chapter 3 survival distributions and life tables exam m notes f05

7/22/2019 Chapter 3 Survival Distributions and Life Tables Exam M Notes F05

1/26

Chapter 3: Survival Distributions and Life Tables

Distribution function of X :

F X (x) = Pr( X x)

Survival function s(x):

s(x) = 1 F X (x)

Probability of death between age x andage y:

Pr( x < X z) = F X (z) F X (x)= s(x) s(z)

Probability of death between age x and

age y given survival to age x:

Pr( x < X z |X > x ) = F X (z) F X (x)

1 F X (x)

= s(x) s(z)

s(x)

Notations:

t q x = Pr[ T (x) t]= prob. ( x) dies within t years= distribution function of T (x)

t px = Pr[ T (x) > t ]= prob. ( x) attains age x + t= 1 tq x

t |u q x = Pr[ t < T (x) t + u]= t+ u q x tq x= t px t+ u px= t px uq x + t

Relations with survival functions:

t px = s(x + t)

s(x)

t q x = 1 s(x + t)

s(x)

Curtate future lifetime ( K (x) greatestinteger in T (x)):

Pr[K (x) = k] = Pr[ k T (x) < k + 1]= k px k+1 px

= k px q x + k= k| q x

Force of mortality (x):

(x) = f X (x)

1 F X (x)=

s (x)s(x)

Relations between survival functions andforce of mortality:

s(x) = exp x

0

(y)dy

n px = exp x + n

x(y)dy

Derivatives:

ddt t

q x = t px (x + t) = f T (x ) (t)

ddt t

px = t px (x + t)

ddt

T x = lxddt

Lx = dx

ddt

ex = (x)ex 1

Mean and variance of T and K :

E [T (x)] complete expectation of life

ex =

0

t px dt

E [K (x)] curtate expectation of life

ex

=

k=1 k p

x

V ar[T (x)] = 2

0

t t px dt e2x

V ar[K (x)] =

k=1

(2k 1) k px e2x

Total lifetime after age x: T x

T x =

0 lx + t dtExam M - Life Contingencies - LGD c 1


2/26

Total lifetime between age x and x + 1 : Lx

Lx = T x T x +1

=1

0lx + t dt =

1

0lx t px dt

Total lifetime from age x to x + n: n Lx

n Lx = T x T x + n =n 1

k=0

Lx + k

=n

0

lx + t dt

Average lifetime after x: ex

ex = T x

lx

Average lifetime from x to x + 1 : ex : 1

ex : 1 = Lxlx

Median future lifetime of (x): m (x)

P r [T (x) > m (x)] = s(x + m(x))

s(x) =

12

Central death rate: m x

mx = lx lx +1

Lx

n mx = lx lx + n

n LxFraction of year lived between age x andage x + 1 by dx : a(x)

a(x) =

1

0 t t px (x + t) dt1 0 t px (x + t) dt

= Lx lx +1lx lx +1

Recursion formulas:

E [K ] = ex = px (1 + ex +1 )E [T ] = ex = px (1 + ex +1 ) + q x a(x)

ex = ex : n + n px ex + nex = ex : n + n px ex + n

E [K (m + n)] = ex : m + n= ex : m + m px ex + m : n

E [T (m + n)] = ex : m + n= ex : m + m px ex + m : n

Exam M - Life Contingencies - LGD c 2


3/26

Chapter 4: Life Insurance

Whole life insurance: Ax

E [Z ] = Ax =

0 vt t px x (t)dt

V ar[Z ] = 2Ax ( Ax )2

n-year term insurance: A1x : n

E [Z ] = A1x : n =n

0

vt t px x (t)dt

V ar[Z ] = 2

A1x : n ( A

1x : n )

