TIA MLC Formula Sheet
Content
A. Prerequisite Review
CALCULUS REVIEW
d
dx
ex = ex
UNIFORM DISTRIBUTION
X ∼ uniform a to b
f (x) =
1
b−a
D {f (g(x))} = f 0 (g(x)) · g 0 (x)
F (x) =
x−a
b−a
(uv)0 = u0 · v + u · v 0
R
R
udv = uv − vdu
E[X] =
a+b
2
R
ex dx = ex + c
Var[X] =
(b−a)2
12
CONDITIONAL PROBABILITY
Pr(A ∩ B)
Pr(A|B) =
Pr(B)
f (x, y)
fX (x|y) =
fY (y)
R∞
fY (y) = −∞ f (x, y) dx
v = (1 + i)−1 = e−δ
v 2 = (1 + i)−2 = e−2δ
v n = (1 + i)−n = e−nδ
E[X] = EY [EX [X|Y ]]
Var[X] = EY VarX [X|Y ] + VarY EX [X|Y ]
NORMAL APPROXIMATION
EXPONENTIAL DISTRIBUTION
all x
E[X] =
X
Exponential distribution is memoryless
xp(x)
all x
E[g(X)] =
X
f (x) =
g(x) · p(x)
all x
Var[X] = E[X 2 ] − (E[X])2
X
F (x) = Pr[X ≤ x] =
p(y)
1
1 −µx
e
µ
1x
−µ
F (x) = 1 − e
Var[Y ] = (a − b)2 · p(1 − p)
CONTINUOUS RANDOM VARIABLES
R
x f (x) dx = 1
R
E[g(X)] = x g(x) f (x) dx
Rx
F (x) = Pr[X ≤ x] = −∞
f (y) dy
f (x) =
d
F (x)
dx
v → v2
i → 2i + i2
Central Limit Theorem:
d → 2d − d2
Y = X1 + X2 + · · · + Xn
i
δ
Xi ’s are iid RVs(µx , σx2 )
→
2i+i2
2δ
Y ≈ Normal(E[Y ] = nµx , Var[Y ] = nσx2 )
ANNUITY CERTAINS
E[X] = µ
95th percentile of Y :
Var[X] = µ2
MIXED RANDOM VARIABLES
CONVERT BETWEEN i, d, and δ
Example:
(1)
X=a
p(a)
d
X=b
p(b)
0
i
1
(2)
a<X<b
f (x)
Rb
p(a) + a f (x) dx + p(b) = 1
R
E[X k ] = ak · p(a) + ab xk f (x) dx + bk · p(b)
d
=i
1−d
i
=d
1+i
δ = ln(1 + i)
i
1 − vn
= an
δ
δ
a
¨n − nv n
(Ia)n =
i
n − an
(Da)n =
i
Z ∞
1
t
a
¯∞ =
v dt =
δ
0
Z ∞
1
t
¯a)∞ =
(I¯
tv dt = 2
δ
0
Z n
a
¯n − nv n
¯a)n =
(I¯
tv t dt =
δ
Z0 n
n−a
¯n
t
¯
(D¯
a)n =
(n − t)v dt =
δ
0
a
¯n =
y = µy + σy × Φ−1 (0.95)
y≤x
Bernoulli Shortcut for the RV:
(
a p(a) = p
Y =
b p(b) = 1 − p
DOUBLE THE FORCE OF INTEREST
1 + i → (1 + i)2
median = mean
DISCRETE RANDOM VARIABLES
X
p(x) = 1
DISCOUNTING
(1)
(2)
ACCUMULATING
COVARIANCE
1 + i = eδ
Var(aX + bY ) = a2 Var(X) + b2 Var(Y )
+2ab Cov(X, Y )
(1 + i)2 = e2δ
(1 + i)n = enδ
Cov(X, Y ) = E(XY ) − E(X)E(Y )
If X and Y are independent
then covariance is 0.
c
2012
The Infinite Actuary
B. Survival Models and Life Tables
RANDOM VARIABLES
LIFE TABLES
Tx ∼ future lifetime of (x)
lx − expected number of survivors at (x)
TEMPORARY LIFE EXPECTANCY
Z n
◦
ex:n = E[Wx ] =
t px dt
0
Kx ∼ number of completed future years by
(x) prior to death.
n dx
− expected number of deaths between
ages x and x + n
Kx = bTx c
ex:n = E[Kx ∧ n] =
E[Wx2 ] = 2
k px
Out of ω births, one dies every year until they
are all dead.
Future lifetime is uniformly distributed from 0
to ω − x.
k=1
n
Z
n
X
DE MOIVRE’S LAW
`x = k (ω − x)
t t px dt
0
Wx ∼ future lifetime or n years whichever is
less.
Wx = Tx ∧ n
FORCE OF MORTALITY
Intuitively, µx+t dt is the probability that a life
age x + t will die in the next instant.
CUMULATIVE DISTRIBUTION
Probability of (x) dying before age x + t.
Fx (t) = Pr(Tx ≤ t)
d
t qx
◦
d `
− dt
x+t
`x+t
∗
n px
d
p
dt t x
= (n px )k
= t px µx+t
= −t px µx+t
Probability of a (x) attaining age x + t.
ACTUARIAL NOTATION
PROBABILITY DENSITY FUNCTION
◦
◦
= Sx (t)
t+u px
t qx
= t px · u px+t
n px
Age doesn’t matter.
ex =
Future lifetime is distributed exponentially
with mean 1/µ.
t qx
= 1 − t px
t|u qx
t| qx
t|u qx
= Sx (t) − Sx (t + u)
= t px − t+u px
= t+u qx − t qx
= t px · u qx+t
◦
for all x
= e−nµ = (px )n
1
µ
1
µ2
Recursion
ex = px (1 + ex+1 )
◦
ex:n = ex (1 − n px )
ex:n = ex (1 − n px )
Discrete Constant Force − Kx ∼ geometric
fKx (k) = pk · q
ex = E[Kx ] =
COMPLETE EXPECTATION OF LIFE
Average future lifetime of (x).
Z ∞
◦
ex = E[Tx ] =
t px dt
Z ∞0
E[Tx2 ] = 2
t t px dt
Var[Kx ] =
p
q2
=
m
ω−x
ω−x−n
ω−x
ω−x
2
half way to omega
= n px (n) + n qx n
2
◦
ex:n
a(x) =
1
2
Discrete DML
=
µx =
`x+t = `x e−µt
k=1
For integral x, ex is the expected number of
future birthdays.
=
1
ω−x
MODIFIED DML
Sx (t) = e−µt
◦
= µx
n
ω−x
=
◦
k| qx
Var[Tx ] =
− probability that (x) will survive t years
and die within the following u years.
is the PDF for Kx
n px
fx (t) = Fx0 (t) = −Sx0 (t) = t px µx+t
CURTATE EXPECTATION OF LIFE
∞
X
ex = E[Kx ] =
k px
n qx
ω−x−t
ω−x
1
ω−x
qx =
CONSTANT FORCE
ex =
− probability that (x) dies within t year
t qx = Fx (t)
◦
n|m qx
Intuitively, the probability that (x) dies at age
x + t.
t px − probability that (x) will attain age
x + t.
t px
Sx (t) =
ex:m+n = ex:m +m px ex+m:n
◦
◦
ex ≈ px 1+ ex+1 + qx 12
µx+t = µ
Sx (t) = Pr(Tx > t) = 1 − Fx (t)
◦
ex = ex:n +n px ex+n
If µ∗x+t = k µx+t :
d
q
dt t x
SURVIVAL DISTRIBUTION
fx (t)
Sx (t)
=
= dtp
t x
Rn
n px = exp − 0 µx+t dt
µx+t =
◦
1
ω−x
µx =
Recursion
p
q
a
ω−x
`x = (ω − x)a
a
Sx (t) = ω−x−t
ω−x
a
ω−x−n
n px =
ω−x
◦
ex =
ω−x
a+1
FRACTIONAL AGES
Uniform Distribution of Deaths (UDD)
- Use linear interpolation.
Constant Force
- Use exponential interpolation.
0
c
2012
The Infinite Actuary
SELECT MORTALITY
[age] = age selected
Read across and then down the table.
GOMPERTZ LAW
µx = Bcx
c > 1, B > 0
i
B
cx (ct − 1)
t px = exp − ln c
h
MAKEHAM LAW
µx = A + Bcx
c > 1, B > 0, A ≥ −B
h
i
B
cx (ct − 1)
t px = exp(−At) · exp − ln c
c
2012
The Infinite Actuary
C. Insurance Benefits
PURE ENDOWMENT
TERM INSURANCE
(
v Tx 0 ≤ Tx ≤ n
1
¯
Z x:n =
0
Tx > n
Z n
¯1 =
A
v t t px µx+t dt
x:n
MLC discount factor
(
0
0 ≤ Tx ≤ n
Z x:n1 =
v n Tx > n
0
n Ex
=
Ax:n1
=
vn
¯x
¯x = A
¯1 + n| A
A
x:n
Z n+m
¯
v t t px µx+t dt
n|m Ax =
n px
Var[Z x:n1 ] = 2n Ex − (n Ex )2
= v 2n n px n qx
n
Bernoulli shortcut
(
1
Z x:n
WHOLE LIFE INSURANCE
1
Ax:n
v Kx +1 Kx = 0, 1, . . . , n − 1
=
0
Kx = n, n + 1, . . .
n−1
X
=
v k+1 k| qx
0
Zx = v Kx +1 , Kx = 0, 1, 2, . . .
∞
X
Ax =
v k+1 k| qx
k=0
N-YEAR DEFERRED WHOLE LIFE
(
0
0 ≤ Tx ≤ n
¯
Z
=
x
n|
v Tx Tx > n
Z ∞
¯
v t t px µx+t dt
n| Ax =
n
¯ = n Ex · A
¯x+n
(
0
Kx = 0, 1, . . . , n − 1
n| Zx =
v Kx +1 Kx = n, n + 1, . . .
1
Ax = Ax:n
+ n| Ax
n+m−1
X
v k+1 k| qx
n|m Ax =
k=n
VARYING TERM INSURANCE
Z n
¯ 1 =
I¯A
tv t t px µx+t dt
x:n
Z0 n
¯A
¯ 1 =
D
(n − t)v t t px µx+t dt
x:n
Ax = vqx + vpx Ax+1
Draw a pain curve.
Ax:n = vqx + vpx Ax+1:n−1
For constant force:
(IA)x = vqx + vpx (Ax+1 + (IA)x+1 )
= n Ex · Ax+n
INCREASING WHOLE LIFE
¯ = Tx v Tx , Tx ≥ 0
I¯Z
x
Z ∞
¯ =
I¯A
tv t t px µx+t dt
x
0
Kx = 0, 1, 2, . . .
