Pension Funds Lecture’s notes (Syllabus)mmartynas/PROGRAMOS/Mag_medziaga-2012-09-07... · Chapter1 Pensionsystems Period when individual is able to work is shorther than the period

Pension FundsLecture’s notes (Syllabus)

by Aldona Skučaite

2011 - 2012

2

Course content

1. Main features of pension systems (see ([1]), ([4]))

(a) Three pension pillars

(b) Classification of pensions

(c) Methods for calculation (determinations) of pension size

(d) Necessity of solidarity between generations

(e) Pension systems in European Union and USA.

2. Demographical and financial models of pension systems (see ([2]), p.p.259-274; 511-520; 533-550 )

(a) Pure and relative probabilities

(b) Lexis diagram

(c) Funding models of pension systems: terminal funding; fundingmethods; individual / aggregate funding methods

3. Pension annuities (see ([3]), p.p. 2-20; 58-63; 137-148; 152-158; 186-195)

(a) Deterministic vs. stochastic approach

(b) Longevity risk: cohort and period mortality tables

(c) Extrapolation methods for construction of projected mortality ta-bles

(d) Single entry mortality table (age adjustment method)

(e) Parametric methods

(f) Lee Carter method and its modifications

4. Longevity risk (see ([3]), p.p. 267-303; 318-333)

(a) Coefficient of variation and other risk measures

(b) Risk and mortality scenario

(c) Pooling and non pooling parts of risk

(d) Methods for management of Longevity risk: hedging, reinsurance,longevity bonds, etc.

Chapter 1

Pension systems

Period when individual is able to work is shorther than the period ofconsumption. During periods when individual is unable to work for remu-neration (is unable to earn for living) he / she must either relay on thefinancial support from others or to ensure his financial situation by savings(investing) and / or participating in some scheme thus acquiring the rightfor pension benefits. To save for retirement means to delay consumption oftoday for the sake of the future.

Typical problems that my arise when saving for retirement:

1. Unknown period of retirement (including longevity risk)

2. Unknown level of prices and living standards

3. Unknown needs (e.g. health status)

4. Inability to save (invest) today (dependents, mortgage, low salary)

1.1 Three pension pillars. Classification of pen-sions

Probably in most modern countries pension benefits are provided viathree main pillars:

Ist pillar - Pension provided via public scheme (social insurance, govern-ment, etc.)

IInd pillar - Pension provided via employer

IIIrd pillar - Individual savings / investments

3

4 CHAPTER 1. PENSION SYSTEMS

Intergenerational solidarity is essential part of almost every system. Sol-idarity between generations may be achieved:

• In informal way, through different generations in one family

• Through labor market (employer)

• Through taxation system in particular state

• Through capital markets

Pension systems may be classified according many peculiarities, e.g..

• Basis for provision of pension benefits

Citizenship - pension benefits are provided for citizens of country

Living conditions - pension benefits are provided only for those un-able to maintain lowest acceptable living standard

Employment - pension benefits are provided for those who were em-ployed in labour market for specified period

Profession - pension benefits are provided for specific professionalsonly

Individual contracts - pension benefits are provided for those whoconcluded individual contracts

• Method for determination of pension size:

Defined benefit (DB)

Defined Contribution (DC)

Flat rate

Means tested

• Administration of scheme

Public

Private

• Method of financing

PAYG

Funded system

1.2. PENSION SYSTEMS IN EU AND USA 5

1.2 Pension systems in EU and USA

1.3 Sample exercises1 Exercise. Prepare short (5-10 min.) presentation about pension systemin specific country. Main accents:

• Combination of (cooperation between) three main pillars

• Methods of financing

• Basis for provision of pension benefit

• Other essential information

6 CHAPTER 1. PENSION SYSTEMS

Chapter 2

Demographical and financialmodels of pension systems

2.1 Population models

2.1.1 Closed population with one possibility of with-drawal

Lifetime of newborn is considered to be continous random variable X,determined in [0, ω], where 0 < ω <∞ - maximum age of individual.

2.1.1 Definintion. F (x) = P [X ≤ x]; x ≥ 0 is d.f. of X and s(x) =F (x) = 1− F (x) = P [X > x] - X is called survival function.

Obvious that F (0) = 0 and s(0) = 1.

It is easy to see that:

P (x < X ≤ z) = F (z)− F (x) = s(x)− s(z), (2.1.1)

P (x < X ≤ z|X > x) =F (z)− F (x)

1− F (x)=s(x)− s(z)

s(x). (2.1.2)

For a person aged (x), (0 ≤ x ≤ ω) future life time is defined by T (x).We will use notation:

tqx = P (T (x) ≤ t); t ≥ 0

tpx = 1− tqx = P (T (x) > t); t > 0;

t|uqx = P (t < T (x) ≤ t+ u) =t+u qx −t qx =t px −t+u px; (2.1.3)

7

8CHAPTER 2. DEMOGRAPHICAL AND FINANCIAL MODELS OF PENSION SYSTEMS

andxp0 = s(x). (2.1.4)

Alternative expressions:

tpx =x+tp0

xp0

=s(x+ t)

s(x); (2.1.5)

tqx = 1− s(x+ t)

s(x);

t|uqx =s(x+ t)− s(x+ t+ u)

s(x).

and:

t|uqx =s(x+ t)

s(x)· s(x+ t)− s(x+ t+ u)

s(x+ t)= tpx · uqx+t. (2.1.6)

If function F (x) is differentiable then F ′(x) = f(x) and:

P (x < X ≤ x+M x|X > x) =F (x+ M x)− F (x)

1− F (x)≈ f(x)

1− F (x)M x. (2.1.7)

Let’s defineµx =

f(x)

1− F (x)=−s′(x)

s(x), (2.1.8)

µx - is called force of mortality at age x.