2

m-year deferred whole life: m | Ax

E [Z ] = m | Ax =

m

vt t px x (t)dt

m | Ax = Ax A1x : m

n-year pure endowment: A 1x : n

E [Z ] = A 1x : n = vn n px n E x

V ar[Z ] = 2A 1x : n ( A 1

x : n )2 = v2n n px n q x

n-year endowment insurance: Ax : n

E [Z ] = Ax : n =n

0vt t px x (t)dt + vn n px

V ar[Z ] = 2Ax : n ( Ax : n )2

Ax : n = A1x : n + A 1

x : n

m-yr deferred n-yr term: m |n Ax

m |n Ax = m E x A 1x + m : n= A1x : m + n A

1x : m

= m | Ax m + n | Ax

Discrete whole life: Ax

E [Z ] = Ax =

k=0

vk+1 k px q x + k

=

k=0

vk+1 k| q x

V ar[Z ] = 2Ax (Ax )2

Discrete n-year term: A1x : n

E [Z ] = A1x : n =n 1

k=0

vk+1 k px q x + k

V ar[Z ] = 2A1x : n (A1x : n )

2

Discrete n-year endowment: Ax : n

E [Z ] = Ax : n =n 1

k=0

vk+1 k px q x + k + vn n px

V ar[Z ] = 2Ax : n ( Ax : n )2

Ax : n = A1x : n + A 1

x : n

Recursion and other relations:

Ax =

A

1

x : n + n | Ax2Ax = 2A1x : n + n |2Ax

n | Ax = n E x Ax + nAx = A1x : n + n E x Ax + nAx = vq x + vpx Ax +1

2Ax = v2q x + v2 px 2AxA1x : n = vq x + vpx A

1x +1: n 1

m |n Ax = vpx (m 1)|n Ax +1Ax : 1 = v

Ax : 2 = vq x + v2 px



4/26

Varying benet insurances:

(I A)x =

0

t + 1 vt t px x (t)dt

(I A)1x : n =n

0

t + 1 vt t px x (t)dt

(IA)x =

0

t vt t px x (t)dt

(IA)1x : n =n

0

t vt t px x (t)dt

(D A)1x : n =

n

0 (n t )vt

t px x (t)dt

(DA )1x : n =n

0

(n t)vt t px x (t)dt

(IA)x = Ax + vpx (IA)x +1= vq x + vpx [(IA)x +1 + Ax +1 ]

(DA )1x : n = nvq x + vpx (DA ) 1

x +1: n 1

(IA)1x : n + ( DA )1x : n = n A

1x : n

(I A)1x : n + ( D A)1x : n = ( n + 1) A

1x : n

(IA)1x : n + ( DA )1x : n = ( n + 1) A

1x : n

Accumulated cost of insurance:

n kx =A1x : nn E x

Share of the survivor:

accumulation factor = 1n E x

= (1 + i)nn px

Interest theory reminder

a n = 1 vn

i

a n = 1 vn

=

i

a n

a = 1

, a =

1i

, a = 1d

(Ia ) n = a n nv n

i

(Da ) n = n a n

(Ia ) = 1

2(n + 1) a n = ( Ia ) n + ( Da ) n

s 1 = i

d = iv

(Ia ) = 1

id =

1 + ii2

Doubling the constant force of interest

1 + i (1 + i)2

v v2

i 2i + i2

d 2d d2

i

2i + i2

2

Limit of interest rate i = 0 :