Pr(SBP is sufficient) =
µ
µ+δ
µ/δ
(IA)x = Ax + vpx (IA)x+1
1
= vqx h
(IA)x:n
i
1
1
+vpx (IA)x+1:n−1
+ Ax+1:n−1
1
1
(DA)x:n
= nvqx + vpx (DA)x+1:n−1
=
¯x =
A
e−n(µ+δ)
µ
µ+δ
Assuming UDD:
¯x =
A
CONSTANT FORCE
n Ex
CONTINUOUS VS. DISCRETE
¯1 =
A
x:n
i 1
A
δ x:n
¯x:n =
A
i 1
A
δ x:n
(m)
Ax
Ax =
vq
vq+d
¯1 = A
¯x (1 − n Ex ) A 1 = Ax (1 − n Ex )
A
x:n
x:n
µ
−n(µ+δ)
¯
A
=
e
n| x
µ+δ
2
1
1
¯
¯
¯
(I A)x = µ Ax
(IA)x = vq
(Ax )2
i
A
δ x
=
+ n Ex
i
Ax
i(m)
¯ = i (IA)
IA
x
δ
x
1
i
1
¯
I A x:n = δ (IA)x:n
1
i
1
¯
DA
x:n = δ (DA)x:n
¯ = i (IA) − i
I¯A
x
δ
δ
x
1
d
−
1
δ
Ax
0
¯A
¯ 1 = nA
¯1
¯ 1 + D
I¯A
x:n
x:n
x:n
1
1
1
(IA)x:n
+ (DA)x:n
= (n + 1)Ax:n
n| Ax
(IZ)x = (Kx + 1)v Kx +1 ,
PERCENTILES
k=0
¯x = v Tx , Tx ≥ 0
Z
Z ∞
¯x =
A
v t t px µx+t dt
n| Ax
RECURSION
ENDOWMENT INSURANCE
(
Tx
0 ≤ Tx ≤ n
¯x:n = v
Z
vn
Tx > n
¯x:n = A
¯1 + n Ex
A
(x:n
v Kx +1 Kx = 0, 1, . . . , n − 1
Zx:n =
vn
Kx = n, n + 1, . . .
1
Ax:n = Ax:n
+ n Ex
Assuming death occurs in the middle of the
year:
DE MOIVRE’S LAW
APV is an annuity-certain for the number of
years of insurance remaining divided by ω − x.
All of the following formulas follow that.
aω−x
a
¯
¯x = ω−x
A
Ax =
ω−x
ω−x
¯a
I¯
(Ia)ω−x
ω−x
¯ =
I¯A
(IA)x =
x
ω−x
ω−x
a
¯n
ω−x
¯a
I¯
n
¯ 1 =
I¯A
x:n
ω−x
¯ a)n
¯A
¯ 1 = (D¯
D
x:n
ω−x
¯1 =
A
x:n
1
Ax:n
=
¯x = (1 + i)1/2 · Ax
A
(m)
Ax
= (1 + i)(m−1)/2m · Ax
an
ω−x
1
(IA)x:n
=
1
(DA)x:n
=
(Ia)n
ω−x
(Da)n
ω−x
(IA)x = Ax + 1| Ax + 2| Ax + · · ·
c
2012
The Infinite Actuary
D. Life Annuities
WHOLE LIFE ANNUITY
N-YEAR DEFERRED WLA
Twin – Whole Life Insurance
No Twin
(
0
¯
Y
=
n| x
a
¯Tx − a
¯n
1 − v Tx
Y¯x = a
¯Tx =
δ
Z ∞
¯x
1−A
a
¯x =
a
¯x =
t Ex dt
δ
0
¯x
n| a
=a
¯x − a
¯x:n
¯x = 1 − δ¯
A
ax
¯x
n| a
= n Ex · a
¯x+n
¨x
n| a
= n Ex · a
¨x+n
Var Y¯x =
2A
¯x
¯x
− A
2
δ2
1 − v Kx +1
Y¨x = a
¨Kx +1 =
d
∞
X
1 − Ax
a
¨x =
a
¨x =
k Ex
d
k=0
Twin – N-Year Endowment Insurance
(
a
¯Tx 0 ≤ Tx ≤ n
Y¯x:n =
n < Tx
a
¯n
¯x:n
1−Z
=
δ
Z n
¯x:n
1−A
a
¯x:n =
a
¯x:n =
t Ex dt
δ
0
¯x:n = 1 − δ¯
A
ax:n
2A
¯x:n − A
¯x:n 2
¯
Var Yx:n =
δ2
1 − Zx:n
Y¨x:n =
d
a
¨x:n =
1 − Ax:n
d
a
¨x:n =
n−1
X
k=0
a
¨x:n = a
¨x − n Ex a
¨x+n
h
Var Y¨x:n
i
=
2A
x:n
k Ex
a
¨x =
¯x (1 − n Ex )
a
¯x:n = a
¯a = (¯
I¯
ax )2
x
0
1
vq + d
a
¨x:n = a
¨x (1 − n Ex )
DE MOIVRE’S LAW
DEFERRED TEMPORARY ANNUITY
a
¨x = 1 + vpx a
¨x+1
Use the most important identity.
No twin.
ax = vpx a
¨x+1
n+m
=
v t t px dt
¯a
¯a
I¯
+ D¯
= n¯
ax:n
x:n
x:n
¯x+1
a
¯x = a
¯x:1 + vpx a
ANNUITIES PAYABLE M-THLY
¯x:n
=a
¯x:n+m − a
¯x:1 + vpx a
¯x+1:n−1
a
¯x:n = a
Always true:
= n| a
¯x − n+m| a
¯x
a
¨x:n = 1 + vpx a
¨x+1:n−1
Ax
= n Ex · a
¯x+n:m
(I¨
a)x = 1 + vpx (¨
ax+1 + (I¨
a)x+1 )
¨x:n
Ax:n = 1 − d(m) a
(I¨
a)x = a
¨x + vpx (I¨
a)x+1
¨x+1:n−1 + (I¨
a)x+1:n−1
(I¨
a)x:n = 1 + vpx a
Assuming UDD:
For annuity-due same 3 formulas as above but
replace a
¯ with a
¨.
N-YR CERTAIN AND LIFE AFTER
(
a
¯n
0 ≤ Tx ≤ n
¯
Yx:n =
a
¯Tx n < Tx
(m)
(m)
= 1 − d(m) a
¨x
(m)
(m)
(m)
(D¨
a)x:n = n + vpx (D¨
a)x+1:n−1
= α(m) a
¨x − β(m) (0 Ex − ∞ Ex )
(m)
a
¨x:n
= α(m) a
¨x:n − β(m) (0 Ex − n Ex )
a
¯x:n = a
¯n + n| a
¯x
ADJUSTED FORCE OF MORTALITY
¨n + n| a
¨x
a
¨x:n = a
If µ∗x+t = µx+t + c and δ ∗ = δ − c, then
α(m) ≈ 1
= n Ex
(m)
a
¨x
β(m) ≈
m−1
2m
s¨x:n =
a
¨x:n
n Ex
≈a
¨x −
m−1
2m
−
m2 −1
12m2
(µx + δ)
−for other annuities, write them in terms of
a whole life annuity
ACCUMULATED APV
a
¯x:n
n Ex
= α(m) n| a
¨x − β(m) (n Ex − ∞ Ex )
Woolhouse (3 terms)
a
¯∗x = a
¯x
s¯x:n =
= α(m)· symbol w/o (m)
−β(m) (“start” - “end”)
a
¨x
(m)
¨x
n| a
INCREASING WHOLE LIFE
¯a
I¯Y¯ x = I¯
T ≥0
Tx x
¯
a
¯x − I¯A
x
¯a =
I¯
x
δ
¯ =a
¯a
I¯A
¯x − δ I¯
x
x
(m)
symbol with
(I¨
a)x:n = a
¨x:n + vpx (I¨
a)x+1:n−1
∗
n Ex
¯1 = 1 − δ¯
ax:n − n Ex
A
x:n
1
µ+δ
RECURSION
n
N-YEAR TEMPORARY ANNUITY
a
¯x =
(I¨
a)x:n + (D¨
a)x:n = (n + 1)¨
ax:n
¯x
n|m a
Yx = aKx = a
¨Kx +1 − 1 ⇒ ax = a
¨x − 1
CONSTANT FORCE
(I¨
a)x = (¨
ax )2
µ/δ
µ
Pr(SBP is insufficient) =
µ+δ
Z
Ax = 1 − d¨
ax
h i
2 A − (A )2
x
x
Var Y¨x =
d2
0 ≤ Tx ≤ n
n < Tx
VARYING TEMPORARY
Z n
¯a
I¯
=
tv t t px dt
x:n
Z0 n
¯a
(n − t)v t t px dt
=
D¯
x:n
sx:n =
ax:n
n Ex
(I¨
ax ) − (Ia)x = a
¨x
− (Ax:n )2
d2
ax:n = a
¨x:n − 1 + n Ex
c
2012
The Infinite Actuary
E. Premium Calculation
LOSS AT ISSUE
ENDOWMENT INSURANCE
SEMICONTINUOUS INSURANCE
GROSS PREMIUMS
L ∼ loss at issue random variable
Fully Continuous:
Benefit paid at the moment of death and premiums paid at beginning of the year.
¯
¯x = Ax
P A
a
¨x
¯1
A
x:n
¯1
P A
x:n =
a
¨x:n
¯
¯x:n = Ax:n
P A
a
¨x:n
Lg = PV of future benefits
+ PV of future expenses
− PV of future premiums
L = PV of benefits − PV of premiums
E[L] = APVFB − APVFP
¯x:n − P¯ a
E[L] = S A
¯x:n
2
¯
2A
¯x:n − A
¯x:n 2
Var[L] = S + Pδ
Fully Discrete:
E[L] = SAx:n − P a
¨x:n
2
2
P
2A
Var[L] = S + d
x:n − (Ax:n )
EQUIVALENCE PRINCIPLE
E[L] = 0 ⇒ APVFB = APVFP
PERCENTILE PREMIUMS
N-YEAR TERM INSURANCE
Whole Life and Endowment Insurance:
Var[Z]
(1 − SBP)
1
¯x + δ
P¯ A
¯
¯x
P A
¯x =
A
¯x + δ
P¯ A
¯1
P¯ A
x:n
1
P x:n
=
Whole Life and Endowment Insurance Only:
¯
Ax:n
¯x = Ax
P¯ A
Px:n =
a
¯x
a
¨x:n
a
¯x =
¯1
P¯ A
x:n
=
¯1
A
x:n
Group of policies:
a
¯x:n
E[S] + Φ−1 (1 − x%)
2
KEY PREMIUM IDENTITIES
¯x = 1 − δ
P¯ A
a
¯x
¯
¯x = δ Ax
P¯ A
¯x
1−A
Px:n =
Px:n
Ax:n
1
−d
a
¨x:n
dAx:n
=
1 − Ax:n
a
¨x:n =
New Equivalence Premium:
E[Lg ] = 0 ⇒ APVFB + APVFE = APVFP
APV of Benefits + APV of Expenses
APV Annuity
2
r
Var(Z)
Var[Lg ] = S + E + G−e
d
G=
APV benefit
benefit premium =
APV annuity
Var[L] =
E[Lg ] = APVFB + APVFE − APVFP
¯x:n − P¯ A
¯ 1
= P¯ A
x:n
1
Ax:n
a
¨x:n
Px:n
=
Px:n + d
N-PAY WHOLE LIFE
¯x
A
¯ ¯
n P Ax =
a
¯x:n
n Px
=
N-YEAR DEFERRED WLA
¯x
n| a
P¯ n| a
¯x =
a
¯x:n
¨x
n| a
=
Var[L] = p · 2 Ax
Fully Continuous Whole Life:
Premiums paid m-thly instead of annually.
Px
(m)
n Px
µ
µ+δ
µ/δ
Ax
=
=
(m)
a
¨x
Ax:n
DE MOIVRE’S LAW
(m)
a
¨x:n
Ax
= (m)
a
¨x:n
Use key premium identities and most important identity.
¨x
n| a
a
¨x:n
(P-P) / P
n Px
1
− P x:n
P x:n1
Fully Continuous:
E[L] = SAx − P a
¨x
2
P
2 A − (A )2
Var[L] = S + d
x
x
CONSTANT FORCE
¯x = µ
P¯ A
Px = vq
1
¯1
P x:n
P¯ A
= vq
x:n = µ
Pr(benefit prem is sufficient) =
(m)
Ax
a
¨x:n
WHOLE LIFE
Fully Discrete:
FRACTIONAL PREMIUMS
(m)
Px:n
P
¯x − P¯ a
E[L] = S A
¯x
2
¯
2A
¯x − A
¯x 2
Var[L] = S + Pδ
Single policy:
Find the policy you must “win” on and then
solve for the premium it will take to win on
that death.
