuity-due
• An n-year temporary life annuity-due pays a unit amount at the start of years 1 up to n (so
at times 0 to n − 1) if the policy holder is alive.
• The present value of the payments made will be
1 + v + v2 + . . . + vK = a
¨K+1
if the policy holder dies before n years, and
1 + v + v 2 + . . . + v n−1 = a
¨n
if the insured lives at least n years.
• Thus, the present value is
Y =
n−1
X
k=0
1[K=k]
k
X
v j + 1[K≥n] a
¨n =
j=0
n−1
X
¨n
1[K=k] a
¨k+1 + 1[K≥n] a
k=0
• Another way of writing Y , after rearrangement of the summation, is
Y =
n−1
X
1[K≥k] v k
k=0
• The net single premium is (new notation):
a
¨x:n = E(Y ) =
n−1
X
k=0
¨n ·
a
¨k+1 · k px qx+k + a
or equivalently, using the other way of writing Y ,
a
¨x:n = E(Y ) =
n−1
X
k=0
33
v k k px .
n px
• Using commutation functions
a
¨x:n
∞
X
v x+k lx+k
=
v x lx
k=0
−
∞
∞
X
v x+k lx+k
k=n
v x lx
Nx X v x+n+j lx+n+j
Nx − Nx+n
=
−
=
,
x
Dx j=0
v lx
Dx
where we change the summation index, replacing k by n + j.
• As for the whole-life annuity-due, We can relate Y to the present value of an n-year standard
endowment (i.e. where the sum insured is paid at the end of year of death if withing n years,
but otherwise at the end of yera n) given by
Z=
n−1
X
1[K=k] v k+1 + 1[K≥n] v n
k=0
with NSP Ax:n .
• Using a
¨j =
1−v j
d
Y
n−1
X
=
1[K=k]
k=0
n−1
X
1−Z
1
[1 − (
1[K=k] v k+1 + 1[K≥n] v n )] =
d
d
k=0
=
• This gives
1 − v k+1
1 − vn
+ 1[K≥n]
d
d
a
¨x:n = E(Y ) = E(
1−Z
1 − Ax:n
)=
d
d
• Letting n → ∞ gives the above relation between the whole life contracts.
• Exercise: Calculate the NSP for a twenty year temporary life annuity-due for (30) and for
(50), using the provided commutation tables (and i = 0.05), and also do this for a forty year
temporary life annuity-due, and discuss the results.
– We can calculate these directly from commutation columns
N30 − N50
D30
N30 − N70
=
D30
N50 − N70
=
D50
N50 − N90
=
D50
39698231 − 11424323.8
= 12.861,
2198405.5
a
¨30:20 =
=
a
¨30:40
= 17.160,
a
¨50:20
a
¨50:40
= 12.108,
= 14.575.
– For equal time periods, the price is greater for (30) than (50). This is to be expected, since
a younger person is more likely to survive to receive all 20 payments. However, the risk of
death between 50 and 70 is not all that greater than the risk between 30 and 50, so the
difference between a
¨30:20 and a
¨50:20 is small.
34
– On the other hand, the risk of death between 50 and 90 is far higher than the risk of death
between 30 and 70, so the difference between a
¨30:40 and a
¨50:40 is somewhat larger.
– Indeed, we saw last lecture that a
¨50 , the NSP for a whole life annuity-due, is 14.636, which
is very close to 14.575. This is because the chance of a 50 year old making it past 90 (and
thus getting more than 40 payments) is very small.
4.3
Deferred annuity-due
• For an n-year deferred annuity-due payments only begin at the start of year n + 1 (end of
year n, so time n) and are only made if the insured is alive.
• The present value is Y = v n + v n+1 + . . . + v K if K ≥ n, and Y = 0 if K < n.
• The NSP is (new notation):
E(Y ) = n| a
¨x =
n px
· vn · a
¨x+n = a
¨x − a
¨x:n
• The first formula is obtained by reversing the summation as in previous sections, and so
writing
∞
∞
X
X
n
r
Y =v
v
1[K=n+j]
r=0
j=r
Then,
E(Y ) = v
n
= vn
∞
X
r=0
∞
X
v r n+r px
v r n pxr px+n
r=0
n
= v n px
∞
X
v r r px+n
r=0
= v n n px a
¨x+n .