Using properties of f(x) and F (x) we get:

µx ≥ 0. (2.1.9)

Moreover:

d(ln(s(y))) =s′(y)

s(y)dy = −µydy, (2.1.10)

Integrating we get:

−∫ x+n

x

µydy = lns(x+ n)

s(x)= ln npx, (2.1.11)

or equivalently:

npx = e−∫ x+n

x µydy (2.1.12)

2.1. POPULATION MODELS 9

and

xp0 = s(x) = e−∫ x0 µsds. (2.1.13)


F (x) = 1− s(x) = 1− e−∫ x0 µsds (2.1.14)

andf(x) = F ′(x) = e−

∫ x0 µsds · µx = xp0µx. (2.1.15)

Analogical formulas for random variable T (x) with d.f. G(t) and p.d.f.g(t) are:

tqx = P (T (x) ≤ t), (2.1.16)

G(t) = tqx, (2.1.17)

g(t) = G′(t) =d

dttqx.

Then

g(t) =d

dttqx =

d

dt(1− s(x+ t)

s(x)) = −s

′(x+ t)

s(x)=s(x+ t)

s(x)· (−s

′(x+ t)

s(x+ t)).

(2.1.18)

g(t) = tpxµx+t; t ≥ 0 (2.1.19)is probability that (x) will die in interval: (t; t+ dt). Then:∫ ∞

0tpxµx+t = 1. (2.1.20)

(since g(t) - is density of random variable).Using

d

dttqx =

d

dt(1− tpx) = − d

dttpx = tpxµx+t. (2.1.21)

we getlimn→∞ npx = 0, (2.1.22)

thenlimn→∞

(− ln(npx)) =∞ (2.1.23)

and finally

limn→∞

∫ x+n

x

µydy =∞. (2.1.24)


2.1.1.1 Mortality table

Suppose l0 is set of newborns (e.g. l0 = 100000). For a newborn j =1, 2, ..., l0 his / her surival function is the same - s(x). Let, L(x) - set ofindividuals from l0 who attained age x. Then:

L(x) =

l0∑j=1

Ij, (2.1.25)

where

Ij =

{1, if individual attained age x;0, otherwise. (2.1.26)


E(Ij) = s(x) (2.1.27)

and then

E(L(x)) =

l0∑j=1

E(Ij) = l0s(x). (2.1.28)

We will definelx:lx = l0s(x). (2.1.29)

L(x) has binomial distribution with parameters n = l0, p = s(x).

Let random variable nDx - number of individuals from l0 who died be-tween (x, x+ n]. Consider E(nDx) = ndx, then:

ndx = l0[s(x)− s(x+ n)] = lx − lx+n. (2.1.30)

2.1.2 Theorem. Suppose T (x) - continous random variable with d.f. G(t)and G(0) = 0; G′(t) = g(t). Let z(t) - nonnegative, monotonic and differen-tiable function. Assume that E(z(t)) exist. Then

E[z(t)] =

∫ ∞0

z(t)g(t)dt = z(0) +

∫ ∞0

z′(t)(1−G(t))dt. (2.1.31)

Proof. see Bowers, p.p. 62-63.


Complete expectation of life :

oex= E[T (x)] =

∫ ∞0

ttpxµx+tdt. (2.1.32)

Using Theorem (2.1.2) with z(t) = t ir G(t) = 1− tpx, we get:

oex=

∫ ∞0

tpxdt. (2.1.33)

Let Lx - total expected number of years lived x and x+ 1 by individualsfrom l0:

Lx =

∫ 1

0

tlx+tµx+tdt+ lx+1, (2.1.34)

Easy to see that:

Lx = −∫ 1

0

tdlx+t + lx+1 = −tlx+t|10 +

∫ 1

0

lx+tdt+ lx+1

=

∫ 1

0

lx+tdt. (2.1.35)

Central death rate at age x:

mx =

∫ 1

0lx+tµx+tdt∫ 1

0lx+tdt

=lx − lx+1

Lx. (2.1.36)

2.1.2 Main equations describing development of popu-lation

lx = l0e−

∫ x0 µsds;

lx = l0 −∫ x

0

lsµsds;

lx = ly −∫ x

y

lsµsds; x > y > 0; (2.1.37)

lx = lye−

∫ xy µsds; x > y > 0.


2.1.3 Closed population with several possibilities of with-drawal (decrement). Pure and relative probabili-ties

Again let T (x) - p.f. of future lifetime of (x) . Let discrete randomvariable J = J(x) describe reason of withdrawal from population (e.g. J = 1- death; J = 2 - lapse, etc.). Define by f(t, j) - joint p.d.f. of variables T (x)and J , by h(j) - marginal p.d.f. of J ; and g(t) - marginal p.d.f. of T (x).

Probability to withdraw due to reason j we will define by:

tq(j)x =

∫ t

0

f(s, j)ds; t ≥ 0; j = 1, 2, ...,m. (2.1.38)

Then h(j) =∫∞

0f(s, j)ds = ∞q

(j)x ; t ≥ 0; j = 1, 2, ...,m.

By analogy:

g(t) =m∑j=1

f(t, j); G(t) =

∫ t

0

g(s)ds. (2.1.39)

Let tq(τ)x - probability to withdraw due to any reason. Then:

tq(τ)x = P (T ≤ t) = G(t) =

∫ t

0

g(s)ds;

tp(τ)x = 1− tq

(τ)x ;

µ(τ)x+t =

g(t)

1−G(t)=

1

tp(τ)x

d

dttq

(τ)x = (2.1.40)

= − 1

tp(τ)x

d

dttp

(τ)x = − d

dtln tp

(τ)x .

Note that formulas (2.1.40) are identical to formulas of closed populationwith one possibility of withdrawal.

Moreover,

f(t, j)dt = P (t < T ≤ t = dt, J = j) (2.1.41)= P ((t < T ≤ t = dt) ∩ J = j|T > t) · P (T > t).