¯x
Var[L] = 2 A
1
Px:n + d
p
Var[S] = 0
COMPARE VARIANCE
Two identical policies (whole life or endowment insurance)
P1 → L1
= Ax+n
Px:n − n Px
= 1 − Ax+n
P x:n1
1
Px:n − P x:n
P x:n1
=1
P2 → L2
Var[L2 ]
=
Var[L1 ]
P2 + d
P1 + d
2
c
2012
The Infinite Actuary
F. Reserves
#1 PROSPECTIVE METHOD
#4 PAID UP
#8 (P-P) / P
#11 RESERVE CREATION FORMULA
All Policies and All Premiums. All other methods only work when premiums are determined
using the equivalence principle.
Whole Life, Endowment Insurance, Term Insurance and Limited Pay Whole Life Only
Accumulated Difference in Premiums equals
Difference in Reserves
Useful when death benefit is a function of the
reserve.
Reserve = APV of the future benefit times
the percentage of the future benefit that you are not funding with
future premiums
Remember
Reserve = APVFB − APVFP
Example – fully continuous whole life:
¯ ¯
¯
¯ ¯ ¯x+t
t V Ax = Ax+t − P Ax a
Example – full continuous whole life:
"
#
¯x
P¯ A
¯
¯
¯
1
−
V
A
=
A
x
t
x+t
¯x+t
P¯ A
1
P
1
= s¨x:n
Example
Px:n − Px
= n Vx:n − n Vx = 1 − n Vx
P x:n1
#9 ANNUITY FORM SPINOFF
All Policies under EP.
#5 ANNUITY FORM
Whole Life and Endowment Insurance Only
Reserve = AAVPP − AAVPB
Whole Life and Endowment Insurance Only
Example – fully discrete whole life:
t Vx:n
= Px:n s¨x:t − t kx
ACCUMULATED COST OF INS
t kx
=
=
t Ex
dx (1+i)t−1 +dx+1 (1+i)t−2 +···+dx+t−1
lx+t
For one year:
1 kx
=
Example – fully discrete endowment ins.:
a
¨x+t:n−t
t Vx:n = 1 −
a
¨x:n
1
Ax:t
Thinking in terms of life tables:
t kx
Reserve = 1 minus (the annuity at the dot divided by the annuity at issue)
qx
px
#6 LIFE INSURANCE FORM
Whole Life and Endowment Insurance Only
Reserve = insurance at the dot minus the insurance at issue divided by one minus the insurance at issue
Example – fully discrete whole life:
Ax+t − Ax
t Vx =
1 − Ax
#3 DIFFERENCE IN PREMIUM
Whole Life, Endowment Insurance, Term Insurance and Limited Pay Whole Life Only
Reserve = accumulated difference in premium
you want to charge and premium
you are actually charging
Example – fully continuous whole life:
¯ ¯
¯ ¯
¯ ¯
a
¯x+t
t V Ax = P Ax+t − P Ax
#7 BENEFIT PREMIUM FORM
Whole Life and Endowment Insurance Only
1. cost of providing ensuing year’s DB
2. rest is for reserve creation
tV
= P s¨t t−1
X
−
vqx+h (bh+1 − h+1 V ) (1 + i)t−h
h=0
#2 RETROSPECTIVE METHOD
Example – fully discrete endowment ins.:
Each premium accounts for two items:
x:n
DB = Reserve: t V = P s¨t
DB = 1 + Reserve:
tV
= P s¨t −
t−1
X
qx+h (1 + i)t−h−1
h=0
a+b Vx = 1 − (1 − a Vx ) (1 − b Vx+a )
DB = 1 + Reserve and qx+h = q :
#10 RESERVE RECURSION
Start with the terminal reserve from the previous year, collect the premium and put those
in the bank. If the policyholder dies pay the
death benefit. If the policyholder lives setup
the next reserve.
Example - fully discrete 20-year endowment insurance on (40):
10 V40:20 + P40:20 (1+i) = q50 +p50 11 V40:20
NAR version: = q50 1 − 11 V40:20 + 11 V40:20
tV
= (P − vq) s¨t
VARIANCE LOSS
For Whole Life and Endowment Insurance:
¯ 2
Var(t L) = S + Pδ
Var(v U )
EP ⇒
=
Var(v U )
(1 − SBP)2
Example – fully discrete endowment ins.:
Var(t L) =
S+
EP ⇒
=
P
d
2
2A
x+t:n−t
2A
x+t:n−t
2
− Ax+t:n−t
2
− Ax+t:n−t
(1 − Ax:n )2
Reserve = premium at the dot minus the premium at issue divided by the premium at the dot plus delta (or d)
CONSTANT FORCE
Example – fully discrete whole life:
Under EP, level premium whole life and term
insurance reserves are 0.
t Vx
=
Px+t − Px
Px+t + d
c
2012
The Infinite Actuary
GROSS PREMIUM RESERVES
MODIFIED RESERVES
To find the gross premium, G, solve new EP
α − lower modified prem in the first year
APVFP = APVFB + APVFE
β − higher modified prem in the renewal years
Expense Reserves:
FPT Reserves:
kV
e
= APVFE − APVFL
kV
e
= AAVPL − AAVPE
1
α = Ax:1
β = benefit premium for insurance issued
to (x + 1)
Gross Premium Reserve:
kV
g
= kV
n
+ kV
1V
m
=0
tV
m
= (t − 1) reserve for insurance issued
to (x + 1)
e
Can also use recursion for expense and gross
premium reserves.
GAIN
SEMI-CONTINUOUS RESERVES
Assuming UDD
i
¯
t V A x = δ t Vx
i
1
¯1
t V Ax:n = δ t V x:n
i
1
1
¯
t V Ax:n = δ t V x:n + t V x:n
INTERIM BENEFIT RESERVES
Exact Method:
(h V + πh ) (1 +
i)s
v 1−s
= s qx+h ·
+s px+h · h+s V
Approximation: use linear interpolation between initial reserve and the terminal reserve.
Gain = Actual Profit − Expected Profit
Profit = (t V + Gt − et ) (1 + i)
−qx+t (St + Et ) − px+t · t+1 V
Analysis of Surplus:
expenses: (et − e0t ) (1 + i) + (Et − Et0 ) qx+t
interest: (i0 − i) (t V + Gt − e0t )
0
mortality: qx+t − qx+t
(St + Et0 − t+1 V )
Order matters. Use actual experience if you
have already accounted for that source.
POLICY ALTERATIONS
Reduced Paid-Up:
RP U =
THIELE’S DIFFERENTIAL EQN
t CVx
Ax+t
Extended Term (solve for n):
d
V
dt t
= δt · t V +Gt −et −(St + Et − t V ) µ[x]+t
t+h V − h Gt − et − (St + Et ) µ[x]+t
tV ≈
1 + h µ[x]+t + δt
t CVx
1
= Ax+t:n
For endowment insurance if the cash value is
large enough to cover the remaining term, then
solve for reduced pure endowment benefit, P E
t CVx
1
= Ax+t:n−t
+ P E · n−t Ex+t
c
2012
The Infinite Actuary
G. Markov Chains
NOTATION
ij
t px
MULTIPLE DECREMENT MODEL
− probability that a subject in state i at
time x will be in state j (where j may
equal i) at time x + t
ii
t px
exit
1
− probability that a subject in state i at
time x stays in state i continuously until time x + t
ii
t px
exit
2
≤ t pii
x
ANNUITIES
THIELE’S DIFFERENTIAL EQN
Pay 1 per year continuously while in state j:
Z ∞
a
¯ij
e−δt t pij
x =
x dt
d
V (i)
dt t
0
(i)
= δt t V (i) − Bt
n
X
(ij)
−
µij
+ t V (j) − t V (i)
x+t St
j=0,j6=i
If payable at the start of the year:
∞
X
a
¨ij
v k k pij
x =
x
k=0
Euler’s method:
t−h V
active
0
(i)
(i)
= t V (i) (1 − δt h) + hBt
n
X
(ij)
+h
µij
+ t V (j) − t V (i)
x+t St
j=0,j6=i
INSURANCE
DISCRETE MARKOV CHAINS
P(t) is the transition matrix at time t
ij
t px
= ij entry of
ii
t px
ii
ii
= pii
x · px+1 · · · px+t−1
P(x)
·
P(x+1)
exit
m
· · · P(x+t−1)
0j
t px
CONTINUOUS MARKOV CHAINS
µij
x
ii
t px
=
1 − t p00
x
R
= exp − 0t µ0•
x+s ds
h→0
Probability of transition from one state to another depends on the model.
KOLMOGOROV’S FORWARD EQNS
X
ik kj
ij jk
d
pij =
t px µx+t − t px µx+t
dt t x
k=0,k6=j
d
pij
dt t x
PERMANENT DISABILITY MODEL
Healthy
0
µ01
x
µ02
x
Disabled
1
=
prob “move
−
into j”
at time x + t
Euler’s method turns a continuous Markov
Chain into a discrete chain with time increments of h and transition “probabilities”
(
hµij
i 6= j
ij
x+t
p
=
h x+t
1 − hµi•
x+t i = j
=
Rt
00
0 s px
j6=k
If all transition forces are constant and no reentry into a state
1
a
¯ii
x = i•
µ +δ
For the multiple decrement model with constant transitions:
µ0j
¯0j
A
x = 0•
µ +δ
RESERVES
tV
(i)
− rate of payment of benefit while the
policyholder is in state i
(ij)
− lump sum benefit payable instantaneously at time t on transition from
state i to state j
St
PREMIUMS
APV of benefit
APV of annuity
− reserve at duration t for a subject in
state i at that time
(i)
Bt
benefit premium =
01
t px
prob “move
out of j”
at time x + t
µ12
x
Dead
2
0
CONSTANT FORCE AND NO RE-ENTRY
ij
h px
for i 6= j
h
R
t i•
= exp − 0 µx+s ds
= lim
00
t px
µ0j
x
µ0•
x
Pay 1 on each future transfer into k:
Z ∞
X
ij jk
¯ik =
A
e−δt
t px µx+t dt
x
11
µ01
x+s t−s px+s ds
c
2012
The Infinite Actuary
H. Multiple Decrement Models
MULTIPLE DECREMENT MODEL
FORCE OF DECREMENT - 2
(τ )
exit
1
µx+t =
(j)
µx+t
exit
2
active
0
=
(τ )
d
q
dt t x
(τ )
t px
(j)
− prob. of decrement due to any cause
m
X
(τ )
(j)
t qx =
t qx
(j)
Ax =
(τ )
j
v k k−1 px qx+k−1
FORCES OF DECREMENT - 1
(τ )
(1)
(2)
=
(τ )
0 s px
Rt
cause
1
(m)
(1)
qx
(2)
(j)
· µx+s ds
Pr(J = j | T = t) =
µx+t
qx
alive
0
(j)
t qx
(j)
=
(j)
¯x
A
=
µx
(τ )
µx
R∞
0
exit
1
µ01
x+t
exit
2
active
0
cause
3
(τ )
(j)
v t t px µx+t dt
CONSTANT TRANSITION FORCES
(j)
(τ )
µ(j)
t qx = (τ ) 1 − t px
µ
(j)
(τ ) t
(j)
t qx
1 − px
t qx =
(τ )
µ(j)
+δ
1
a
¯x = (τ )
µ
+δ
(j)
(j)
(τ )
µ
¯
A
1 − n Ex
= (τ )
1
x:n
=
µ(τ )
µ
0 is redundant so we use:
(j)
µx+t
1
2
X
(i)
dx
Constant Force and UDDMDT:
(j) (τ )
(τ ) qx /qx
p0 (j)
x = px
UDDAST:
For 2 UDDASTs use the midpoint:
(1)
qx = q 0 (1)
1 − 12 q 0 (2)
x
x
q 0 (1)
x
alive
0
For 3 or more UDDASTs use integration:
0 (j)
tp x
(j)
· µx+t = q 0 (j)
x - factor out of integral
For 3 UDDASTs (if you like memorizing):
(1)
0 (3) + 1 q 0 (2) q 0 (3)
qx = q 0 (1)
1 − 12 q 0 (2)
x +q x
x
x
3 x
Hybrids:
cause
1
Draw a picture. For 2 competing UDDASTs
use the midpoint of the interval.