• Using commutation functions:
¨x
n| a
=a
¨x − a
¨x:n =
Nx Nx − Nx+n
Nx+n
−
=
Dx
Dx
Dx
• This is the basic component of a pension. Typically a pension will also make a payment
upon the death of the holder prior to age x + n, so they combine deferred annuities-due with
term insurance.
4.4
Immediate life annuity
• Immediate life annuities have payments at the end of the years (or other time units), rather
than at the start (as is the case for life annuity-due contracts).
• The final payment is at end of year K (death occurs in year K + 1) for a whole life immediate
annuity and at end of year n for an n-year temporary immediate life annuity if K ≥ n.
35
• The NSP ax (whole life) and ax:n (term) can be calculated using
ax = a
¨x − 1,
ax:n = a
¨x:n+1 − 1.
• So,
ax =¨
ax − 1 =
ax:n = a
¨x:n+1
4.5
N x − Dx
Nx+1
Nx
−1=
=
,
Dx
Dx
Dx
Nx+1 − Nx+1+n
Nx − Nx+n+1
−1=
−1=
Dx
Dx
Summary of net single premiums for life annuities
Whole-due
Term-due
a
¨x:n
Deferred-due
¨x
n| a
Whole-immediate
Term-immediate
1 − Ax
d
1 − Ax:n
=
d
a
¨x =
=a
¨x − a
¨x:n =
ax = a
¨x − 1
¨x:n+1 − 1
ax:n = a
36
Ax:n − Ax
d
Nx
Dx
Nx − Nx+n
Dx
Nx+n
Dx
Nx+1
Dx
Nx+1 − Nx+1+n
Dx
5
Calculating premiums
5.1
Introduction
• In previous lectures we have concentrated on the expected present value of a contract with
a single payment at the end of life (life insurance), or a contract with annual payments (life
annuity). Each of these expectations is expressed as a net single premium, considered as the
fair price in today’s money for that contract.
• Most life insurance contracts are not bought by paying one lump sum but by regular premiums, so we now have to add in the complication of present values of premiums paid.
• We consider annual premiums, paid at the start of the year (other time units can of course
be used with similar mathematics).
• We consider the so-called loss to the insurer L, which is the random quantity that is the
difference between the present value of the benefits and the present value of the premium
payments under the contract.
• For a fair price, we want Ex (L) = 0, in which case the premiums are called net premiums.
• Premiums which are constant over time are called level.
5.2
Whole life and term insurance
• If (x) buys a whole life insurance with unit payout at the end of the year of death, and pays
level net annual premiums Px for the full length of the contract (so till death), then it is
effectively an exchange of a whole life annuity-due for a whole life insurance.
• In this case
L=
∞
X
k=0
¨k+1 = v K+1 − Px a
1[K=k] v k+1 − Px a
¨K+1
where K is random. The first term is the present value of the insurance payout; the second
is the present value of the premiums paid at the start of each year.
• Requiring that E(L) = 0 leads to
E v K+1
A
= x
Px =
a
¨x
E a
¨K+1
• Example. Suppose that a person takes out a whole-life insurance policy which pays amount
25000 at the end of year of death. Assume an interest rate is i = 0.05. For a person aged
x = (40), calculate the net premium for this policy assuming level annual payments at the
start of each year. Repeat for x = (60).
– This is an example from previous lectures. Using commutation functions, the net present
values for a unit payout at the end of year of death are:
M40
275145.23
= 0.2080
=
D40
1322891.9
M60
186711.648
=
= 0.4259.
=
D60
438355.9
A40 =
A60
37
If the payout is 25000, the NSPs are 25000 × 0.2080 = 5200.00 and 25000 × 0.4259 =
10647.50 respectively
– The NSP for unit payments made at the start of each year (whole life annuity due) are:
N40
22002680.1
=
= 16.632
D40
1322891.9
N60
=
= 12.055.
D60
a
¨40 =
a
¨60
The net premiums thus need to be (per unit of payment received),
0.2080
= 0.0125, ⇒ 25000 × 0.0125 = 312.65,
16.632
0.4259
= 0.0353, ⇒ 25000 × 0.0353 = 883.24.
=
12.055
P40 =
P60
– The net annual premium for (60) is much higher than for (40), reflecting that there will be
fewer years for which the accumulated premiums can reach the fair price for the payout.