Then in analogy with (2.1.19) we may define the force of decrement dueto cause j:

µjx+t =f(t, j)

1−G(t)=f(t, j)

tp(τ)x

, (2.1.42)


then:f(t, j)dt = tp

(τ)x µjx+tdt, j = 1, 2, ...,m; t ≥ 0. (2.1.43)

Differentiating (2.1.38) and using (2.1.40) we obtain:

µ(j)x+t =

1

tp(τ)x

d

dttq

(j)x . (2.1.44)

Then from (2.1.38), (2.1.39) and (2.1.40) we get:

tq(τ)x =

∫ t

0

g(s)ds =

∫ t

0

m∑j=1

f(s, j)ds = (2.1.45)

=m∑j=1

∫ t

0

f(s, j)ds =m∑j=1

tq(τ)x .

And:

µ(τ)x+t =

m∑j=1

µ(j)x+t. (2.1.46)

So:

f(t, j) = tp(τ)x µ

(j)x+t

g(t) = tp(τ)x µ

(τ)x+t

h(j) = ∞q(j)x

h(j|T = t) =f(t, j)

g(t)=

tp(τ)x µ

(j)x+t

tp(τ)x µ

(τ)x+t

=µ

(j)x+t

µ(τ)x+t

(2.1.47)

tq(j)x =

∫ t

0sp

(τ)x µ

(j)x+sds

2.1.3.1 Multiple decrement table

Suppose we investigate group of individuals l(τ)a , each of them may with-

draw due to reasons µ(τ)y ; y ≥ a. Then:

l(τ)x = l(τ)

x exp[−∫ x

a

µ(τ)y dy], (2.1.48)

so total number of withdrawals is:


d(τ)x = l(τ)

x − l(τ)x+1 = l(τ)

x (1−l(τ)x+1

l(τ)x

) =

= l(τ)x (1− exp[−

∫ x+1

x

µ(τ)y ])) = l(τ)

x (1− p(τ)x ) = l(τ)

x q(τ)x ,(2.1.49)

and:

µ(τ)x = − 1

l(τ)x

dl(τ)x

dx. (2.1.50)

Suppose l(j)x , j = 1, 2, ...,m, - subset of individuals aged (x) from l(τ)a who

will withdraw due to reasonj, then:

l(τ)x =

m∑j=1

l(j)x (2.1.51)

and

µ(j)x = lim

h→0

l(j)x − l(j)x+1

hl(τ)x

= − 1

l(τ)x

dl(j)x

dx. (2.1.52)

So:

µ(τ)x = − 1

l(τ)x

d

dx

m∑j=1

l(j)x =m∑j=1

µ(j)x . (2.1.53)

l(j)x − l(j)x+1 = d(j)

x =

∫ x+1

x

l(τ)y µ(j)

y dy. (2.1.54)

Dividing (2.1.54) by l(τ)x we obtain:

d(j)x

l(τ)x

= q(j)x =

∫ x+1

xy−xp

(τ)x µ(j)

y dy, (2.1.55)

where q(j)x - probability to withdraw due to reason j when there are m

possible causes of withdrawal.

2.1.3.2 Associated single decrement table

Now we will introduce so called "pure" (independent, net, uncom-peting) probabilities of withdrawal, that is probabilities of decrement in casethere are no other causes of decrement:


tp′(j)x = exp[−

∫ t

0

µ(j)x+sds] (2.1.56)

tq′(j)x = 1− tp

′(j)x (2.1.57)

Now we will not require:

limt→∞ tp

′(j)x = 0.

But from∫ x

0µ

(τ)x+t =∞ it is obvious that

∫ x0µ

(j)x+t =∞ for at least one j.

Obviuos that

tp(τ)x = exp[−

∫ t

0

(µ(1)x+s + µ

(2)x+s + ...+ µ

(j)x+s)] =

j∏i=1

tp′(i)x , (2.1.58)

sotp

(τ)x ≤ tp

′(j)x ∀j,

andtp

(τ)x µ

(j)x+t ≤ tp

′(j)x µ

(j)x+t. (2.1.59)

Integrating (2.1.59) we get:

q(j)x =

∫ 1

0tp

(τ)x µ

(j)x+tdt ≤

∫ 1

0tp

′(j)x µ

(j)x+tdt = q

′(j)x . (2.1.60)

Let j = 2 and t = 1, then from (2.1.45) ir (2.1.59) we have:

p(τ)x = p

′(1)x · p′(2)

x = (1− q′(1)x ) · (1− q′(2)

x ); ⇒⇒ 1− q(τ)

x = (1− q′(1)x ) · (1− q′(2)

x ); ⇒⇒ q(τ)

x = q′(1)x + q

′(2)x − q′(1)

x · q′(2)x ; ⇒ (2.1.61)

⇒ q(1)x + q(2)

x = q′(1)x + q

′(2)x − q′(1)

x · q′(2)x .

Assume that deaths are uniformly distributed during the year, e.g.

tq(j)x = t · q(j)

x ; j = 1, 2...,m; 0 ≤ t ≤ 1,

then we have:

tq(τ)x = t · q(τ)

x = 1− p(τ)x .


From (2.1.44):

µ(j)x+t · tp(τ)

x =d

dttq

(j)x = q(j)

x .

Then

µ(j)x+t =

q(j)x

tp(τ)x

=q

(j)x

1− tq(τ)x

,

so

q′(j)x = 1− exp[−

∫ 1

0

µ(j)x+tdt] = 1− exp[−

∫ 1

0

q(j)x

1− tq(τ)x

dt]

= 1− exp[q

(j)x

q(τ)x

ln(1− tq(τ)x )|10] = 1− exp[

q(j)x

q(τ)x

ln(1− q(τ)x ).

2.2 Demographical models of pension funds

2.2.1 Lexis diagram

See ([2]), p.p. 511-518.

2.2.1.1 Stationary and stable populations

See ([2]), p.p. 518 - 525

2.3 Funding methods in pension funds

Assume that all members enter pension fund at age a, retirement age - - r.Survival function - s(x) with s(a) = 1. For a ≤ x < r, there may be severalcauses of withdrawal, for x ≥ r - the only cause of withdrawal is death.