Axi=0 1

A1x : ni=0 n q x

n | Axi=0 n px

Ax : ni=0 1

m |n Axi=0

m |n q x(IA)x

i=0 1 + ex(IA)x

i=0 ex



5/26

Chapter 5: Life Annuities

Whole life annuity: ax

ax = E [a T ] =

0 at t px (x + t)dt

=

0

vt t px dt =

0

t E x dt

V ar[a T ] =2Ax ( Ax )2

2

n-year temporary annuity: ax : n

ax : n =

n

0 vt

t px dt =

n

0t E x dt

V ar[Y ] =2Ax : n ( Ax : n )2

2

n-year deferred annuity: n | ax

n | ax =

n

vt t px dt =

n

t E x dt

n | ax = vn n px ax + n = n E x ax + n

V ar[Y ] = 2

v2n n px (ax + n 2ax + n ) (n | ax )2

n-yr certain and life annuity: ax : n

ax : n = ax + a n ax : n= a n + n | ax = a n + n E x ax + n

Most important identity1 = ax + Ax

ax = 1 Ax

Ax : n = 1 ax : n

2Ax : n = 1 (2 ) 2ax : n

ax = 1 Ax

d

ax : n = 1 Ax : n

d

1 = dax : n + Ax : n

Recursion relations

ax = ax : 1 + vpx ax +1ax = ax : n + n | axax = 1 + vpx ax +1

2ax = 1 + v2 px 2ax +1ax : n = 1 + vpx ax +1: n 1a (m )x : n = a

(m )x n E x a

(m )x + n

ax = vpx + vpx ax +1ax : n = vpx + vpx ax +1: n 1

(I a)x = 1 + vpx [(I a)x +1 + ax +1 ]= ax + vpx (I a)x +1

Whole life annuity due: ax

ax = E [a K +1 ] =

k=0

vk k px

V ar[a K +1 ] =2Ax (Ax )2

d2

n-yr temporary annuity due: ax : n

ax : n = E [Y ] =n 1

k=0

vk k px

V ar[Y ] =2Ax : n (Ax : n )2

d2

n-yr deferred annuity due: n | ax

n | ax = E [Y ] =

k= n

vk k px

= ax ax : n= n E x ax + n

n-yr certain and life due: ax : n

ax : n = ax + a n ax : n

= a n +

k= n

vk k px

= a n + n | ax



6/26

Whole life immediate: ax

ax = E [a K ] =

k=1

vk k px

ax =

1 (1 + i)Axi

m-thly annuities

a (m )x = 1 A(m )x

d(m )

V ar[Y ] =2A(m )x (A(

m )x )2

(d(m ) )2

a (m )x = a(m )x

1m

a (m )x : n

= a (m )x : n

1

m(1 n E x )

Accumulation function:

sx : n = ax : n

n E x=

n

0

1n t E x + t

Limit of interest rate i = 0 :

axi=0 ex

axi=0 1 + ex

axi=0 ex

ax : ni=0 ex : n

ax : ni=0 1 + ex : n 1

ax : ni=0 ex : n



7/26

Chapter 6: Benet Premiums

Loss function:Loss = PV of Benets PV of PremiumsFully continuous equivalence premiums(whole life and endowment only):

P ( Ax ) =Axax

P ( Ax ) = Ax1 Ax

P ( Ax ) = 1ax

V ar[L] = 1 +P

22Ax ( Ax )2

V ar[L] =2Ax ( Ax )2

( ax )2

V ar[L] =2Ax ( Ax )2

(1 Ax )2

Fully discrete equivalence premiums(whole life and endowment only):

P (Ax ) = Ax

ax= P x

P (Ax ) = dAx1 Ax

P (Ax ) = 1ax

d

V ar[L] = 1 + P d

22Ax (Ax )2

V ar[L] =2Ax (Ax )2

(dax )2

V ar[L] =2Ax (Ax )2

(1 Ax )2

Semicontinuous equivalence premiums:

P ( Ax ) =Axax

m-thly equivalence premiums:

P (m )# = A#a (m )#

h-payment insurance premiums:

h P ( Ax ) =A

xax : h

h P ( Ax : n ) =Ax : nax : h

h P x = Axax : h

h P x : n = Ax : n

ax : h

Pure endowment annual premium P 1x : n :it is the reciprocal of the actuarial accumulatedvalue sx : n because the share of the survivor whohas deposited P 1x : n at the beginning of each yearfor n years is the contractual $1 pure endow-ment, i.e.