– Finally we emphasize here that, for (40), the NSP of a whole-life due annuity with annual
payments of 312.65 is 312.65 × 16.632 = 5200. The NSP of a whole life insurance paying
25000 at the end of year of death is 25000 × 0.2080 = 5200.00. The point is to get these
to match.
• Exercise: Using d¨
ax + Ax = 1, derive the following equalities and give interpretations for
them:
1
= d + Px
and
Px = dAx + Px Ax
a
¨x
5.2.1
Effect on risk to the insurer
• Remember that a
¨K+1 = 1 − v K+1 /d, so
L=v
Px
−
1 − v K+1 =
d
K+1
Px
1+
d
v K+1 −
Px
d
• Easy to verify that E(L) = 0.
• This leads to
Var(L) =
Px
1+
d
2
Var v K+1
• The variance of the loss to the
insurer when a whole life policy is paid in a single payment (at
time 0) is simply Var v K+1 , because there is no uncertaintty in the premiums being paid.
• So paying in instalments increases the risk to the insurer when considered in terms of the
variance, possibly by a considerable amount.
38
5.3
Term insurance
• The same method is used to calculate the (level) premiums for an n-year term life insurance
policy that pays a unit amount at the end of the year of death.
• Example: Consider a 10 year term life insurance policy for a life aged 40 paying C at the end
of year of death (if during the 10 year term). The policy is to be paid with level net annual
premiums Π, paid annually in advance while the policy is in force.
– The loss to the insurer is

 1[K=k] Cv k+1 − Π¨
ak+1
for k = 0, 1, . . . , 9
L=
 1[K≥10] −Π¨
a10
for k ≥ 10
The first component is the present value of the payout C made at the end of year of life,
minus the present value of the premiums paid. The second component is the value of the
premiums paid, but with no payout.
– The corresponding probabilities (not explicitly needed) are
P (K = k) = k p40 q40+k , and P (K ≥ 10) = 10 p40 .
– These are quantities for which we already have formulae. The insurer is providing a 10year term life insurance which has expected present value CA140:10 , in return for a 10-year
temporary life annuity-due with expected present value Π¨
a40:10 . Thus
a40:10
E40 (L) = CA140:10 − Π¨
Setting E(L) = 0 leads to equal annual premiums
A140:10
Π=C×
a
¨40:10
• This illustrates that, generally, the level net annual premium for an n-year term insurance
paying a unit amount is
A1
1
= x:n
Px:n
a
¨x:n
• Using commutation tables,
1
Px:n
=
Mx −Mx+n
Dx
Nx −Nx+n
Dx
=
Mx − Mx+n
Nx − Nx+n
1
can be derived for any given lifetime model by calculating A1x:n and using the earlier
• Px:n
results a
¨x:n = (1 − Ax:n )/d and Ax:n = A1x:n + Ax:n1 , leading to
1
Px:n
=
dA1x:n
1 − A1x:n − Ax:n1
where Ax:n1 is calculated as in previous lectures.
39
• Example. Consider a 10 year term life insurance policy paying C = 100000 on a life aged 40,
with interest rate i = 0.05. Suppose the lifetime model is given by the life tables. Find the
net annual premium Π for this policy.
– We have
Π=C
A140:10
M40 − M50
D40
275 145.230 − 236 544.628
=C
= 364.90.
·
=C
a
¨40:10
D40
N40 − N50
22 608.1 − 11 424323.8
– This value (paid yearly for ten years) might seem quite small. However, the insurance
company only has to give money out if (40) dies within ten years, and
10 p40
=
l50
= 0.961
l40
i.e. less than 4% of such policies involve any cost to the insurance company at all.
• It may be attractive to pay (level net) annual premiums Πx over a period different from the
period over which benefit payments (may) take place. To derive the premiums follow the
same steps as above. For example (and check),
– Whole life insurance paid by premiums for m years (if alive):
Πx =
Mx
Nx − Nx+m
– n-year (term) life insurance paid by premiums for m < n years (if alive):
Πx =
Mx − Mx+n
Nx − Nx+m
– n-year deferred life insurance paid by premiums for m years (if alive):
Πx =
5.4
Mx+n
Nx − Nx+m
Endowments
• Remember that a unit-payment pure endowment pays out at the end of year n if the policy
holder is alive. Annual premiums Px:n1 will be paid at the start of years 1 to n (if alive).