Let

l(x, u) = n(u)s(x); u = t− x+ a, (2.3.1)

where n(u) - density of new entrants at time u = t−x+a - time of joiningpension fund.

Suppose that at t = 0 person is aged x (a ≤ x < r). Salary rate for suchperson is w(x), let g(t) - inflation related factor, then annual salary rate of

2.3. FUNDING METHODS IN PENSION FUNDS 17

(x) at time t is:w(x)g(t); a ≤ x < r.

Total annual salary at time t:

W(t) =

∫ r

a

l(x, t− x+ a)w(x)g(t)dx.

Suppose pension replacement factor is f (calculated as percentage of lastsalary), then:

fw(r)g(t).

Assume pension is indexed by factor h(x); h(r) = 1, so annual pensionbenefit:

fw(r)g(t− x+ r)h(x); x ≥ r.

2.3.1 Terminal funding

Actuarial present value of continuous pension annuity:

ahr =

∫ ∞r

e−δ(x−r)h(x)s(x)

s(r)dx,

where δ - interest rate.Then Normal cost rate, or Required contribution rate) for individ-

uals attaining age r between t and t+ dt is:

PT(t) = fw(r)g(t)l(r, t− r + a)ahr

2.3.2 Main actuarial functions of pension fund

2.3.2.1 Accrual functions of Actuarial liability

LetM(x) - nondecreasing, right continuous function and 0 ≤M(x) ≤ 1; x ≥a. Assume that M(a) = 0; M(x) = 1; x ≥ r.

For terminal funding:

M(x) =

{0, x < a;1, x ≥ a.

Suppose there exist function m(x):


M(x) =

∫ x

a

m(y)dy;x ≥ a.

Assume that

m(x) =

continuous, a < x < r;right continuous, x = a;left continuous, x = r;0, x > r.

Then

M ′(x) = m(x), a < x < r.

2.3.2.2 Actuarial present value of future pensions for active mem-bers (aA)(t)

(aA)(t) =

∫ r

a

e−δ(r−x)PT (t+ r − x)dx. (2.3.2)

Notice that:

∂

∂tPT (t+ r − x) = − ∂

∂xPT (t+ r − x).

Then:

d

dt(aA)(t) =

∫ r

a

e−δ(r−x) ∂

∂tPT (t+ r − x)dx

= −∫ r

a

e−δ(r−x) ∂

∂xPT (t+ r − x)dx

= −e−δ(r−x)PT (t+ r − x)|ra + δ

∫ r

a

e−δ(r−x)PT (t+ r − x)dx

= e−δ(r−a)PT (t+ r − a)−PT (t) + δ(aA)(t).

2.3.2.3 Normal cost rate - P(t)

P(t) =

∫ r

a

e−δ(r−x)PT (t+ r − x)m(x)dx, (2.3.3)

It is easy to show that:


∫ u+r−a

u

PT (u+ r − a)m(a+ t− u)dt = PT (u+ r − a)

∫ u+r−a

u

m(a+ t− u)dt =

= PT (u+ r − a)(M(r)−M(a)) = PT (u+ r − a).

2.3.2.4 Accrued actuarial liability - (aV)(t)

(aV)(t) =

∫ r

a

e−δ(r−x)PT (t+ r − x)M(x)dx (2.3.4)

Integrating (2.3.3) by parts obtain:

P(t) =

∫ r

a

e−δ(r−x)PT (t+ r − x)dM(x)

= e−δ(r−x)PT (t+ r − x)M(x)|rx=a −

δ

∫ r

a

e−δ(r−x)PT (t+ r − x)M(x)dx+

∫ r

a

e−δ(r−x) ∂

∂tPT (t+ r − x)M(x)dx

= PT (t)− δ(aV )(t) +d

dt(aV )(t),

or equivalently:

P(t) + δ(aV)(t) = PT (t) +d

dt(aV)(t).

2.3.2.5 Actuarial present value of future normal costs - (Pa)(t)

(Pa)(t) =

∫ r

a

e−δ(r−x)PT(t+ r − x)

∫ r

x

m(y)dydx (2.3.5)

or

(Pa)(t) =

∫ r

a

e−δ(r−x)PT(t+ r − x)(1−M(x))dx

From (2.3.2), (2.3.4) ir (2.3.5) we obtain:

(aV)(t) = (aA)(t)− (Pa)(t),

or

(aA)(t) = (aV)(t) + (Pa)(t).


2.3.3 Funding methods

2.3.3.1 Individual funding methods

Different ways to choose m(x) and M(x)

2.3.3.1.1 Accrued benefit cost method

m(x) =1

r − a; M(x) =

x− ar − a

.

or

m(x) =w(x)∫ r

aw(y)dy

; M(x) =

∫ xaw(y)dy∫ r

aw(y)dy

.

2.3.3.1.2 Entry age accrual cost method

π

∫ r

a

e−δ(y−a)s(y)dy = e−δ(r−a)s(r)ahr .

(aV )(x) = e−δ(r−x) s(r)

s(x)ahr −

π︷︸︸︷(e−δ(r−a)s(r)ahr∫ rae−δ(y−a)s(y)dy

)∫ r

x

e−δ(y−x) s(y)

s(x)dy

= (aA)(x)−

(aA)(x)︷︸︸︷e−δ(r−x) s(r)

s(x)ahr ·

eδ(r−x)e−δ(r−a)∫ rxe−δ(y−x)s(y)dy∫ r

ae−δ(y−a)s(y)dy

= (aA)(x)

(1−

∫ rxe−δys(y)dy∫ r

ae−δys(y)dy

)= (aA)(x)M(x),

so

M(x) = 1−∫ rxe−δys(y)dy∫ r

ae−δys(y)dy

=

∫ xae−δys(y)dy∫ r

ae−δys(y)dy

and:

m(x) =e−δxs(x)∫ raeδys(y)dy

.