q 0 (2)
x
alive
0
ASSET SHARE
Recursion:
[k AS + G(1 − ck ) − ek ] (1 + i)
(d)
0 (2)
0 (m)
= t p0 (1)
x · tp x · · · tp x
(w)
(τ )
= qx+k + qx+k · k+1 CV + px+k · k+1 AS
q 0 (3)
x
alive
0
(τ )
t px
cause
2
+δ
ABSOLUTE RATE OF DECREMENT
R
t (j)
0 (j)
0 (j)
t q x = 1 − t p x = 1 − exp − 0 µx+s ds
exit
m
−
t qx
µ02
x+t
µ0m
x+t
dx
(τ )
`x
(3)
qx
1 − t px
(τ )
cause
2
(τ )
µx+t
If all forces are a constant multiple of the total
force for all t
¯(j)
A
x
=
q 0 (j)
x =
Rates of Decrement
k=1
µ0j
x+t
Probabilities of Decrement
i6=j
j=1
∞
X
Super secret approximation:
(j)
d
q
dt t x
(τ )
t px
(τ )
qx
Each cause is independent of the other causes.
(j)
(j)
qx − prob. of decrement due to cause #j
MDT vs. AST
µx+t = µx+t + µx+t + · · · µx+t
R
(τ )
t (τ )
t px = exp − 0 µx+s ds
(j)
t qx
exit
m
ABSOLUTE RATE OF DECREMENT
cause
3
Notice how in the rate of decrement “worlds”
there is only one cause of decrement. For the
rates of decrement we have 3 tables and for the
multiple decrement “world” we have one table
with 3 competing decrements.
0 (j)
tq x
n-th Asset Share using Asset Share Creation:
(d)
(w)
n−1
X
G(1 − ck ) − ek − vqx+k − vqx+k · k+1 CV
k=0
(τ )
n−k Ex+k
For additional premium of ∆G:
∆AS =
n−1
X
∆G(1 − ck )
k=0
(τ )
n−k Ex+k
(j)
≥ t qx
=
n−1
X
∆G(1 − ck )k px v k
k=0
(τ )
n Ex
(τ )
c
2012
The Infinite Actuary
I. Multiple Life Models
JOINT-LIFE STATUS
LAST-SURVIVOR STATUS
CONTINGENT PROBABILITIES
REVERSIONARY ANNUITY
Txy = min[Tx , Ty ]
Txy = max[Tx , Ty ]
1
n q xy
(y) gets money after (x) dies:
Failure on the first death.
Txy = Tx + Ty − Txy
t qxy
t pxy
− probability first death of (x) and (y)
occurs within t years
that for (xy) = that for (x) + that for (y)
−that for (xy)
− probability that (x) will attain x + t
and (y) will attain y + t
t qxy
t pxy
= t px · t py
t qxy
= 1 − t pxy
(work with p’s)
fxy (t) = t pxy · µx+t:y+t
µx+t:y+t = µx+t + µy+t
Z n
n qxy =
t pxy µx+t:y+t dt
◦
0
∞
Z
exy =
t pxy
0
∞
X
exy =
dt
k pxy
k=1
0
¯xy =
A
t qxy
= t qx · t qy
t pxy
− probability at least one of (x) or (y)
survives for t years
(work the the q’s)
t pxy
= 1 − t qxy
t pxy
= t px + t py − t pxy
∞
v
Other versions:
a
¯x|y:n = a
¯y:n − a
¯xy:n
a
¯x:n |y = a
¯y − a
¯xy:n
0
2
n q xy
In other words:
a
¯u|v = a
¯v − a
¯uv
n
Z
=
t py
µy+t · t qx dt
= n q xy1 + n q xy2
1
n q xy
= n q xy2 + n qx · n py
Probability (x) dies more than n years after
death of (y):
= n+1 qxy − n qxy
n| qxy
= n| qx + n| qx − n| qxy
◦
◦
2
· ∞ q x+n:y
n px
◦
t
t pxy
µx+t:y+t dt
a
¯xy = a
¯x + a
¯y − a
¯xy
CONTINGENT INSURANCE
Z ∞
¯1 =
A
v t · t px µx+t · t py dt
xy
¯x + A
¯y − A
¯xy
¯xy = A
A
1
Axy:n
=
1
Ax:n
+
1
Ay:n
−
−d
0
1
Axy:n
c
¯ 2=
A
xy
¯xy = 1 − δ¯
A
axy
1
a
¨ xy
− probability that (y) dies after (x) and
before n years from now
If (x) and (y) independent:
Z n
1
n q xy =
t px µx+t · t py dt
n qy
n| qxy
0
Pxy =
s
0
fxy (t) = t px µx+t + t py µy+t − t pxy µx+t:y+t
◦
0
2
n q xy
a
¯x|y = a
¯y − a
¯xy
COMMON SHOCK
Find
total
total
total
the total force for each status:
force on (x) = µ∗x + λ
force on (y) = µ∗y + λ
force on (xy) = µ∗x + µ∗y + λ
Pr[(x) dies first] =
µ∗
x
∗
µ∗
x +µy +λ
Pr[(y) dies first] =
µ∗
y
∗ +λ
µ∗
+µ
x
y
exy = ex + ey − exy
exy = pxy (1 + ex+1:y+1 )
Z ∞
a
¯xy =
v t t pxy dt
Z
− probability that last death happens
within t years
1
n q xy
− probability (x) dies before (y) and before n years from now
Z n Z ∞
=
fxy (s, t) dt ds
∞
Z
v t · t py µy+t · t qx dt
0
EXACTLY ONE STATUS
¯ 2=
A
xy
∞
Z
Pr[Tx = Ty ] =
¯x =
A
µ∗
x +λ
µ∗
x +λ+δ
¯y =
A
µ∗
y +λ
µ∗
y +λ+δ
λ
∗
µ∗
x +µy +λ
¯y+t dt
v t · t px µx+t · t py A
0
[1]
t pxy
− probability exactly 1 of (x) and (y) live
t years
¯y = A
¯ 1 +A
¯ 2
A
xy
xy
[1]
t pxy
= t px + t py − 2 t pxy
SBP for a payment of 1 at the death of (x) if
he dies more than n years after the death of
(y):
[1]
¯x + a
¯y − 2¯
axy
a
¯xy = a
n Ex
¯ 2
·A
x+n:y
CONSTANT FORCE
µx +µy
µx +µy +δ
¯xy =
A
a
¯xy =
1
n q xy
1
µx +µy +δ
=
¯1 =
A
xy
µx
q
µx +µy n xy
µx
µx +µy +δ
c
2012
The Infinite Actuary
DE MOIVRE’S LAW
◦
exx =
ω−x
3
◦
◦
◦
exy = y−x px · eyy +y−x qx · ey
2
n q xy
=
¯1 =
A
xy
1
q
2 n xy
a
¯y:ω−x
ω−x
c
2012
The Infinite Actuary
J. Other Topics
PENSION MATHEMATICS
PROFIT SIGNATURE
Replacement ratio:
pension income in year after retirement
R=
salary in the year before retirement
(
Πt =
Profit(0)
t−1 px · Profit(t)
t=0
t>0
where t−1 px is the probability that policy is in
force at beginning of year t.
Salary Scale:
sy
salary received in year of age y to y + 1
=
sx
salary received in year of age x tp x + 1
v(t) − current market price on a t year zerocoupon bond that pays $1 at time t.
Solve for j such that:
n
X
Πt vjt = 0
t=0
− think of it as the discount function.
yt − t year spot rate of interest
v(t) =
1
(1 + yt )t
PROFIT MARGIN
f (t, t+k) − forward rate from time t to t + k
NPV
Pa
¨x:n
a
¨(x)y =
k px
v(k)
k=0
A risk is diversifiable if:
q
P
Var[ N
i=1 Xi ]
lim
=0
N →∞
N
DISCOUNTED PAYBACK PERIOD
Find the smallest m such that:
m
X
Πt vrt ≥ 0
UNIVERSAL LIFE - Notation
− Expenses Incurred
+ Interest Earned on Premium less Expenses
ECt − expense charge for year t
The profit for year 0 is the negative of the expenses incurred at time 0.
Annual Profit =
Previous Rsv accumulated with interest
+ Premium Collected
− Expected Cost of Benefits
Type A:
Annual Profit =
ADBt = F A − AVt
DBt = F A
Previous Rsv (usually AV)
+ Premium Collected
− Expenses Incurred
ADBt = F A
DBt = F A + AVt
+ Interest Earned on
Prev Rsv plus Premium less Expenses
− Expected Cost of Benefits
1
∗
CoIt = ADBt Ax+t−1:1
= ADBt · vq qx+t−1
− Expected Surrender Benefits
− Expected Cost of Ending Rsv
CoItA =
∗
[F A−(AVt−1 +Pt −ECt −CoItA )(1+ict )]vq qx+t−1
∗
[F A−(AVt−1 +Pt −ECt )(1+ict )]vq qx+t−1
∗
1−vq qx+t−1
(1+ic
t)
Type B:
CoItB
= FA ·
∗
vq qx+t−1
CORRIDOR FACTORS
To qualify as life insurance the death benefit
must be at least a certain multiple (γt for year
t) of the account value.
ADBtc = (γt − 1)AVt
t=0
F A − face amount
AVt − account value at end of year t
CVt − cash value at end of year t
DBt − death benefit for year t
ADBt − additional death benefit for year t
CoIt − cost of insurance for year t
SCt − surrender charge for year t
ict − credited interest rate in year t
It − credited interest in year t
vq − discount factor used in CoI calc
∗
qx+t−1
−mort. rate used in CoI calc for year t
ANNUAL PROFIT
The profit for year 0 is the negative of the expenses incurred at time 0.