P 1x : n sx : n = 1 (1)

P minus P over P problems:The difference in magnitude of level benet pre-miums is solely attributable to the investmentfeature of the contract. Hence, comparisons of

the policy values of survivors at age x + n maybe done by analyzing future benets:

( n P x P 1x : n )sx : n = Ax + n = n P x P 1x : n

P 1x : n

(P x : n n P x )sx : n = 1 Ax + n = P x : n n P x

P 1x : n

(P x : n P 1x : n )sx : n = 1 = P x : n P 1x : n

P 1x : n

Miscellaneous identities:

Ax : n = P ( Ax : n )

P ( Ax : n ) +

Ax : n = P x : nP x : n + d

ax : n = 1

P ( Ax : n ) +

ax : n = 1P x : n + d



8/26

Chapter 7: Benet Reserves

Benet reserve tV :The expected value of the prospective loss attime t.

Continuous reserve formulas:

Prospective: tV ( Ax ) = Ax + t P ( Ax )ax + t

Retrospective: tV ( Ax ) = P ( Ax )sx : t A1x : n

t E xPremium diff.: tV ( Ax ) = P ( Ax + t ) P ( Ax ) ax + t

Paid-up Ins.: tV ( Ax ) = 1 P ( Ax )

P ( Ax + t ) Ax + t

Annuity res.: tV ( Ax ) = 1 ax + t

ax

Death ben.: tV ( Ax ) =Ax + t Ax

1 Ax

Premium res.: tV ( Ax ) =P ( Ax + t ) P ( Ax )

P ( Ax + t ) +

Discrete reserve formulas:

k V x = Ax + k P k ax + k

k V x : n = P x + k : n k P x : n ax + k : n k

k V x : n = 1 P x : n

P x + k : n kAx + k : n k

t V x : n = P x : n sx : n tkx

k V x = 1 ax + k

ax

k V x = P x + k P x

P x + k + d

k V x = Ax + k Ax

1 Ax

h-payment reserves:ht V = Ax + t : n t h P ( Ax : n )ax + t : h t

hk V x : n = Ax + k : n k h P x : n ax + k : h k

hk V ( Ax : n ) = Ax + k : n k h P ( Ax : n )ax + k : h k

hk V

1( m )x : n = Ax + k : n k h P

(m )x : n a

(m )x + k : h k

Variance of the loss function

V ar[ t L] = 1 + P

22Ax + t ( Ax + t )2

V ar[ t L] =2Ax + t ( Ax + t )2

(1 Ax )2 assuming EP

V ar[ t L] = 1 +P

22Ax + t : n t ( Ax + t : n t )

2

V ar[ t L] =2Ax + t : n t ( Ax + t : n t )

2

(1 Ax : n )2 assuming EP

Cost of insurance: funding of the accumu-lated costs of the death claims incurred betweenage x and x + t by the living at t , e.g.

4kx = dx (1 + i)3 + dx +1 (1 + i)2 + dx +2 (1 + i) + dx +

lx +4

=A1x : 44E x

1kx = dxlx +1

= q x px

Accumulated differences of premiums:

( n P x P 1x : n )sx : n = n

n V x n V 1

x : n )= Ax + n 0 = Ax + n

( n P x P x )sx : n = nn V x n V x= P x ax + n

(P x : n P x )sx : n = n V x : n n V x= 1 n V x

( m P x : n m P x )sx : m = mm V x : n mm V x= A

x + m : n m Ax + m

Relation between various terminal re-serves (whole life/endowment only):

m + n + pV x = 1 (1 m V x )(1 n V x + m )(1 pV x + m + n )