If


π

∫ r

a

e−δ(y−a)s(y)w(y)dy = (aA)(a) = e−δ(r−a)s(r)ahr ,

then

M(x) = 1−∫ rxe−δys(y)w(y)dy∫ r

ae−δys(y)w(y)dy

and

m(x) =e−δxs(x)w(x)∫ r

ae−δys(y)w(y)dy

.

2.3.3.2 Aggregate funding methods

(aU)(t) = (aV)(t)− (aF)(t). (2.3.6)

d

dt(aF)(t) = (aC)(t) + δ(aF)(t)−PT(t), aF)(0) = const (2.3.7)

Assume:

(aC)(t) = P(t) + λ(t)(aU)(t), (2.3.8)

Leta(t) =

(Pa)(t)

P(t)

and:

λ(t) = 1/a(t)

Then:

(aC)(t) = P(t) +(aV)(t)− (aF)(t)

a(t)

=(Pa)(t) + (aV)(t)− (aF)(t)

a(t)=

(aA)(t)− (aF)(t)

a(t),

because (aA)(t) = (Pa)(t) + (aV)(t), so

(aC)(t)a(t) = (aA)(t)− aF)(t).


From (2.3.7) and (2.3.8):

d

dt(aF)(t) = (aC)(t) + δ(aF)(t)−PT(t) (2.3.9)

= P(t) +(aU)(t)

a(t)+ δ(aF)(t)−PT(t)

Remember:

P(t) + δ(aV)(t) = PT (t) +d

dt(aV)(t)

so

d

dt(aV)(t) = P(t) + δ(aV)(t)−PT (t) (2.3.10)

Subtract (2.3.9) from (2.3.10):

− d

dt(aU)(t) = −δ(aU)(t) +

(aU)(t)/a(t)︷︸︸︷(aC)(t)− P (t)

= (1

a(t)− δ)(aU)(t),

sod

dt(aU)(t) = −(

1

a(t)− δ)(aU)(t).

Then: ∫ t

0

ddu

(aU)(u)

(aU)(u)= −

∫ t

0

(1

a(u)− δ)du

and finally:

(aU)(t) = (aU)(0) exp [−∫ t

0

(1/a(u)− δ)du],

or

(aF)(t) = (aV)(t)−

−((aV)(0)− (aF)(0)) exp [−∫ t

0

(1/a(u)− δ)du].

We need to show that exp [−∫ t

0(1/a(u)− δ)du]→ 0, t→∞.


Suppose a(u) < a∞| = 1/δ, so

1/a(u)− δ ≥ ε > 0,

then:

exp [−∫ t

0

(1/a(u)− δ)du]→ 0, t→∞

and:

(aF)(t)→ (aV)(t).

2.3.3.3 Actuarial methods for retired members

(rA)(t) =

∫ ∞r

l(x, t− x+ a)fw(r)g(t− x+ r)ahxdx (2.3.11)

(rA)(t) =

∫ ∞r

n(t− x+ a)fw(r)g(t− x+ r)[

∫ ∞x

e−δ(y−x)h(y)s(y)dy]dx

(2.3.12)

(rV)(t) = (rA)(t)

B(t) =

∫ ∞r

l(x, t− x+ a)fw(r)g(t− x+ r)h(x)dx,

or:

B(t) =

∫ ∞r

n(x, t− x+ a)g(t− x+ r)fw(r)s(x)h(x)dx, (2.3.13)


d

dt(B(t)) =

∫ ∞r

fw(r)s(x)h(x)∂

∂t[n(x, t− x+ a)g(t− x+ r)]dx

= −∫ ∞r

fw(r)s(x)h(x)∂

∂x[n(x, t− x+ a)g(t− x+ r)]dx

= −fw(r)s(x)h(x)n(x, t− x+ a)g(t− x+ r)|x=∞x=r +∫ ∞

r

fw(r)n(x, t− x+ a)g(t− x+ r)[

s′(x) = −s(x)µx︷︸︸︷s′(x) h(x) + s(x)h′(x)]dx

=

"replacement effect"︷︸︸︷fw(r)l(r, t− r + a)g(t)−

∫ ∞r

fw(r)l(x, t− x+ a)g(t− x+ r)h(x)µxdx

"adjustment effect"︷︸︸︷+

∫ ∞r

fw(r)l(x, t− x+ a)g(t− x+ r)h′(x)dx =

= I − II + III

(rV)(t) = (rA)(t) =

∫ ∞r

n(t−x+a)fw(r)g(t−x+r)[

∫ ∞x

e−δ(y−x)h(y)s(y)dy]dx

Then:

d

dt(rV)(t) = fw(r)

∫ ∞r

[

∫ ∞x

e−δ(y−x)h(y)s(y)dy]∂

∂t[n(t− x+ a)g(t− x+ r)]dx

= −fw(r)

∫ ∞r

[

∫ ∞x

e−δ(y−x)h(y)s(y)dy]∂

∂x[n(t− x+ a)g(t− x+ r)]dx

= −fw(r)[

∫ ∞x

e−δ(y−x)h(y)s(y)dy][n(t− x+ a)g(t− x+ r)]|∞x=r +

fw(r)δ

∫ ∞r

[

∫ ∞x

e−δ(y−x)h(y)s(y)dy][n(t− x+ a)g(t− x+ r)]dx−

fw(r)

∫ ∞r

s(x)h(x)n(t− x+ a)g(t− x+ r)dx

= PT(t) + δ(rV)(t)−B(t)

so,

PT(t) + δ(rV)(t) =d

dt(rV)(t) + B(t) (2.3.14)

2.4. SAMPLE EXERCISES 25

2.3.3.4 Mutual functions for active and retired members

A(t) = (aA)(t) + (rA)(t)

V(t) = (aV)(t) + (rV)(t)

Remember that

P(t) + δ(aV)(t) = PT (t) +d

dt(aV)(t) (2.3.15)

from (2.3.14) and (2.3.15) we get:

P(t) + δV(t) = B(t) +d

dtV(t)

2.4 Sample exercises

2 Exercise. Assume pension fund was established 2 years ago at t = 0, sonow it is time moment t = 2 (end of 2nd year). At time t = 0 cohort ofemployees that joined pension plan was:

Age (x) No of employees30 100035 100040 1000

Assume that population is closed. For 30 ≤ x ≤ 65, then s(x) =e−µx(x−a); µx = 0, 0007+0, 00005·100.04·x. For 65 ≤ x ≤ 110, qx = GHx

1+GHx ;G =0.0000023;H = 1.134

Suppose at time t = 0: w(30) = 1000;w(35) = 1025;w(40) = 1050,annual salary increase is 0,5%. Let i = 0, 04, r = 65, initial pension benefit- 60% (of last salary), pension is indexed by 0,5% annually.