CoItA =
k
0
∞
X
DBt = AVt + ADBt
Type A:
Profit Margin =
v(t)
(1 + f (t, t + k)) =
v(t + k)
Z ∞
¯ y=
A(x)
v(t) t px µx+t dt
UNIVERSAL LIFE - Annual Profit
Type B:
INTERNAL RATE OF RETURN
INTEREST RATE RISK
UNIVERSAL LIFE - COI
UNIVERSAL LIFE - Roll Forward
ADBtf = F A − AVt
Account Value Roll Forward:
ADBt = max[ADBtc , ADBtf ]
Starting AV (AVt−1 )
∗
CoIt = ADBt · vq qx+t−1
+ Premium (Pt )
− Expense Charge (ECt )
− Mortality Charge (CoIt )
+ Credited Interest (It )
= Ending Account Value (AVt )
AVtA =
∗
(AVt−1 +Pt −ECt −F A vq qx+t−1
)(1+ict )
∗
1−vq qx+t−1
(1+ic
t)
AVtB = same as numerator for AVtA
CVt = max[AVt − SCt , 0]
− Expected Cost of Ending Rsv
c
2012
The Infinite Actuary
CALCULUS REVIEW
d
dx
ex = ex
UNIFORM DISTRIBUTION
X ∼ uniform a to b
f (x) =
1
b−a
D {f (g(x))} = f 0 (g(x)) · g 0 (x)
F (x) =
x−a
b−a
(uv)0 = u0 · v + u · v 0
R
R
udv = uv − vdu
E[X] =
a+b
2
R
ex dx = ex + c
Var[X] =
(b−a)2
12
CONDITIONAL PROBABILITY
Pr(A ∩ B)
Pr(A|B) =
Pr(B)
f (x, y)
fX (x|y) =
fY (y)
R∞
fY (y) = −∞ f (x, y) dx
v = (1 + i)−1 = e−δ
v 2 = (1 + i)−2 = e−2δ
v n = (1 + i)−n = e−nδ
E[X] = EY [EX [X|Y ]]
Var[X] = EY VarX [X|Y ] + VarY EX [X|Y ]
NORMAL APPROXIMATION
EXPONENTIAL DISTRIBUTION
all x
E[X] =
X
Exponential distribution is memoryless
xp(x)
all x
E[g(X)] =
X
f (x) =
g(x) · p(x)
all x
Var[X] = E[X 2 ] − (E[X])2
X
F (x) = Pr[X ≤ x] =
p(y)
1
1 −µx
e
µ
1x
−µ
F (x) = 1 − e
Var[Y ] = (a − b)2 · p(1 − p)
CONTINUOUS RANDOM VARIABLES
R
x f (x) dx = 1
R
E[g(X)] = x g(x) f (x) dx
Rx
F (x) = Pr[X ≤ x] = −∞
f (y) dy
f (x) =
d
F (x)
dx
v → v2
i → 2i + i2
Central Limit Theorem:
d → 2d − d2
Y = X1 + X2 + · · · + Xn
i
δ
Xi ’s are iid RVs(µx , σx2 )
→
2i+i2
2δ
Y ≈ Normal(E[Y ] = nµx , Var[Y ] = nσx2 )
ANNUITY CERTAINS
E[X] = µ
95th percentile of Y :
Var[X] = µ2
MIXED RANDOM VARIABLES
CONVERT BETWEEN i, d, and δ
Example:
(1)
X=a
p(a)
d
X=b
p(b)
0
i
1
(2)
a<X<b
f (x)
Rb
p(a) + a f (x) dx + p(b) = 1
R
E[X k ] = ak · p(a) + ab xk f (x) dx + bk · p(b)
d
=i
1−d
i
=d
1+i
δ = ln(1 + i)
i
1 − vn
= an
δ
δ
a
¨n − nv n
(Ia)n =
i
n − an
(Da)n =
i
Z ∞
1
t
a
¯∞ =
v dt =
δ
0
Z ∞
1
t
¯a)∞ =
(I¯
tv dt = 2
δ
0
Z n
a
¯n − nv n
¯a)n =
(I¯
tv t dt =
δ
Z0 n
n−a
¯n
t
¯
(D¯
a)n =
(n − t)v dt =
δ
0
a
¯n =
y = µy + σy × Φ−1 (0.95)
y≤x
Bernoulli Shortcut for the RV:
(
a p(a) = p
Y =
b p(b) = 1 − p
DOUBLE THE FORCE OF INTEREST
1 + i → (1 + i)2
median = mean
DISCRETE RANDOM VARIABLES
X
p(x) = 1
DISCOUNTING
(1)
(2)
ACCUMULATING
COVARIANCE
1 + i = eδ
Var(aX + bY ) = a2 Var(X) + b2 Var(Y )
+2ab Cov(X, Y )
(1 + i)2 = e2δ
(1 + i)n = enδ
Cov(X, Y ) = E(XY ) − E(X)E(Y )
If X and Y are independent
then covariance is 0.
c
2012
The Infinite Actuary
B. Survival Models and Life Tables
RANDOM VARIABLES
LIFE TABLES
Tx ∼ future lifetime of (x)
lx − expected number of survivors at (x)
TEMPORARY LIFE EXPECTANCY
Z n
◦
ex:n = E[Wx ] =
t px dt
0
Kx ∼ number of completed future years by
(x) prior to death.
n dx
− expected number of deaths between
ages x and x + n
Kx = bTx c
ex:n = E[Kx ∧ n] =
E[Wx2 ] = 2
k px
Out of ω births, one dies every year until they
are all dead.
Future lifetime is uniformly distributed from 0
to ω − x.
k=1
n
Z
n
X
DE MOIVRE’S LAW
`x = k (ω − x)
t t px dt
0
Wx ∼ future lifetime or n years whichever is
less.
Wx = Tx ∧ n
FORCE OF MORTALITY
Intuitively, µx+t dt is the probability that a life
age x + t will die in the next instant.
CUMULATIVE DISTRIBUTION
Probability of (x) dying before age x + t.
Fx (t) = Pr(Tx ≤ t)
d
t qx
◦
d `
− dt
x+t
`x+t
∗
n px
d
p
dt t x
= (n px )k
= t px µx+t
= −t px µx+t
Probability of a (x) attaining age x + t.
ACTUARIAL NOTATION
PROBABILITY DENSITY FUNCTION
◦
◦
= Sx (t)
t+u px
t qx
= t px · u px+t
n px
Age doesn’t matter.
ex =
Future lifetime is distributed exponentially
with mean 1/µ.
t qx
= 1 − t px
t|u qx
t| qx
t|u qx
= Sx (t) − Sx (t + u)
= t px − t+u px
= t+u qx − t qx
= t px · u qx+t
◦
for all x
= e−nµ = (px )n
1
µ
1
µ2
Recursion
ex = px (1 + ex+1 )
◦
ex:n = ex (1 − n px )
ex:n = ex (1 − n px )
Discrete Constant Force − Kx ∼ geometric
fKx (k) = pk · q
ex = E[Kx ] =
COMPLETE EXPECTATION OF LIFE
Average future lifetime of (x).
Z ∞
◦
ex = E[Tx ] =
t px dt
Z ∞0
E[Tx2 ] = 2
t t px dt
Var[Kx ] =
p
q2
=
m
ω−x
ω−x−n
ω−x
ω−x
2
half way to omega
= n px (n) + n qx n
2
◦
ex:n
a(x) =
1
2
Discrete DML
=
µx =
`x+t = `x e−µt
k=1
For integral x, ex is the expected number of
future birthdays.
=
1
ω−x
MODIFIED DML
Sx (t) = e−µt
◦
= µx
n
ω−x
=
◦
k| qx
Var[Tx ] =
− probability that (x) will survive t years
and die within the following u years.
is the PDF for Kx
n px
fx (t) = Fx0 (t) = −Sx0 (t) = t px µx+t
CURTATE EXPECTATION OF LIFE
∞
X
ex = E[Kx ] =
k px
n qx
ω−x−t
ω−x
1
ω−x
qx =
CONSTANT FORCE
ex =
− probability that (x) dies within t year
t qx = Fx (t)
◦
n|m qx
Intuitively, the probability that (x) dies at age
x + t.
t px − probability that (x) will attain age
x + t.
t px
Sx (t) =
ex:m+n = ex:m +m px ex+m:n
◦
◦
ex ≈ px 1+ ex+1 + qx 12
µx+t = µ
Sx (t) = Pr(Tx > t) = 1 − Fx (t)
◦
ex = ex:n +n px ex+n
If µ∗x+t = k µx+t :
d
q
dt t x
SURVIVAL DISTRIBUTION
fx (t)
Sx (t)
=
= dtp
t x
Rn
n px = exp − 0 µx+t dt
µx+t =
◦
1
ω−x
µx =
Recursion
p
q
a
ω−x
`x = (ω − x)a
a
Sx (t) = ω−x−t
ω−x
a
ω−x−n
n px =
ω−x
◦
ex =
ω−x
a+1
FRACTIONAL AGES
Uniform Distribution of Deaths (UDD)
- Use linear interpolation.
Constant Force
- Use exponential interpolation.
0
c
2012
The Infinite Actuary
SELECT MORTALITY
[age] = age selected
Read across and then down the table.
GOMPERTZ LAW
µx = Bcx
c > 1, B > 0
i
B
cx (ct − 1)
t px = exp − ln c
h
MAKEHAM LAW
µx = A + Bcx
c > 1, B > 0, A ≥ −B
h
i
B
cx (ct − 1)
t px = exp(−At) · exp − ln c
c
2012
The Infinite Actuary
C. Insurance Benefits
PURE ENDOWMENT
TERM INSURANCE
(
v Tx 0 ≤ Tx ≤ n
1
¯
Z x:n =
0
Tx > n
Z n
¯1 =
A
v t t px µx+t dt
x:n
MLC discount factor
(
0
0 ≤ Tx ≤ n
Z x:n1 =
v n Tx > n
0
n Ex
=
Ax:n1
=
vn
¯x
¯x = A
¯1 + n| A
A
x:n
Z n+m
¯
v t t px µx+t dt
n|m Ax =
n px
Var[Z x:n1 ] = 2n Ex − (n Ex )2
= v 2n n px n qx
n
Bernoulli shortcut
(
1
Z x:n
WHOLE LIFE INSURANCE
1
Ax:n
v Kx +1 Kx = 0, 1, . . . , n − 1
=
0
Kx = n, n + 1, . . .
n−1
X
=
v k+1 k| qx
0
Zx = v Kx +1 , Kx = 0, 1, 2, . . .
∞
X
Ax =
v k+1 k| qx
k=0
N-YEAR DEFERRED WHOLE LIFE
(
0
0 ≤ Tx ≤ n
¯
Z
=
x
n|
v Tx Tx > n
Z ∞
¯
v t t px µx+t dt
n| Ax =
n
¯ = n Ex · A
¯x+n
(
0
Kx = 0, 1, . . . , n − 1
n| Zx =
v Kx +1 Kx = n, n + 1, . . .
1
Ax = Ax:n
+ n| Ax
n+m−1
X
v k+1 k| qx
n|m Ax =
k=n
VARYING TERM INSURANCE
Z n
¯ 1 =
I¯A
tv t t px µx+t dt
x:n
Z0 n
¯A
¯ 1 =
D
(n − t)v t t px µx+t dt
x:n
Ax = vqx + vpx Ax+1
Draw a pain curve.
Ax:n = vqx + vpx Ax+1:n−1
For constant force:
(IA)x = vqx + vpx (Ax+1 + (IA)x+1 )
= n Ex · Ax+n
INCREASING WHOLE LIFE
¯ = Tx v Tx , Tx ≥ 0
I¯Z
x
Z ∞
¯ =
I¯A
tv t t px µx+t dt
x
0
Kx = 0, 1, 2, . . .