9/26

Chapter 8: Benet Reserves

Notations:b j : death benet payable at the end of year of death for the j -th policy year j 1: benet premium paid at the beginning of the j -th policy yearbt : death benet payable at the moment of death t : annual rate of benet premiums payable continuously at tBenet reserve:

h V =

j =0

bh + j +1 v j +1 j px + h q x + h + j

j =0

h + j v j j px + h

t V =

0

bt + u vu u px + t x (t + u)du

0 t+ u vu u px + t du

Recursion relations:

h V + h = v q x + h bh +1 + v px + h h+1 V (h V + h )(1 + i) = q x + h bh +1 + px + h h+1 V (h V + h )(1 + i) = h+1 V + q x + h (bh +1 h+1 V )

Terminology:policy year h+1 the policy year from time t = h to time t = h + 1h V + h initial benet reserve for policy year h + 1h V terminal benet reserve for policy year hh +1 V terminal benet reserve for policy year h + 1

Net amount at Risk for policy year h + 1

Net Amount Risk bh +1 h+1 V

When the death benet is dened as a function of the reserve:For each premium P , the cost of providing the ensuing years death benet , based on the net amount atrisk at age x + h, is : vq x + h (bh +1 h+1 V ). The leftover, P vq x + h (bh +1 h+1 V ) is the source of reservecreation. Accumulated to age x + n, we have:

n V =n 1

h =0

[P vq x + h (bh +1 h+1 V )] (1 + i)n h

= P s n n 1

h =0

vq x + h (bh +1 h+1 V )(1 + i)n h

If the death benet is equal to the benet reserve for the rst n policy years

n V = P s n

If the death benet is equal to $1 plus the benet reserve for the rst n policy years

n V = P s n

n 1

h =0 vq x + h (1 + i)n h



10/26

If the death benet is equal to $1 plus the benet reserve for the rst n policy years and q x + h q constant

n V = P s n vq s n = ( P vq )s n

Reserves at fractional durations:

( h V + h )(1 + i)s = s px + h h+ s V + s q x + h v1 s

UDD ( h V + h )(1 + i)s = (1 s q x + h ) h + s V + s q x + h v1 s

= h+ s V + s q x + h (v1 s h+ s V )h + s V = v1 s 1 s q x + h + s bh +1 + v1 s 1 s px + h + s h+1 V

UDD h+ s V = (1 s)( h V + h ) + s(h +1 V )i.e. h+ s V = (1 s)( h V ) + ( s)(h +1 V ) + (1 s)(h )

unearned premium

Next year losses:h losses incurred from time h to h + 1

E [h ] = 0V ar[h ] = v2(bh +1 h+1 V )2 px + h q x + h

The Hattendorf theorem

V ar[ h L] = V ar[h ] + v2 px + h V ar[ h +1 L]= v2(bh +1 h+1 V )2 px + h q x + h + v2 px + h V ar[ h +1 L]

V ar[ h L] = v2(bh +1 h+1 V )2 px + h q x + h

+ v4(bh +2 h+2 V )2 px + h px + h +1 q x + h +1+ v6(bh +3 h+3 V )2 px + h px + h +1 px + h +2 q x + h +2 +



11/26

Chapter 9: Multiple Life Functions

Joint survival function:

sT (x )T (y) (s, t ) = P r [T (x) > s &T (y) > t ]

t pxy = sT (x )T (y)(t, t )= P r [T (x) > t and T (y) > t ]

Joint life status:

F T (t) = P r [min(T (x), T (y)) t]= tq xy= 1 t pxy

Independant livest pxy = t px t pyt q xy = tq x + tq y tq x tq y

Complete expectation of the joint-life sta-tus:

exy =

0

t pxy dt

PDF joint-life status:

f T (xy )(t) = t pxy xy (t)

xy (t) =f T (xy ) (t)

1 F T (xy )(t) =

f T (xy ) (t)t pxy

Independant lives

xy (t) = (x + t) + (y + t)f T (xy ) (t) = t px t py [(x + t) + (y + t)]

Curtate joint-life functions:

k pxy = k px k py [IL]k q xy = kq x + kq y kq x kq y [IL]