Assume:

m1(x) =1

r − a

m2(x) =w(x)∫ r

aw(y)dy

m3(x) =e−δxs(x)∫ r

ae−δys(y)dy

m4(x) =e−δxs(x)w(x)∫ r

ae−δys(y)w(y)dy

Assume that pension is accrued by level percentage π of current salary.Find π and (aV (2)).

3 Exercise. There are three decrement causes from pension fund: 1 - death;2 - change of employer; 3 - disability. Assume:

µ1(x) = 0, 0007 + 0, 00005 · 100,04x

µ2(x) = 0, 005

µ3(x) = 0.00007x

1 Part. Construct multiple decrement table (calculate q(1)x ; q

(2)x ; q

(3)x ; d

(1)x ; d

(2)x ; d

(3)x

and l(τ)55 , if x = 55, 56, ...., 65, l(τ)

x = 1000):Age (x) q

(1)x q

(2)x q

(3)x l

(τ)x d

(1)x d

(2)x d

(3)x

55 100056...65

2 Part. Active members of pension fund are entitled for pension benefit.

Assume that a65 = 11, 45 (first payment at age 65). Pension benefit - 30%of last salary, initial salary w(55) = 12000, annual salary increase - 0.5%.

Method of accrual - Entry age accrual cost method, interest rate - i = 4%.Premiums into fund are paid at the end of year (so, first payment - at age56; last - at age 65).

Calculate:


a) Percentage π1, if funding is based on level salary percentage throughoutcareer.

b) Percentage π1, if funding is based on level premiums (in monetaryterms) throughout career.

c) In both cases calculate P (x) and compare them.


Chapter 3

Pension annuities

3.1 Early actuarial modelsJan de Witt (1671) model:

ax = a1| · qx + a2| ·1| qx + a3| ·2| qx + ....

E. Halley (1693) model and mortality table of city of Breslau:

ax =ω−x−1∑t=1

(1 + i)−tLx+t

Lx

3.2 Life annuities

3.2.1 Life annuities: deterministic vs stochastic approach

See ([3]), p.p. 2-39.

3.2.1.1 Deterministic approach

Let S - annuity premium, b - amount of pension (paid at the end of eachyear), lx - number of individuals aged (x) at t = 0. Then, assuming thatlxV0 = lxS:

lx+tVt = lx+t−1Vt−1(1 + i)− lx+tbt, t = 1, 2, ...ω − x. (3.2.1)

Or:Vt =

lx+t−1

lx+t

Vt−1(1 + i)− b,

29

30 CHAPTER 3. PENSION ANNUITIES

and:

Vt − Vt−1 = Vt−1i+lx+t−1 − lx+t

lx+t

Vt−1(1 + i)− b.

So we obtain:

S =ω−x∑t=1

blx+t

lx(1 + i)−t.

Or, using survival probabilities:

S = bω−x∑t=1

tpx(1 + i)−t.

Finally:

S = bω−x∑t=1

at|tpxqx+t, (3.2.2)

where:at| =

1− (1 + i)−t

i

We may also write:

S = b · E(aKx) = bax,

where Kx = Int(Tx).

On the other hand we may show that:

Vt = bax+t.

Vt, depending on situation, may be regarded as assets or liabilities.

3.2.1.1.1 Mutuality effect From (3.2.1):

lx+t−1Vt−1(1 + i) = lx+tVt + lx+tbt, t = 1, 2, ...ω − x.

Moreover (3.2.1) may be presented:

Vt = Vt−1(1 + i)(1 +lx+t−1 − lx+t

lx+t

)− b

= Vt−1(1 + i)(1 + θx+t)− b,

3.3. INTRODUCING LONGEVITY RISK: COHORT VS PERIOD MORTALITY TABLES31

where:θx+t =

lx+t−1 − lx+t

lx+t

,

or:θx+t =

1

px+t−1

− 1.

θx+t is called (interest from mutuality, or mortality drag).

3.2.1.1.2 Solidarity

3.2.1.1.3 Tontine annuities

3.2.1.2 Stochastic approach

3.2.1.3 Random present value of life annuity

Y = aKx|

Possible values of such random variable:

y0 = 0

y1 = (1 + i)−1

....

yω−x = (1 + i)−1 + (1 + i)−2 + ...+ (1 + i)−(ω−x)

Then:P(aKx| = yh

)= P (Kx = h) .

3.2.1.4 Portfolio results: stochastic approach

3.3 Introducing longevity risk: Cohort vs Pe-riod mortality tables

See ([3]), p.p. 45-62Cohort tablesPeriod tablesSelect mortalitySurvival function, life expectancy, markers or life tables in age

continuous context.


3.4 Forecasting mortality under longevity riskSee ([3]), p.p. 137-172.

3.4.1 Extrapolation of annual death probabilities

3.4.1.1 Reduction factors

Rx(t− t′) > 0; t′ - Reduction factors:

qx(t) = qx(t′) ·Rx(t− t′), t > t′.

3.4.1.2 Exponential formula

qx(t) = qx(t′)rt−t

′

x , t.y.