Pr(SBP is sufficient) =
µ
µ+δ
µ/δ
(IA)x = Ax + vpx (IA)x+1
1
= vqx h
(IA)x:n
i
1
1
+vpx (IA)x+1:n−1
+ Ax+1:n−1
1
1
(DA)x:n
= nvqx + vpx (DA)x+1:n−1
=
¯x =
A
e−n(µ+δ)
µ
µ+δ
Assuming UDD:
¯x =
A
CONSTANT FORCE
n Ex
CONTINUOUS VS. DISCRETE
¯1 =
A
x:n
i 1
A
δ x:n
¯x:n =
A
i 1
A
δ x:n
(m)
Ax
Ax =
vq
vq+d
¯1 = A
¯x (1 − n Ex ) A 1 = Ax (1 − n Ex )
A
x:n
x:n
µ
−n(µ+δ)
¯
A
=
e
n| x
µ+δ
2
1
1
¯
¯
¯
(I A)x = µ Ax
(IA)x = vq
(Ax )2
i
A
δ x
=
+ n Ex
i
Ax
i(m)
¯ = i (IA)
IA
x
δ
x
1
i
1
¯
I A x:n = δ (IA)x:n
1
i
1
¯
DA
x:n = δ (DA)x:n
¯ = i (IA) − i
I¯A
x
δ
δ
x
1
d
−
1
δ
Ax
0
¯A
¯ 1 = nA
¯1
¯ 1 + D
I¯A
x:n
x:n
x:n
1
1
1
(IA)x:n
+ (DA)x:n
= (n + 1)Ax:n
n| Ax
(IZ)x = (Kx + 1)v Kx +1 ,
PERCENTILES
k=0
¯x = v Tx , Tx ≥ 0
Z
Z ∞
¯x =
A
v t t px µx+t dt
n| Ax
RECURSION
ENDOWMENT INSURANCE
(
Tx
0 ≤ Tx ≤ n
¯x:n = v
Z
vn
Tx > n
¯x:n = A
¯1 + n Ex
A
(x:n
v Kx +1 Kx = 0, 1, . . . , n − 1
Zx:n =
vn
Kx = n, n + 1, . . .
1
Ax:n = Ax:n
+ n Ex
Assuming death occurs in the middle of the
year:
DE MOIVRE’S LAW
APV is an annuity-certain for the number of
years of insurance remaining divided by ω − x.
All of the following formulas follow that.
aω−x
a
¯
¯x = ω−x
A
Ax =
ω−x
ω−x
¯a
I¯
(Ia)ω−x
ω−x
¯ =
I¯A
(IA)x =
x
ω−x
ω−x
a
¯n
ω−x
¯a
I¯
n
¯ 1 =
I¯A
x:n
ω−x
¯ a)n
¯A
¯ 1 = (D¯
D
x:n
ω−x
¯1 =
A
x:n
1
Ax:n
=
¯x = (1 + i)1/2 · Ax
A
(m)
Ax
= (1 + i)(m−1)/2m · Ax
an
ω−x
1
(IA)x:n
=
1
(DA)x:n
=
(Ia)n
ω−x
(Da)n
ω−x
(IA)x = Ax + 1| Ax + 2| Ax + · · ·
c
2012
The Infinite Actuary
D. Life Annuities
WHOLE LIFE ANNUITY
N-YEAR DEFERRED WLA
Twin – Whole Life Insurance
No Twin
(
0
¯
Y
=
n| x
a
¯Tx − a
¯n
1 − v Tx
Y¯x = a
¯Tx =
δ
Z ∞
¯x
1−A
a
¯x =
a
¯x =
t Ex dt
δ
0
¯x
n| a
=a
¯x − a
¯x:n
¯x = 1 − δ¯
A
ax
¯x
n| a
= n Ex · a
¯x+n
¨x
n| a
= n Ex · a
¨x+n
Var Y¯x =
2A
¯x
¯x
− A
2
δ2
1 − v Kx +1
Y¨x = a
¨Kx +1 =
d
∞
X
1 − Ax
a
¨x =
a
¨x =
k Ex
d
k=0
Twin – N-Year Endowment Insurance
(
a
¯Tx 0 ≤ Tx ≤ n
Y¯x:n =
n < Tx
a
¯n
¯x:n
1−Z
=
δ
Z n
¯x:n
1−A
a
¯x:n =
a
¯x:n =
t Ex dt
δ
0
¯x:n = 1 − δ¯
A
ax:n
2A
¯x:n − A
¯x:n 2
¯
Var Yx:n =
δ2
1 − Zx:n
Y¨x:n =
d
a
¨x:n =
1 − Ax:n
d
a
¨x:n =
n−1
X
k=0
a
¨x:n = a
¨x − n Ex a
¨x+n
h
Var Y¨x:n
i
=
2A
x:n
k Ex
a
¨x =
¯x (1 − n Ex )
a
¯x:n = a
¯a = (¯
I¯
ax )2
x
0
1
vq + d
a
¨x:n = a
¨x (1 − n Ex )
DE MOIVRE’S LAW
DEFERRED TEMPORARY ANNUITY
a
¨x = 1 + vpx a
¨x+1
Use the most important identity.
No twin.
ax = vpx a
¨x+1
n+m
=
v t t px dt
¯a
¯a
I¯
+ D¯
= n¯
ax:n
x:n
x:n
¯x+1
a
¯x = a
¯x:1 + vpx a
ANNUITIES PAYABLE M-THLY
¯x:n
=a
¯x:n+m − a
¯x:1 + vpx a
¯x+1:n−1
a
¯x:n = a
Always true:
= n| a
¯x − n+m| a
¯x
a
¨x:n = 1 + vpx a
¨x+1:n−1
Ax
= n Ex · a
¯x+n:m
(I¨
a)x = 1 + vpx (¨
ax+1 + (I¨
a)x+1 )
¨x:n
Ax:n = 1 − d(m) a
(I¨
a)x = a
¨x + vpx (I¨
a)x+1
¨x+1:n−1 + (I¨
a)x+1:n−1
(I¨
a)x:n = 1 + vpx a
Assuming UDD:
For annuity-due same 3 formulas as above but
replace a
¯ with a
¨.
N-YR CERTAIN AND LIFE AFTER
(
a
¯n
0 ≤ Tx ≤ n
¯
Yx:n =
a
¯Tx n < Tx
(m)
(m)
= 1 − d(m) a
¨x
(m)
(m)
(m)
(D¨
a)x:n = n + vpx (D¨
a)x+1:n−1
= α(m) a
¨x − β(m) (0 Ex − ∞ Ex )
(m)
a
¨x:n
= α(m) a
¨x:n − β(m) (0 Ex − n Ex )
a
¯x:n = a
¯n + n| a
¯x
ADJUSTED FORCE OF MORTALITY
¨n + n| a
¨x
a
¨x:n = a
If µ∗x+t = µx+t + c and δ ∗ = δ − c, then
α(m) ≈ 1
= n Ex
(m)
a
¨x
β(m) ≈
m−1
2m
s¨x:n =
a
¨x:n
n Ex
≈a
¨x −
m−1
2m
−
m2 −1
12m2
(µx + δ)
−for other annuities, write them in terms of
a whole life annuity
ACCUMULATED APV
a
¯x:n
n Ex
= α(m) n| a
¨x − β(m) (n Ex − ∞ Ex )
Woolhouse (3 terms)
a
¯∗x = a
¯x
s¯x:n =
= α(m)· symbol w/o (m)
−β(m) (“start” - “end”)
a
¨x
(m)
¨x
n| a
INCREASING WHOLE LIFE
¯a
I¯Y¯ x = I¯
T ≥0
Tx x
¯
a
¯x − I¯A
x
¯a =
I¯
x
δ
¯ =a
¯a
I¯A
¯x − δ I¯
x
x
(m)
symbol with
(I¨
a)x:n = a
¨x:n + vpx (I¨
a)x+1:n−1
∗
n Ex
¯1 = 1 − δ¯
ax:n − n Ex
A
x:n
1
µ+δ
RECURSION
n
N-YEAR TEMPORARY ANNUITY
a
¯x =
(I¨
a)x:n + (D¨
a)x:n = (n + 1)¨
ax:n
¯x
n|m a
Yx = aKx = a
¨Kx +1 − 1 ⇒ ax = a
¨x − 1
CONSTANT FORCE
(I¨
a)x = (¨
ax )2
µ/δ
µ
Pr(SBP is insufficient) =
µ+δ
Z
Ax = 1 − d¨
ax
h i
2 A − (A )2
x
x
Var Y¨x =
d2
0 ≤ Tx ≤ n
n < Tx
VARYING TEMPORARY
Z n
¯a
I¯
=
tv t t px dt
x:n
Z0 n
¯a
(n − t)v t t px dt
=
D¯
x:n
sx:n =
ax:n
n Ex
(I¨
ax ) − (Ia)x = a
¨x
− (Ax:n )2
d2
ax:n = a
¨x:n − 1 + n Ex
c
2012
The Infinite Actuary
E. Premium Calculation
LOSS AT ISSUE
ENDOWMENT INSURANCE
SEMICONTINUOUS INSURANCE
GROSS PREMIUMS
L ∼ loss at issue random variable
Fully Continuous:
Benefit paid at the moment of death and premiums paid at beginning of the year.
¯
¯x = Ax
P A
a
¨x
¯1
A
x:n
¯1
P A
x:n =
a
¨x:n
¯
¯x:n = Ax:n
P A
a
¨x:n
Lg = PV of future benefits
+ PV of future expenses
− PV of future premiums
L = PV of benefits − PV of premiums
E[L] = APVFB − APVFP
¯x:n − P¯ a
E[L] = S A
¯x:n
2
¯
2A
¯x:n − A
¯x:n 2
Var[L] = S + Pδ
Fully Discrete:
E[L] = SAx:n − P a
¨x:n
2
2
P
2A
Var[L] = S + d
x:n − (Ax:n )
EQUIVALENCE PRINCIPLE
E[L] = 0 ⇒ APVFB = APVFP
PERCENTILE PREMIUMS
N-YEAR TERM INSURANCE
Whole Life and Endowment Insurance:
Var[Z]
(1 − SBP)
1
¯x + δ
P¯ A
¯
¯x
P A
¯x =
A
¯x + δ
P¯ A
¯1
P¯ A
x:n
1
P x:n
=
Whole Life and Endowment Insurance Only:
¯
Ax:n
¯x = Ax
P¯ A
Px:n =
a
¯x
a
¨x:n
a
¯x =
¯1
P¯ A
x:n
=
¯1
A
x:n
Group of policies:
a
¯x:n
E[S] + Φ−1 (1 − x%)
2
KEY PREMIUM IDENTITIES
¯x = 1 − δ
P¯ A
a
¯x
¯
¯x = δ Ax
P¯ A
¯x
1−A
Px:n =
Px:n
Ax:n
1
−d
a
¨x:n
dAx:n
=
1 − Ax:n
a
¨x:n =
New Equivalence Premium:
E[Lg ] = 0 ⇒ APVFB + APVFE = APVFP
APV of Benefits + APV of Expenses
APV Annuity
2
r
Var(Z)
Var[Lg ] = S + E + G−e
d
G=
APV benefit
benefit premium =
APV annuity
Var[L] =
E[Lg ] = APVFB + APVFE − APVFP
¯x:n − P¯ A
¯ 1
= P¯ A
x:n
1
Ax:n
a
¨x:n
Px:n
=
Px:n + d
N-PAY WHOLE LIFE
¯x
A
¯ ¯
n P Ax =
a
¯x:n
n Px
=
N-YEAR DEFERRED WLA
¯x
n| a
P¯ n| a
¯x =
a
¯x:n
¨x
n| a
=
Var[L] = p · 2 Ax
Fully Continuous Whole Life:
Premiums paid m-thly instead of annually.
Px
(m)
n Px
µ
µ+δ
µ/δ
Ax
=
=
(m)
a
¨x
Ax:n
DE MOIVRE’S LAW
(m)
a
¨x:n
Ax
= (m)
a
¨x:n
Use key premium identities and most important identity.
¨x
n| a
a
¨x:n
(P-P) / P
n Px
1
− P x:n
P x:n1
Fully Continuous:
E[L] = SAx − P a
¨x
2
P
2 A − (A )2
Var[L] = S + d
x
x
CONSTANT FORCE
¯x = µ
P¯ A
Px = vq
1
¯1
P x:n
P¯ A
= vq
x:n = µ
Pr(benefit prem is sufficient) =
(m)
Ax
a
¨x:n
WHOLE LIFE
Fully Discrete:
FRACTIONAL PREMIUMS
(m)
Px:n
P
¯x − P¯ a
E[L] = S A
¯x
2
¯
2A
¯x − A
¯x 2
Var[L] = S + Pδ
Single policy:
Find the policy you must “win” on and then
solve for the premium it will take to win on
that death.