• The loss to the insurer is

 1[K=k] −Px:n1 a
¨k+1
for k = 0, 1, . . . , n − 1
L=
 1[K≥n] v n − P 1 a
for k ≥ n
x:n ¨n
• The benefit paid out by the insurance company is the NSP of a unit payment pure endowment,
¨x:n .
namely Ax:n1 . The NSP of an n-year annuity due (i.e. the income to the company) is a
Therefore to make these match we need a premium Px:n1 as follows.
¨x:n = 0 so
• E(L) = 0 ⇔ Ax:n1 − Px:n1 a
Px:n1 =
Ax:n1
Dx+n
=
a
¨x:n
Nx − Nx+n
40
• This is the level net annual premium for a unit payment to the insured.
• The level net annual premium for a standard endowment is
1
+ Px:n1 =
Px:n = Px:n
Ax:n
Mx − Mx+n + Dx+n
=
a
¨x:n
Nx − Nx+n
• One may want to buy a life annuity paid by level net annual premiums, for example an n-year
deferred immediate life annuity paid for by m ≤ n annual premiums (in advance, if alive) this is a natural basic scenario for a pension. Following the same steps as above, the annual
premium will be
Nx+n+1
Πx =
Nx − Nx+m
• Note that an n-year deferred immediate life annuity has first payment at the end of year
n + 1, hence is the same contract as an n + 1-year deferred life annuity-due, which is the
same as a whole life annuity due minus an n + 1-year (term, temporary) life annuity due.
41
6
Further topics
6.1
Expenses
• Insurance companies cannot sell contracts at net premium price, due to expenses they need
to cover and to ensure they can deal with risks involved due to the random payments. And
they may want to make some profit.
• We wish to calculate the adequate premium P a , also called the ‘expense loaded premium’,
which is the premium which (only just) covers the costs and the benefits.
• There are many ways in which expenses can be taken into account, we consider three main
types for a policy which offers a 1 unit benefit.
Aquisition Expenses Usually denoted α, these are incurred at the time of writing the
policy, and cover things like agent’s commission, medical examinations, and writing
contracts. Their total enters the calculation of P a as a lump sum, paid at time 0.
Collection Expenses These are charged at the beginning of every year in which a premium
will be collected. They are generally proportional to P a , so their expected present value
is βP a multiplied by the relevant annuity-due.
Administration Expenses These are charged at the start of every year in which the policy
in is effect, whether or not premiums are paid. They are typically proportional to policy
benefits, so their present value is γC, multiplied by an annuity due (which may not
match the annuity-due for the premiums).
• Note that the premiums P a are, as before, assumed to be constant, so all expenses are
included in the overall computation in order to determine P a .
• Denoting the corresponding components of the adequate (annual) premium by P α , P β and
P γ , and the level net annual premium by P , we can write
Pa = P + Pα + Pβ + Pγ
Of course, not all of these are included in every policy.
6.2
Illustration: standard endowment
• We illustrate this by considering a standard endowment with term n years for a life aged x
and paying benefit of 1 unit, with (level expense loaded) annual premium paid at the start
of every year that the policy is in force.
• Remember the net annual premium for this contract:
Px:n =
Ax:n
a
¨x:n
a
• We want to calculate the adequate premium Px:n
, the value of the premiums that will cover
a
the benefits and expenses. The expected present value of these premiums is Px:n
·a
¨x:n . This
must match the combined total expected value of the benefits paid and the expenses incurred.
42
• The idea is that we compute the present values of any expenses incurred, and lump these in
with the benefit paid to the insured. We then balance total expenditure with total income
to arrive at the level net premium.
• There are four components of expenditure:
1. Ax:n , the expected present value of the benefits;
2. α, the expected present value of the acquisition expenses;
a
3. βPx:n
·a
¨x:n , the expected present value of the collection expenses;
4. γ¨
ax:n , the expected present value of the administration expenses.
• Using, as before, the concept of zero expected loss to the insurer leads to
a
a
Px:n
·a
¨x:n = Ax:n + α + βPx:n
·a
¨x:n + γ¨
ax:n
α
a
a
Px:n
= Px:n +
+ βPx:n
+γ
a
¨x:n
Ax:n + α + γ¨
ax:n
a
⇒ Px:n
=
.