3.4.1.3 More general exponential formula

qx(t) = ax + bx · ctx,

3.4.2 Projected mortality tables

kp↗x (t) =

k−1∏j=0

(1− qx+j(t+ j)) ;

e↗x (t) =ω−x∑k=1

kp↗x (t) + 1/2.

e↑x(t) =ω−x∑k=1

kp↑x(t) + 1/2;

3.4.2.1 Single entry projected table

qxmin+h(τ)(τ+xmin+h(τ)), qxmin+h(τ)+1(τ+xmin+1+h(τ)), ...qx+h(τ)(τ+x+h(τ)).

h(τ) is called Rueff adjustment:

h(τ) =

≥ 0, τ < τ ;0, τ = τ ;≤ 0, τ > τ .

kp[τ ;h(τ)]↗x

k−1∏j=0

[1− qx+h(τ)+j(τ + x+ h(τ) + j)]

3.4. FORECASTING MORTALITY UNDER LONGEVITY RISK 33

3.4.3 Parametric methods

3.4.4 Exposure to risk and Crude mortality rate

ETRxt - "exposed to risk at age x in year t".

ETRxt ≈−Lxt · qx(t)ln(1− qx(t))

.

µx(t) =Dxt

ETRxt

= mx(t).

mx(t) - Crude death rate.

3.4.5 Lee Carter method

(1992)Let:

µx+ξ1(t+ ξ2) = µx(t); 0 ≤ ξ1, ξ2 ≤ 1.

then

mx = µx.

Lee Carter model:

ln(mx(t)) = ax + bxkt + εx,t; (3.4.1)

E(εx,t) = 0;D(εx,t) = σ2ε .

Constraints:tmax∑tmin

kt = 0;xmax∑xmin

βx = 1

ortmax∑tmin

kt = 0;xmax∑xmin

β2x = 1

Objective function

O =xm∑x=x1

tn∑t=t1

(ln mx(t)− αx − βxkt)2 (3.4.2)


3.4.5.1 Singular value decomposition

tn∑t=t1

ln mx(t) = (tn − t1 + 1)αx + βx

tn∑t=t1

κt = (tn − t1 + 1)αx,

So

αx =1

tn − t1 + 1

tn∑t=t1

ln mx(t),

M =

mx1(t1) · · · mx1(tn)... . . . ...

mxm(t1) · · · mxm(tn)

Z = ln M − α =

ln mx1(t1)− αx1 · · · ln mx1(tn)− αx1

... . . . ...ln mxm(t1)− αxm · · · ln mxm(tn)− αxm

Minimize:

OLS(β, k) =xn∑x=x1

tn∑t=t1

(zxt − βxkt)2.

Using singular value decomposition obtain estimation of parameters

3.4.5.2 Newton - Rapson method

Differentiate O (3.4.2) and equate to zero. Then:

tn∑t=t1

(ln mx(t)− αx − βxkt) = 0;x = x1, x2, ..., xm

xm∑x=x1

βx(ln mx(t)− αx − βxkt) = 0; t = t1, t2, ..., tn

tn∑t=t1

kt(ln mx(t)− αx − βxkt) = 0;x = x1, x2, ..., xm


Using Newton - Rapson procedure obtain estimates:

α(k+1)x = α(k)

x +

∑tnt=t1

(ln mx(t)− α(k)x − β(k)

x k(k)t )

tn − t1 + 1

k(k+1)t = k

(k)t +

∑xm

x=x1β

(k)x (ln mx(t)− α(k+1)

x − β(k)x k

(k)t )∑xm

x=x1(β

(k)x )2

β(k+1)x = β(k)

x +

∑tnt=t1

k(k+1)t (ln mx(t)− α(k+1)

x − β(k)x k

(k+1)t )∑tn

t=t1(k

(k+1)t )2

Finally

αx → αx + βxk

kt = (kt − k)∑

βx

βx =βx∑βx,

3.4.6 Mortality forecasts

Random walk with drift:

kt+1 = a+ kt + εt.

3.5 Sample exercises4 Exercise. Let qx = GHx

1+GHx ;G = 0.0000023;H = 1.134; 65 ≤ x ≤ 110

Assume L65 = 100. Find distributions of L70;L80;L100.

Find distribution of a65|, E[a65|] and V ar[a65|].

5 Exercise. Using real Lithuanian mortality data (for example, from www.mortality.org)find Lee Carter estimates of αx, βx and kt.


Chapter 4

Management of Longevity risk

See ([3]) 267 - 342.

4.1 Main types of mortality risk

Process (insurance) risk

Model and parametric risk

Catastrophic risk

4.1.0.1 Main types of actuarial modeling

Static vs dynamic modelling

4.2 Measuring Longevity risk in a static frame-work

Suppose we have several set of assumptions about future mortality, e.g. weconsider the set of possible mortality scenarios A(τ) (specific scenario fromthis set will be denoted by Ah(τ)).

Let Y (Π)t random present value of benefits paid from portfolio. Let Nt -

random number of pensioners from cohort N0 (concrete realization will bedenoted by nt (N0 = n0)). We will denote policy for j individual by indexj = 1, 2, ..., n0. Then total portfolio at time t is:

Πt = {j|T (j)x > t},

37

38 CHAPTER 4. MANAGEMENT OF LONGEVITY RISK

benefits:B

(Π)t =

∑j:j∈Πi

b(j); j = 1, 2, ...,

present value of benefits d.v. t = 1, 2, ...:

Y(j)t = b(j)a

K(j)x0

Y(Π)t =

∑j:j∈Πi

Y(j)t

Let b(j) = b; ∀j, then:B

(Π)t = bNt.

So:

Y Πt =

ω−x0∑h=t+1

BΠh (1 + i)−(h−t) =

ω−x0∑h=t+1

bNh(1 + i)−(h−t)

Suppose real mortality depend on mortality scenario Ah(τ). Then forhomogenous portfolio:

E(Y

(Π)t |Ah(τ), nt

)= ntE

(Y

(1)t |Ah(τ)

).