¯x
Var[L] = 2 A
1
Px:n + d
p
Var[S] = 0
COMPARE VARIANCE
Two identical policies (whole life or endowment insurance)
P1 → L1
= Ax+n
Px:n − n Px
= 1 − Ax+n
P x:n1
1
Px:n − P x:n
P x:n1
=1
P2 → L2
Var[L2 ]
=
Var[L1 ]
P2 + d
P1 + d
2
c
2012
The Infinite Actuary
F. Reserves
#1 PROSPECTIVE METHOD
#4 PAID UP
#8 (P-P) / P
#11 RESERVE CREATION FORMULA
All Policies and All Premiums. All other methods only work when premiums are determined
using the equivalence principle.
Whole Life, Endowment Insurance, Term Insurance and Limited Pay Whole Life Only
Accumulated Difference in Premiums equals
Difference in Reserves
Useful when death benefit is a function of the
reserve.
Reserve = APV of the future benefit times
the percentage of the future benefit that you are not funding with
future premiums
Remember
Reserve = APVFB − APVFP
Example – fully continuous whole life:
¯ ¯
¯
¯ ¯ ¯x+t
t V Ax = Ax+t − P Ax a
Example – full continuous whole life:
"
#
¯x
P¯ A
¯
¯
¯
1
−
V
A
=
A
x
t
x+t
¯x+t
P¯ A
1
P
1
= s¨x:n
Example
Px:n − Px
= n Vx:n − n Vx = 1 − n Vx
P x:n1
#9 ANNUITY FORM SPINOFF
All Policies under EP.
#5 ANNUITY FORM
Whole Life and Endowment Insurance Only
Reserve = AAVPP − AAVPB
Whole Life and Endowment Insurance Only
Example – fully discrete whole life:
t Vx:n
= Px:n s¨x:t − t kx
ACCUMULATED COST OF INS
t kx
=
=
t Ex
dx (1+i)t−1 +dx+1 (1+i)t−2 +···+dx+t−1
lx+t
For one year:
1 kx
=
Example – fully discrete endowment ins.:
a
¨x+t:n−t
t Vx:n = 1 −
a
¨x:n
1
Ax:t
Thinking in terms of life tables:
t kx
Reserve = 1 minus (the annuity at the dot divided by the annuity at issue)
qx
px
#6 LIFE INSURANCE FORM
Whole Life and Endowment Insurance Only
Reserve = insurance at the dot minus the insurance at issue divided by one minus the insurance at issue
Example – fully discrete whole life:
Ax+t − Ax
t Vx =
1 − Ax
#3 DIFFERENCE IN PREMIUM
Whole Life, Endowment Insurance, Term Insurance and Limited Pay Whole Life Only
Reserve = accumulated difference in premium
you want to charge and premium
you are actually charging
Example – fully continuous whole life:
¯ ¯
¯ ¯
¯ ¯
a
¯x+t
t V Ax = P Ax+t − P Ax
#7 BENEFIT PREMIUM FORM
Whole Life and Endowment Insurance Only
1. cost of providing ensuing year’s DB
2. rest is for reserve creation
tV
= P s¨t t−1
X
−
vqx+h (bh+1 − h+1 V ) (1 + i)t−h
h=0
#2 RETROSPECTIVE METHOD
Example – fully discrete endowment ins.:
Each premium accounts for two items:
x:n
DB = Reserve: t V = P s¨t
DB = 1 + Reserve:
tV
= P s¨t −
t−1
X
qx+h (1 + i)t−h−1
h=0
a+b Vx = 1 − (1 − a Vx ) (1 − b Vx+a )
DB = 1 + Reserve and qx+h = q :
#10 RESERVE RECURSION
Start with the terminal reserve from the previous year, collect the premium and put those
in the bank. If the policyholder dies pay the
death benefit. If the policyholder lives setup
the next reserve.
Example - fully discrete 20-year endowment insurance on (40):
10 V40:20 + P40:20 (1+i) = q50 +p50 11 V40:20
NAR version: = q50 1 − 11 V40:20 + 11 V40:20
tV
= (P − vq) s¨t
VARIANCE LOSS
For Whole Life and Endowment Insurance:
¯ 2
Var(t L) = S + Pδ
Var(v U )
EP ⇒
=
Var(v U )
(1 − SBP)2
Example – fully discrete endowment ins.:
Var(t L) =
S+
EP ⇒
=
P
d
2
2A
x+t:n−t
2A
x+t:n−t
2
− Ax+t:n−t
2
− Ax+t:n−t
(1 − Ax:n )2
Reserve = premium at the dot minus the premium at issue divided by the premium at the dot plus delta (or d)
CONSTANT FORCE
Example – fully discrete whole life:
Under EP, level premium whole life and term
insurance reserves are 0.
t Vx
=
Px+t − Px
Px+t + d
c
2012
The Infinite Actuary
GROSS PREMIUM RESERVES
MODIFIED RESERVES
To find the gross premium, G, solve new EP
α − lower modified prem in the first year
APVFP = APVFB + APVFE
β − higher modified prem in the renewal years
Expense Reserves:
FPT Reserves:
kV
e
= APVFE − APVFL
kV
e
= AAVPL − AAVPE
1
α = Ax:1
β = benefit premium for insurance issued
to (x + 1)
Gross Premium Reserve:
kV
g
= kV
n
+ kV
1V
m
=0
tV
m
= (t − 1) reserve for insurance issued
to (x + 1)
e
Can also use recursion for expense and gross
premium reserves.
GAIN
SEMI-CONTINUOUS RESERVES
Assuming UDD
i
¯
t V A x = δ t Vx
i
1
¯1
t V Ax:n = δ t V x:n
i
1
1
¯
t V Ax:n = δ t V x:n + t V x:n
INTERIM BENEFIT RESERVES
Exact Method:
(h V + πh ) (1 +
i)s
v 1−s
= s qx+h ·
+s px+h · h+s V
Approximation: use linear interpolation between initial reserve and the terminal reserve.
Gain = Actual Profit − Expected Profit
Profit = (t V + Gt − et ) (1 + i)
−qx+t (St + Et ) − px+t · t+1 V
Analysis of Surplus:
expenses: (et − e0t ) (1 + i) + (Et − Et0 ) qx+t
interest: (i0 − i) (t V + Gt − e0t )
0
mortality: qx+t − qx+t
(St + Et0 − t+1 V )
Order matters. Use actual experience if you
have already accounted for that source.
POLICY ALTERATIONS
Reduced Paid-Up:
RP U =
THIELE’S DIFFERENTIAL EQN
t CVx
Ax+t
Extended Term (solve for n):
d
V
dt t
= δt · t V +Gt −et −(St + Et − t V ) µ[x]+t
t+h V − h Gt − et − (St + Et ) µ[x]+t
tV ≈
1 + h µ[x]+t + δt
t CVx
1
= Ax+t:n
For endowment insurance if the cash value is
large enough to cover the remaining term, then
solve for reduced pure endowment benefit, P E
t CVx
1
= Ax+t:n−t
+ P E · n−t Ex+t
c
2012
The Infinite Actuary
G. Markov Chains
NOTATION
ij
t px
MULTIPLE DECREMENT MODEL
− probability that a subject in state i at
time x will be in state j (where j may
equal i) at time x + t
ii
t px
exit
1
− probability that a subject in state i at
time x stays in state i continuously until time x + t
ii
t px
exit
2
≤ t pii
x
ANNUITIES
THIELE’S DIFFERENTIAL EQN
Pay 1 per year continuously while in state j:
Z ∞
a
¯ij
e−δt t pij
x =
x dt
d
V (i)
dt t
0
(i)
= δt t V (i) − Bt
n
X
(ij)
−
µij
+ t V (j) − t V (i)
x+t St
j=0,j6=i
If payable at the start of the year:
∞
X
a
¨ij
v k k pij
x =
x
k=0
Euler’s method:
t−h V
active
0
(i)
(i)
= t V (i) (1 − δt h) + hBt
n
X
(ij)
+h
µij
+ t V (j) − t V (i)
x+t St
j=0,j6=i
INSURANCE
DISCRETE MARKOV CHAINS
P(t) is the transition matrix at time t
ij
t px
= ij entry of
ii
t px
ii
ii
= pii
x · px+1 · · · px+t−1
P(x)
·
P(x+1)
exit
m
· · · P(x+t−1)
0j
t px
CONTINUOUS MARKOV CHAINS
µij
x
ii
t px
=
1 − t p00
x
R
= exp − 0t µ0•
x+s ds
h→0
Probability of transition from one state to another depends on the model.
KOLMOGOROV’S FORWARD EQNS
X
ik kj
ij jk
d
pij =
t px µx+t − t px µx+t
dt t x
k=0,k6=j
d
pij
dt t x
PERMANENT DISABILITY MODEL
Healthy
0
µ01
x
µ02
x
Disabled
1
=
prob “move
−
into j”
at time x + t
Euler’s method turns a continuous Markov
Chain into a discrete chain with time increments of h and transition “probabilities”
(
hµij
i 6= j
ij
x+t
p
=
h x+t
1 − hµi•
x+t i = j
=
Rt
00
0 s px
j6=k
If all transition forces are constant and no reentry into a state
1
a
¯ii
x = i•
µ +δ
For the multiple decrement model with constant transitions:
µ0j
¯0j
A
x = 0•
µ +δ
RESERVES
tV
(i)
− rate of payment of benefit while the
policyholder is in state i
(ij)
− lump sum benefit payable instantaneously at time t on transition from
state i to state j
St
PREMIUMS
APV of benefit
APV of annuity
− reserve at duration t for a subject in
state i at that time
(i)
Bt
benefit premium =
01
t px
prob “move
out of j”
at time x + t
µ12
x
Dead
2
0
CONSTANT FORCE AND NO RE-ENTRY
ij
h px
for i 6= j
h
R
t i•
= exp − 0 µx+s ds
= lim
00
t px
µ0j
x
µ0•
x
Pay 1 on each future transfer into k:
Z ∞
X
ij jk
¯ik =
A
e−δt
t px µx+t dt
x
11
µ01
x+s t−s px+s ds
c
2012
The Infinite Actuary
H. Multiple Decrement Models
MULTIPLE DECREMENT MODEL
FORCE OF DECREMENT - 2
(τ )
exit
1
µx+t =
(j)
µx+t
exit
2
active
0
=
(τ )
d
q
dt t x
(τ )
t px
(j)
− prob. of decrement due to any cause
m
X
(τ )
(j)
t qx =
t qx
(j)
Ax =
(τ )
j
v k k−1 px qx+k−1
FORCES OF DECREMENT - 1
(τ )
(1)
(2)
=
(τ )
0 s px
Rt
cause
1
(m)
(1)
qx
(2)
(j)
· µx+s ds
Pr(J = j | T = t) =
µx+t
qx
alive
0
(j)
t qx
(j)
=
(j)
¯x
A
=
µx
(τ )
µx
R∞
0
exit
1
µ01
x+t
exit
2
active
0
cause
3
(τ )
(j)
v t t px µx+t dt
CONSTANT TRANSITION FORCES
(j)
(τ )
µ(j)
t qx = (τ ) 1 − t px
µ
(j)
(τ ) t
(j)
t qx
1 − px
t qx =
(τ )
µ(j)
+δ
1
a
¯x = (τ )
µ
+δ
(j)
(j)
(τ )
µ
¯
A
1 − n Ex
= (τ )
1
x:n
=
µ(τ )
µ
0 is redundant so we use:
(j)
µx+t
1
2
X
(i)
dx
Constant Force and UDDMDT:
(j) (τ )
(τ ) qx /qx
p0 (j)
x = px
UDDAST:
For 2 UDDASTs use the midpoint:
(1)
qx = q 0 (1)
1 − 12 q 0 (2)
x
x
q 0 (1)
x
alive
0
For 3 or more UDDASTs use integration:
0 (j)
tp x
(j)
· µx+t = q 0 (j)
x - factor out of integral
For 3 UDDASTs (if you like memorizing):
(1)
0 (3) + 1 q 0 (2) q 0 (3)
qx = q 0 (1)
1 − 12 q 0 (2)
x +q x
x
x
3 x
Hybrids:
cause
1
Draw a picture. For 2 competing UDDASTs
use the midpoint of the interval.