P r [K = k] = k pxy k+1 pxy= k pxy q x + k :y+ k= k pxy q x + k :y+ k = k|q xy

q x + t :y+ t = q x + k + q y+ k q x + k q y+ k [IL]

exy = E [K (xy)] =

1k pxy

Last survivor status T (xy):

T (xy) + T (xy) = T (x) + T (y)

T (xy) T (xy) = T (x) T (y)f T (xy ) + f T (xy ) = f T (x ) + f T (y)

F T (xy ) + F T (xy ) = F T (x ) + F T (y)t pxy + t pxy = t px + t pyAxy + Axy = Ax + Ayaxy + axy = ax + ayexy + exy = ex + eyexy + exy = ex + ey

n |q xy = n |q x + n | q y n |q xy

Complete expectation of the last-survivorstatus:

exy =

0

t pxy dt

exy =

1k pxy

Variances:

V ar[T (u)] = 2

0 t t pu dt (eu )2V ar[T (xy)] = 2

0

t t pxy dt (exy )2

V ar[T (xy)] = 2

0

t t pxy dt (exy )2

Notes:

For joint-life status, work with ps:n pxy = n px n py

For last-survivor status, work with q s:

n q xy = n q x n q y

Exactly one status:

n p[1]xy = n pxy n pxy

= n px + n py 2 n px n py

= n q x + n q y 2 n q x n q ya [1]xy = ax + ay 2axy



12/26

Common shock model:

sT (x )(t) = sT (x ) (t) sz (t)

= sT (x ) (t) e t

sT (y)(t) = sT (y) (t) sz (t)

= sT (y) (t) e t

sT (x )T (y)(t) = sT (x ) (t) T (y) (t) sz (t)

= sT (x ) (t) sT (y)(t) e t

xy (t) = (x + t) + (y + t) +

Insurance functions:

Au =

k=0

vk+1 k pu q u + k

=

k=0vk+1 P r [K = k]

Axy =

k=0

vk+1 k pxy q x + k :y+ k

Axy =

k=0

vk+1 ( k px q x + k + k py q y+ k

k pxy q x + k :y+ k )

Variance of insurance functions:

V ar[Z ] = 2Au (Au )2

V ar[Z ] = 2Axy (Axy )2

Cov[vT (xy ) , vT (xy ) ] = ( Ax Axy )( Ay Axy )

Insurances:

Ax = 1 axAxy = 1 axyA

xy = 1 a

xy

Premiums:

P x = 1ax

d

P xy = 1axy

d

P xy = 1axy

d

Annuity functions:

au =

0

vt t pu dt

var [Y ] =2Au ( Au )2

2

Reversionary annuities:A reversioanry annuity is payable during the ex-istence of one status u only if another status vhas failed. E.g. an annuity of 1 per year payablecontinuously to ( y) after the death of ( x).

ax |y = ay axy

Covariance of T (xy) and T (xy):

Cov [T (xy), T (xy)] = Cov [T (x), T (y)] + {E [T (x)] E [T (xy)]} {(E [T (y)] E [T (xy)]}

= Cov [T (x), T (y)] + ( ex exy ) ( ey exy )= ( ex exy ) (ey exy ) [IL]



13/26


14/26

Basic relationships:

t p( )x = exp t

0

(1)x (s) + + (m )x (s) ds

t p( )x =m

i=1t p (i )x

t q ( j )x tq ( j )xt p ( j )x t p(

)x

UDD for multiple decrements:

t q ( j )x = t q ( j )

x

t q ( )x = t q ( )

x

q ( j )x = t p( )

x ( j )x (t)

( j )x (t) = q ( j )x

t p( )x

= q ( j )x

1 t q ( )x

t p ( j )x = t p( )x

q( j )x

qxt

Decrements uniformly distributed in theassociated single decrement table:

t q ( j )x = t q ( j )

x

q (1)x = q (1)x 1 1

2q (2)x

q (2)x = q (2)

x 1 12

q (1)x

q (1)x = q (1)x 1 12

q (2)x 12

q (3)x + 13

q (2)x q (3)x

Actuarial present values

A =m

j =1

0

B ( j )x + t vt t p( )x (

j )x (t)dt

Instead of summing the benets for each pos-sible cause of death, it is often easier to writethe benet as one benet given regardless of thecause of death and add/subtract other benetsaccording to the cause of death.