If future lifetimes are i.i.d. :

D(Y

(Π)t |Ah(τ), nt

)= ntD

(Y

(1)t |Ah(τ)

).

Coefficient of variation (CV), or Risk index:

CV(Y

(Π)t |Ah(τ), nt

)=

1√nt

√D(Y

(1)t |Ah(τ)

)E(Y

(1)t |Ah(τ)

) .


limnt→∞

CV(Y

(Π)t |Ah(τ), nt

)= 0.

Now let assign "weights" ρh;∑ρh = 1 to mortality scenarios Ah(τ).

Then

E(Y

(Π)t |nt

)= Eρ

[E(Y

(Π)t |A(τ), nt

)]= ntEρ

[E(Y

(1)t |A(τ)

)]= nt

m∑h=1

E(Y

(1)t |Ah(τ)

)ρh = ntE(Y

(1)t ),

4.2. MEASURING LONGEVITY RISK IN A STATIC FRAMEWORK 39

E(Y(1)t ) =

∑mh=1E

(Y

(1)t |Ah(τ)

)ρh.

Moreover:

D(Y

(Π)t |nt

)= Eρ

[D(Y

(Π)t |nt, A(τ)

)]+Dρ

[E(Y

(Π)t |nt, A(τ)

)]= ntEρ

[D(Y

(1)t |A(τ)

)]+ n2

tDρ

[E(Y

(1)t |A(τ)

)]= nt

m∑h=1

D(Y

(1)t |Ah(τ)

)ρh + n2

tDρ

[E(Y

(1)t |A(τ)

)]= nt

m∑h=1

D(Y

(1)t |Ah(τ)

)ρh + n2

t

m∑h=1

[E(Y

(1)t |Ah(τ)

)− E(Y

(1)t )]2

ρh.

Then

CV(Y

(Π)t |nt

)=

√D(Y

(Π)t

)E(Y

(Π)t

)=

√√√√ 1

nt

Eρ[D(Y(1)t |A(τ))]

E2(Y(1)t )

+Dρ[E(Y

(1)t |A(τ))]

E2(Y(1)t )

=

=√I + II.

I - "pooling risk"; II- "non pooling risk".

It is easy to see that

limnt→∞

CV(Y

(Π)t |nt

)=

√√√√Dρ[E(Y(1)t |A(τ))]

E2(Y(1)t )

,

so non pooling risk does not decrease when size of portfolio increases.


4.3 Other methods for management of mortal-ity risk

4.3.1 Hedging

4.3.2 Reinsurance

4.3.3 Longevity bonds

4.4 Sample exercises

6 Exercise. Suppose annuity a65 is paid, interest rate is i, probabilities ofmortality - q65, q66, ...., q100 = 1 and w = 101. Assume that at t = 0 personis aged 65, so annuity is paid. Let Yt - present value of this annuity at timet. Provide formulas for Y0 and Y3 using probabilities qx and / or px. Provideformulas for E[Y0] and D[Y0].

7 Exercise. Suppose that possible mortality scenarios are:Scenario, Ah: I II III IV V

E(Y |Ah) 13, 514 13, 821 14, 271 14, 450 15, 112D(Y |Ah) 20, 114 25, 006 22, 714 21, 978 24, 215

probability: 0.1 0.1 0.6 0.1 0.1

Calculate E(Y0) and D(Y0), if n = 100

8 Exercise. Suppose mortality is modeled using Heligman - Pollard law,e.g.:

qx =G · (Kx)

1 +G · (Kx); ω = 110; x− age.

A1 A2 A3 A4 A5

G 6, 378E − 07 3, 803E − 06 2, 005E − 06 1, 060E − 06 3, 149E − 06H 1, 14992 1, 12347 1, 13025 1, 13705 1, 11962

"weights" 0.1 0.1 0.6 0.1 0.1

Using i = 4% calculate:e65 = E[K65|Ah] =

∑117−65k=0 k ·k|1 qx

and√D[K65|Ah].

For t = {0; 5}, h = 1, 2, 3, 4, 5, N0 = 1 and l b = 1 calculate:


E(Y

(Π)t |Ah

); D

(Y

(Π)t |Ah

)kurY Π

t =117−65∑h=t+1

Nh(1 + i)−(h−t).

For t = 0 calculate:

E(Y

(Π)t

); D

(Y

(Π)t

).

For N0 = 1, 100, 1000 calculate Coefficient of variation:

CV(Y

(Π)t |nt

)=

√D(Y

(Π)t

)E(Y

(Π)t

)=

√√√√ 1nt

Eρ[D(Y (1)t |A(τ))]

E2(Y (1)t )

+Dρ[E(Y (1)

t |A(τ))]

E2(Y (1)t )

;

define pooling and non pooling risk.

9 Exercise. Suppose that mortality is described using III scenariu (see ex-ercise (8)).

Insurance company uses hedging, e.g. pension annuity is paid for (65)year old individual, moreover upon death of insured person (annuity) reserveis paid for beneficiaries. Assume that all deaths occur and benefits are paidat the end of each year. Other conditions as in exercise (8).

Calculate insurance premium and coefficient of variation. Compare bothinsurance premium and coefficient of variation with the case described inexercise (8), explain differences of product riskiness and marketability inboth cases.


Literature

[1] Barr, N. "Reforming Pensions: Myths, Truths and Policy Choices".International Social Security Review, Vol. 55 No. 2. (2002).

[2] Bowers, N.L.; Gerber, H.U.; Hickman, J.C. "Actuarial Mathematics".The Society of Actuaries, (1986).

[3] Pitacco, E., et. al. "Modelling Longevity Dynamics for Pension andAnnuity Business". Oxford University Press (2009).

[4] Werding, M. "After Another Decade of Reform: Do Pension Systemsin Europe Converge?". CESifo DICE Report No. 1 (2003).

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Pension Funds Lecture’s notes (Syllabus)mmartynas/PROGRAMOS/Mag_medziaga-2012-09-07... · Chapter1 Pensionsystems Period when individual is able to work is shorther than the period