(1 − β)¨
ax:n
• If the annual premiums are not paid for the whole contract period of n years, but instead
for m < n years, only the main premium and the β component are affected, with expected
present value proportional to a
¨x:m , and we get annual premiums
ax:n
Ax:n + α + γ¨
(1 − β)¨
ax:m
• It is simplest to calculate everything in terms of a payment of a benefit of one unit, and then
multiply up as necessary. So, express α as a percentage of the actual benefit to be paid.
• Example. Use the life tables (hence AER is 5%) to compute the adequate annual premium
for the following policy.
– a 25-year standard endowment for (40);
– a sum insured of 50 000 payable at the end of the term or at the end of year of death;
– annual premiums to be paid at the start of each year the policy holder is alive;
– acquisition expenses of α = 1, 000 = 0.02C,
– collection expenses at 10% of the premium (β = 0.1),
– adminstration expenses at 1% of the sum insured (γ = 0.01).
To solve,
– We have
A40:25 =
M40 − M65 + D65
= 0.328500338,
D40
a
¨40:25 =
N40 − N65
= 14.1014925
D40
– The net annual premium (without including any expenses) is
P40:25 = 50000 ×
A40:25
= 50, 000 × 0.02329543 = 1164.77
a
¨40:25
43
– The adequate annual premium is given by
a
P40:25
= 50000
0.328500338 + 0.02 + 0.01 × 14.1014925
= 1928.50.
0.9 × 14.1014925
– Notice here that the α term is expressed per unit of benefit payment.
– In this case the expenses have a large effect on the premiums.
6.3
Illustration: whole life insurance
• Example. Use the provided commutation tables (AER is 5%) to calculate the adequate
annual premium for a whole life insurance of 100, 000 issued to (35), with the benefit payable
at the end of the year of death. Assume that annual premiums are to be paid at the start of
each year for at most 30 years (so till age 65) if the policy holder is alive. Include acquisition
expenses of 1, 200, collection expenses at 15% of the premium and administration expenses
of 100 at the start of each policy year.
– a whole-life insurance for (35);
– a sum insured of C = 100000 payable at the end of year of death;
– annual premiums to be paid until the age of 65;
– acquisition expenses of α = 1200 = 0.012C,
– collection expenses at 15% of the premium (β = 0.15),
– adminstration expenses of 100 collected at the start of each year (γ = 0.001).
These components are as follows.
– The benefit is a whole-life insurance with NSP Ax = Mx /Dx per unit paid, with x = 35.
– The expected present value of a unit premium paid is a
¨x:n , with x = 35 and n = 30. That
is, an annuity-due with length n.
– α = 0.012 per unit of benefit.
Nx
– The NSP of the administration expenses is a
¨x = D
, as this is the NSP of a whole-life due
x
annuity, i.e. with payments at the start of each year until death.
Suppose that the premium per unit of benefit is P a . The equation leading to zero expected
loss is
¨x:n + γ¨
ax
Pa × a
¨x:n = Ax + α + βP a × a
Ax + α + γ¨
ax
⇒ Pa =
(1 − β)¨
ax:n
M35
N35
+ 0.012 + 0.001 × D
35
= D35
N35 −N65
(1 − 0.15) D35
From commutation tables we find
M35
=
D35
N35
=
D35
N35 − N65
=
D35
291913.545
= 0.170923,
1707865.3
29734986.2
= 17.41062,
1707865.3
29734986.2 − 3347929.9
= 15.45031.
1707865.3
44
Thus we obtain
Pa =
0.170923 + 0.012 + 0.001 × 17.41062
0.2003337
=
= 0.01525449.
(1 − 0.15)(15.45031)
13.13276
This is the adequate annual premium for unit benefit. For C = 100000 we need CP a =
1525.45.
6.4
Family income insurance
• A family income insurance is typically an n-year term insurance providing an annuity of 1
unit per year from the end of the year of death of the policy holder until the end of the term.
It is usually paid for with (level net) annual premiums Π at the start of the year, for m ≤ n
years if the policy holder is alive. (Expenses can of course also be included in the premiums
as before.)