Premiums:

P ( )x =

k=0B ( )k+1 v

k+1 k p( )x q (

)x + k

k=0vk k p( )x

P ( j )x =

k=0B ( j )k+1 v

k+1 k p( )x q ( j )x + k

k=0vk k p

( )x



15/26

Chapter 15: Models Including Expenses

Notations:

G expense loaded ( or gross) premium

b face amount of the policyG/b per unit gross premium

Expense policy fee:The portion of G that is independent of b.

Asset shares notations:

G level annual contract premiumk AS asset share assigned to the policy at time t = k

ck fraction of premium paid for expenses at k (i.e. cG is the expenxe premium)ek expenses paid per policy at time t = k

q (d)x + k probability of decrement by death

q (w )x + k probability of decrement by withdrawalk CV cash amount due to the policy holder as a withdrawal benet

bk death benet due at time t = k

Recursion formula:

[ k AS + G(1 ck ) ek ] (1 + i) = q (d)x + k bk+1 + q

(w )x + k k+1 CV + p

( )x + k k+1 AS

= k+1 AS + q (d)x + k (bk+1 k+1 AS ) + q (w)x + k ( k+1 CV k+1 AS )

Direct formula:

n AS =n 1

h =0

G(1 ch ) eh vq (d)x + h bh +1 vq

(w )x + h h +1 CV

n h E ( )x + h



16/26

Constant Force of Mortality

Chapter 3

(x) = > 0, xs(x) = e x

lx = l0e x

n px = e n = ( px )n

ex = 1

= E [T ] = E [X ]

ex : n = ex (1 n px )

V ar[T ] = V ar[x] = 12

mx =

Median[T ] = ln2

= Median[ X ]ex =

pxq x

= E [K ]

V ar[K ] = px(q x )2

Chapter 4

Ax = +

2Ax =

+ 2 A1x : n = Ax (1 n E x )

n E x = e n (+ )

(IA)x = ( + )2

Ax = q q + i

2Ax = q

q + 2 i + i2

A1x : n = Ax (1 n E x )

Chapter 5

ax = 1 +

2ax = 1 + 2

ax = 1 + i

q + i

2ax = (1 + i)2

q + 2 i + i2

ax : n = ax (1 n E x )ax : n = ax (1 n E x )

(IA)x = Ax ax =

( + )2

(I A)x = Ax ax = (1 + i)

( + )(q + i)(IA)x = Ax ax =

q (1 + i)(q + i)2

(Ia )x = ( ax )2 = 1

( + )2

(I a)x = ( ax )2 =1 + iq + i

2

Chapter 6

P x = vq x = P 1x : nP (

Ax ) = =

P (

A

1

x : n )For fully discrete whole life, w/ EP,

V ar[Loss ] = p 2Ax

For fully continuous whole life, w/EP,

V ar[Loss ] = 2Ax

Chapter 7

t V ( Ax ) = 0 , t 0

kV

x = 0 , k = 0 , 1, 2, . . .

For fully discrete whole life, assuming EP,

V ar[k Loss ] = p 2Ax , k = 0 , 1, 2, . . .