q 0 (2)
x
alive
0
ASSET SHARE
Recursion:
[k AS + G(1 − ck ) − ek ] (1 + i)
(d)
0 (2)
0 (m)
= t p0 (1)
x · tp x · · · tp x
(w)
(τ )
= qx+k + qx+k · k+1 CV + px+k · k+1 AS
q 0 (3)
x
alive
0
(τ )
t px
cause
2
+δ
ABSOLUTE RATE OF DECREMENT
R
t (j)
0 (j)
0 (j)
t q x = 1 − t p x = 1 − exp − 0 µx+s ds
exit
m
−
t qx
µ02
x+t
µ0m
x+t
dx
(τ )
`x
(3)
qx
1 − t px
(τ )
cause
2
(τ )
µx+t
If all forces are a constant multiple of the total
force for all t
¯(j)
A
x
=
q 0 (j)
x =
Rates of Decrement
k=1
µ0j
x+t
Probabilities of Decrement
i6=j
j=1
∞
X
Super secret approximation:
(j)
d
q
dt t x
(τ )
t px
(τ )
qx
Each cause is independent of the other causes.
(j)
(j)
qx − prob. of decrement due to cause #j
MDT vs. AST
µx+t = µx+t + µx+t + · · · µx+t
R
(τ )
t (τ )
t px = exp − 0 µx+s ds
(j)
t qx
exit
m
ABSOLUTE RATE OF DECREMENT
cause
3
Notice how in the rate of decrement “worlds”
there is only one cause of decrement. For the
rates of decrement we have 3 tables and for the
multiple decrement “world” we have one table
with 3 competing decrements.
0 (j)
tq x
n-th Asset Share using Asset Share Creation:
(d)
(w)
n−1
X
G(1 − ck ) − ek − vqx+k − vqx+k · k+1 CV
k=0
(τ )
n−k Ex+k
For additional premium of ∆G:
∆AS =
n−1
X
∆G(1 − ck )
k=0
(τ )
n−k Ex+k
(j)
≥ t qx
=
n−1
X
∆G(1 − ck )k px v k
k=0
(τ )
n Ex
(τ )
c
2012
The Infinite Actuary
I. Multiple Life Models
JOINT-LIFE STATUS
LAST-SURVIVOR STATUS
CONTINGENT PROBABILITIES
REVERSIONARY ANNUITY
Txy = min[Tx , Ty ]
Txy = max[Tx , Ty ]
1
n q xy
(y) gets money after (x) dies:
Failure on the first death.
Txy = Tx + Ty − Txy
t qxy
t pxy
− probability first death of (x) and (y)
occurs within t years
that for (xy) = that for (x) + that for (y)
−that for (xy)
− probability that (x) will attain x + t
and (y) will attain y + t
t qxy
t pxy
= t px · t py
t qxy
= 1 − t pxy
(work with p’s)
fxy (t) = t pxy · µx+t:y+t
µx+t:y+t = µx+t + µy+t
Z n
n qxy =
t pxy µx+t:y+t dt
◦
0
∞
Z
exy =
t pxy
0
∞
X
exy =
dt
k pxy
k=1
0
¯xy =
A
t qxy
= t qx · t qy
t pxy
− probability at least one of (x) or (y)
survives for t years
(work the the q’s)
t pxy
= 1 − t qxy
t pxy
= t px + t py − t pxy
∞
v
Other versions:
a
¯x|y:n = a
¯y:n − a
¯xy:n
a
¯x:n |y = a
¯y − a
¯xy:n
0
2
n q xy
In other words:
a
¯u|v = a
¯v − a
¯uv
n
Z
=
t py
µy+t · t qx dt
= n q xy1 + n q xy2
1
n q xy
= n q xy2 + n qx · n py
Probability (x) dies more than n years after
death of (y):
= n+1 qxy − n qxy
n| qxy
= n| qx + n| qx − n| qxy
◦
◦
2
· ∞ q x+n:y
n px
◦
t
t pxy
µx+t:y+t dt
a
¯xy = a
¯x + a
¯y − a
¯xy
CONTINGENT INSURANCE
Z ∞
¯1 =
A
v t · t px µx+t · t py dt
xy
¯x + A
¯y − A
¯xy
¯xy = A
A
1
Axy:n
=
1
Ax:n
+
1
Ay:n
−
−d
0
1
Axy:n
c
¯ 2=
A
xy
¯xy = 1 − δ¯
A
axy
1
a
¨ xy
− probability that (y) dies after (x) and
before n years from now
If (x) and (y) independent:
Z n
1
n q xy =
t px µx+t · t py dt
n qy
n| qxy
0
Pxy =
s
0
fxy (t) = t px µx+t + t py µy+t − t pxy µx+t:y+t
◦
0
2
n q xy
a
¯x|y = a
¯y − a
¯xy
COMMON SHOCK
Find
total
total
total
the total force for each status:
force on (x) = µ∗x + λ
force on (y) = µ∗y + λ
force on (xy) = µ∗x + µ∗y + λ
Pr[(x) dies first] =
µ∗
x
∗
µ∗
x +µy +λ
Pr[(y) dies first] =
µ∗
y
∗ +λ
µ∗
+µ
x
y
exy = ex + ey − exy
exy = pxy (1 + ex+1:y+1 )
Z ∞
a
¯xy =
v t t pxy dt
Z
− probability that last death happens
within t years
1
n q xy
− probability (x) dies before (y) and before n years from now
Z n Z ∞
=
fxy (s, t) dt ds
∞
Z
v t · t py µy+t · t qx dt
0
EXACTLY ONE STATUS
¯ 2=
A
xy
∞
Z
Pr[Tx = Ty ] =
¯x =
A
µ∗
x +λ
µ∗
x +λ+δ
¯y =
A
µ∗
y +λ
µ∗
y +λ+δ
λ
∗
µ∗
x +µy +λ
¯y+t dt
v t · t px µx+t · t py A
0
[1]
t pxy
− probability exactly 1 of (x) and (y) live
t years
¯y = A
¯ 1 +A
¯ 2
A
xy
xy
[1]
t pxy
= t px + t py − 2 t pxy
SBP for a payment of 1 at the death of (x) if
he dies more than n years after the death of
(y):
[1]
¯x + a
¯y − 2¯
axy
a
¯xy = a
n Ex
¯ 2
·A
x+n:y
CONSTANT FORCE
µx +µy
µx +µy +δ
¯xy =
A
a
¯xy =
1
n q xy
1
µx +µy +δ
=
¯1 =
A
xy
µx
q
µx +µy n xy
µx
µx +µy +δ
c
2012
The Infinite Actuary
DE MOIVRE’S LAW
◦
exx =
ω−x
3
◦
◦
◦
exy = y−x px · eyy +y−x qx · ey
2
n q xy
=
¯1 =
A
xy
1
q
2 n xy
a
¯y:ω−x
ω−x
c
2012
The Infinite Actuary
J. Other Topics
PENSION MATHEMATICS
PROFIT SIGNATURE
Replacement ratio:
pension income in year after retirement
R=
salary in the year before retirement
(
Πt =
Profit(0)
t−1 px · Profit(t)
t=0
t>0
where t−1 px is the probability that policy is in
force at beginning of year t.
Salary Scale:
sy
salary received in year of age y to y + 1
=
sx
salary received in year of age x tp x + 1
v(t) − current market price on a t year zerocoupon bond that pays $1 at time t.
Solve for j such that:
n
X
Πt vjt = 0
t=0
− think of it as the discount function.
yt − t year spot rate of interest
v(t) =
1
(1 + yt )t
PROFIT MARGIN
f (t, t+k) − forward rate from time t to t + k
NPV
Pa
¨x:n
a
¨(x)y =
k px
v(k)
k=0
A risk is diversifiable if:
q
P
Var[ N
i=1 Xi ]
lim
=0
N →∞
N
DISCOUNTED PAYBACK PERIOD
Find the smallest m such that:
m
X
Πt vrt ≥ 0
UNIVERSAL LIFE - Notation
− Expenses Incurred
+ Interest Earned on Premium less Expenses
ECt − expense charge for year t
The profit for year 0 is the negative of the expenses incurred at time 0.
Annual Profit =
Previous Rsv accumulated with interest
+ Premium Collected
− Expected Cost of Benefits
Type A:
Annual Profit =
ADBt = F A − AVt
DBt = F A
Previous Rsv (usually AV)
+ Premium Collected
− Expenses Incurred
ADBt = F A
DBt = F A + AVt
+ Interest Earned on
Prev Rsv plus Premium less Expenses
− Expected Cost of Benefits
1
∗
CoIt = ADBt Ax+t−1:1
= ADBt · vq qx+t−1
− Expected Surrender Benefits
− Expected Cost of Ending Rsv
CoItA =
∗
[F A−(AVt−1 +Pt −ECt −CoItA )(1+ict )]vq qx+t−1
∗
[F A−(AVt−1 +Pt −ECt )(1+ict )]vq qx+t−1
∗
1−vq qx+t−1
(1+ic
t)
Type B:
CoItB
= FA ·
∗
vq qx+t−1
CORRIDOR FACTORS
To qualify as life insurance the death benefit
must be at least a certain multiple (γt for year
t) of the account value.
ADBtc = (γt − 1)AVt
t=0
F A − face amount
AVt − account value at end of year t
CVt − cash value at end of year t
DBt − death benefit for year t
ADBt − additional death benefit for year t
CoIt − cost of insurance for year t
SCt − surrender charge for year t
ict − credited interest rate in year t
It − credited interest in year t
vq − discount factor used in CoI calc
∗
qx+t−1
−mort. rate used in CoI calc for year t
ANNUAL PROFIT
The profit for year 0 is the negative of the expenses incurred at time 0.
CoItA =
k
0
∞
X
DBt = AVt + ADBt
Type A:
Profit Margin =
v(t)
(1 + f (t, t + k)) =
v(t + k)
Z ∞
¯ y=
A(x)
v(t) t px µx+t dt
UNIVERSAL LIFE - Annual Profit
Type B:
INTERNAL RATE OF RETURN
INTEREST RATE RISK
UNIVERSAL LIFE - COI
UNIVERSAL LIFE - Roll Forward
ADBtf = F A − AVt
Account Value Roll Forward:
ADBt = max[ADBtc , ADBtf ]
Starting AV (AVt−1 )
∗
CoIt = ADBt · vq qx+t−1
+ Premium (Pt )
− Expense Charge (ECt )
− Mortality Charge (CoIt )
+ Credited Interest (It )
= Ending Account Value (AVt )
AVtA =
∗
(AVt−1 +Pt −ECt −F A vq qx+t−1
)(1+ict )
∗
1−vq qx+t−1
(1+ic
t)
AVtB = same as numerator for AVtA
CVt = max[AVt − SCt , 0]
− Expected Cost of Ending Rsv
c
2012
The Infinite Actuary