• This is equivalent to
– a contract where the insurance company guarantees to pay 1 unit at the end of every year,
which is an immediate annuity of period n, with present value an ;
– and the policy holder pays the premiums Π while alive, which is a temporary annuity-due
with expected present value Π¨
ax:m ,
– and while alive immediately returns the benefit payments of 1 unit, which is an immediate
temporary annuity with expected present value ax:n .
• To calculate the premiums Π:
Π¨
ax:m + ax:n = an
Π=
an − ax:n
a
¨x:m
• (Of course, this follows from a detailed analysis deriving the random loss L to the insurer
and setting E(L) = 0 - check!)
• Formulae required are:
an = v(1 − v n )/(1 − v)
Nx+1 − Nx+1+n
Nx − Nx+n+1
−1=
ax:n = a
¨x:n+1 − 1 =
Dx
Dx
1 − Ax:m
Nx − Nx+m
a
¨x:m =
=
d
Dx
• Example. Consider a 25-year family income policy on a life aged 40 paying 20000 annually
from the end of year of death. The AER is 5% and the life tables are used to model lifetime.
Find the value of each level premium (paid as the start of each year) if i) they are paid at
the start of each year until death or 25 years, ii) they are paid at the start of each year until
death or 10 years.
45
– We have
1 − 1.05−25
= 14.09394456 ,
0.05
(see previous example) and also
a25 =
a40:25 =
a
¨40:25 = 14.1014925
N41 − N66
= 13.3403842.
D40
– Putting this together we obtain
Π = 20, 000 ×
a25 − a40:25
= 1, 068.77 .
a
¨40:25
– If instead m = 10, the only thing that changes is a
¨x:m = 7.996387536 and now we can
obtain the revised premium directly, or by noticing that we must have
¨40:25
14.1014925
ˆ =Π× a
Π
= 1, 884.75 .
=Π×
a
¨40:10
7.996387536
• The obvious question is whether there is any difference between these from the insurance
company point of view (remember: the fact that two polices have equal present values does
not mean they have the same risks attached to them).
– Consider a policy holder who has survived until age 64 – they have 1 q64 = dx /ℓx ≈ 0.01952
(about 2%) and their remaining expected benefit from the policy is
20, 000
× 1 q64 ≈ 371.81 ≪ 1, 068.77
1.05
– So for this year (and perhaps a couple before that) the benefits are worth less than the
premiums.
– So the insured should just cancel the policy, with loss to the company.
– This is a disadvantage of level net premiums.
– Insurance companies have to set up their policies so the premiums and benefits are sensibly
balanced at each stage. One way is to choose m < n (as above). A second way is to bolt
on a pure endowment or deferred annuity to family income insurance - the changes to
premiums for this are easy to price. This suits many customers and adds a large benefit
to the end of the policy - diminishing the risk of premature cancellation.
• An interesting (and popular) variation to a family income policy is with an explicit link to
one recipient, so the policy provides an income for a named person after death of the insured,
normally for as long as the named recipient is alive (or for a maximum term).
• Such a benefit is often part of an occupational pension.
• If the future lifetimes of both people involved can be assumed to be independent random
variables, the calculations are straightforward, with payment at the end of a particular year
taking place if and only if the insured has died before that time and the recipient is still alive,
and the probability of this joint event just the product of the individual events’ probabilities.
If the future lifetimes are dependent (as may well be the case due to e.g. lifestyle similarities)
then more detailed lifetime models are needed to provide probabilities for the joint events.
46
6.5
Reserves
• We assume the contract and premiums to be fixed, we just investigate the consequences of
this contract for the insurer over the period of the contract.
• With net premiums, the expected loss to the insurer at the time of policy issue, t = 0, is
zero. This means that the total present value at t = 0 of the expected future benefits is equal
to the total present value at t = 0 of the expected future premiums.
• Now consider the expected loss to the insurer at a later time t > 0, under the assumption
that the insured is still alive, so T > t.
• For discrete time in years, when using life tables, we consider the expected loss at time
k ≥ 0, conditioned on curtate future lifetime K ≥ k. We focus on the discrete time case, for
continuous time replace k by t in the following notation, the concepts are the same.
• Let the random quantity k L be the difference between the present value at time k of all future
benefit payments and the present value at time k of all future premium payments. The net
premium reserve is the conditional expectation of k L given that K ≥ k.
• The basic notation for the net premium reserve is k V , with additional sub- and superscripts
indicating specific contracts as in the earlier notation for present values and net single premiums.
• Life insurance policies tend to be designed to have positive net premium reserve throughout
the period of the contract, which implies an expected positive future loss to the insurance
company and hence an expected positive benefit for the insured which should be a motivation
to continue the policy.
• Of course, this expected future loss to the insurer is off-set by premiums that have already
been paid, hence they indeed reflect a reserve for the insurer.
• Reserves are used for a variety of important actions, including risk calculations at any moment in time and to provide opportunities for people to change their insurance contracts
(‘alterations’ or ‘conversions’). For example, an insured person may wish to change (’converse’), after several years in a term life insurance, to a whole life insurance. If this is allowed
without incurring further costs, then the reserve just before the conversion can be transferred
to the new contract.
6.6
Net premium reserve for a whole life insurance policy
• Suppose there is a benefit payment of 1 unit at the end of year of death, and net annual
premium Px paid at start of each year while insured is alive.
• Suppose the future year is k, where we are imagining K ≥ k. The present value in year k of
benefits and payments will be:
Benefit
v
v2
v3
..
.
Premium
Px
Px (1 + v)
Px (1 + v + v 2 )
..
.
v K+1−k
Px a
¨K+1−k
For
K=k
K =k+1
K =k+2
..
.
i.e. lives extra years
0
1
2
..
.
K −k
47
• Thus, the random loss to the insurer of all future benefit payment(s) and premiums at the
end of year k, k = 0, 1, . . ., conditioned on the insured being alive at that moment, is
kL
= v K+1−k − Px a
¨K+1−k
• This is the present value, at the end of year k, of the future benefit payment minus the
present value at that time of all future premiums, and all conditional on the insured being
alive at the end of year k.
• The net premium reserve for this policy at the end of year k is the expectation of this loss
(and with new notation):
k Vx
= E(k L|K ≥ k) = E(v K+1−k |K ≥ k) − Px E(¨
aK+1−k |K ≥ k),
remembering that these are also conditional on K ≥ x. We can think about Q = K − k as
the extra years lived by someone aged x + k, where Q ≥ 0, in which case this expectation is
E(k L|K ≥ k) = E(v Q+1 |X ≥ x + k) − Px E(¨
aQ+1 |X ≥ x + k)
k Vx
= Ax+k − Px a
¨x+k ,
for k = 0, 1, . . .
• This derivation follows the same steps as derivation of Ax and a
¨x .
• A number of relationships can be used to re-express k Vx . For example, recall that we calculate
the premium in this case as
Ax
Px =
,
a
¨x
and also recall the relationship
1 − Ax
.
a
¨x =
d
Some manipulation now reveals
k Vx
=1−
a
¨x+k
Nx+k /Dx+k
=1−
.
a
¨x
Nx /Dx
• Such formulae are used to construct tables of net premium reserves for values of k ≥ 0.
• For example, for age x = 40, we obtain the following net premium reserves (for a final benefit
of one unit) using commutation tables:
k
0
1
2
5
20
40
59
k Vx
0
0.0104
0.0211
0.0555
0.2752
0.6322
0.9399
• The derivations for other kinds of policy are similar.
48
– For example, the net premium reserve for an n-year term life insurance paying 1 unit
benefit at the end of year of death if the insured dies within n years, and financed by net
annual premiums paid at the start of each year of the policy while the insured is alive, is,
for k = 0, 1, . . . , n − 1,
1
k V x:n
1
1
1
= Ax+k:n−k
− P x:n
a
¨x+k:n−k = Ax+k:n−k
−
1
Ax:n
a
¨
a
¨x:n x+k:n−k
– Similarly, for an n-year endowment the net premium reserve is, for k = 0, 1, . . . , n − 1,
k Vx:n
= Ax+k:n−k − Px:n a
¨x+k:n−k
• Numerical examples are rather tedious. For full illustrations of the calculations, see Gerber
(section 6.2). Other examples are shown in the project report Mikola.pdf, pages 6-8, which
is available on the Actuarial Mathematics II website - this was the final year project of
Durham student Matthew Mikola, who graduated with a B.Sc. in 2007.
• The principles are similar when expense-loaded premiums are used.
49