For fully continuous whole life, assuming EP,

V ar[ t Loss ] = 2Ax , t 0

Chapter 9For two constant forces, i.e. M acting on ( x)and F acting on ( y), we have:

Axy = M + F

M + F +

axy = 1

M + F +

exy = 1M + F

Axy = q xyq xy + i

axy = 1 + iq xy + i

exy = pxy

q xy



17/26

De Moivres Law

Chapter 3

s(x) = 1 x

lx = l0 x ( x)

q x = (x) = 1 x

n |m q x = m x

n px = x n

xt px (x + t) = q x = (x) = f T (x), 0 t < x

Lx = 1

2(lx + lx +1 )

ex = x

2 = E [T ] = Median[ T ]

ex = x

2

12

= E [K ]

V ar[T ] = ( x)2

12

V ar[K ] = ( x)2 1

12

mx = q x1 12 q x

= 2dx

lx + lx +1

a(x) = E [S ] = 1

2ex : n = n n px +

n2 n

q x

ex : n = ex : n + n2 n

q x

Chapter 4

Ax =a x x

A1x : n = a n x

2Ax =a 2( x )2( x)

Ax =

a x x

A1x : n = a n x

(IA)x =(Ia ) x

x

(IA)x =(Ia ) x

x

(IA)1x : n = (Ia ) n

x

(IA)1x : n = (Ia ) n

x

Chapter 5No useful formulas: use ax = 1 A xd and thechapter 4 formulas.

Chapter 9

exx = x

3 ( MDML with = 2 / ( x))

exx = 2( x)

3exy = y x px eyy + y x q x ey

For two lives with different s, simply translateone of the age by the difference in s. E.g.

Age 30, = 100 Age 15, = 85

Modied De Moivres Law

Chapter 3

s(x) = 1 x

c

lx = l0 x

c

( x)c

(x) = c x

n px = x n

x

c

ex = x

c + 1 = E [T ]

V ar[T ] = ( x)2c(c + 1) 2(c + 2)

Chapter 9

exx = x

2c + 1

ex with = 2c x



18/26

Uniform Distribution of Deaths (UDD)

Chapter 3

t q x = t q x , 0 t 1

(x + t) = q x1 tq x , 0 t 1

lx + t = lx t dx , 0 t 1

s q x + t = sq x1 tq x

, 0 s + t 1

V ar[T ] = V ar[K ] + 112

mx = (x + 12

) = q x

1 12 q x

Lx = lx 12

dx

a(x) = 12

ex : 1 = px + 12

q x

t px (x + t) = q x , 0 t 1

Chapter 4

Ax = i

Ax

A(m )x = ii(m )

Ax

A1x : n = i A1x : n

(I A)1x : n = i

(IA)1x : n

Ax : n = i

A1x : n + A

1x : n =

i

A1x : n + n E x

n | Ax = i

n |Ax

2Ax = 2i + i2

2 2Ax

Chapter 5

a (m )x ax m 1

2ma (m )x = (m)ax (m)

a (m )x : n = (m)ax : n (m)(1 n E x )

with: (m) = id

i(m )d(m ) 1

and (m) = i i(m )

i(m )d(m )

m 12m

a (m )x : n = a(m )x n E x a

(m )x + n

Chapter 6

P ( Ax ) = i

P x

P ( A1x : n ) = i

P 1x : n

P ( Ax : n ) = i

P 1x : n + P

1x : n

P (m )x = P x

(m) (m)(P x + d)

P (m )x : n = P x : n

(m) (m)(P 1x : n + d)

n P (m )x = n P x

(m) (m)(P 1x : n + d)

h P (m ) ( A1x : n ) = i hP 1(m )x : n

Chapter 7

hk V

(m )x : n =

hk V x : n + (m) h P

(m )x : n kV

1x : h

hk V

(m ) ( Ax : n ) = hk V ( Ax : n ) + (m) h P (m ) ( Ax : n ) k V 1x :

hk V ( Ax : n ) =

i

hk V 1

x : h + hk V

1x : h

Chapter 10UDDMDT

t q ( j )x = t q ( j )xq ( j )x =

( j )x (0)

q ( )x = ( )

x (0)

t p ( j )x = t p( )

x

q ( j )xq ( )x

( j )x = q ( j )x1 t q ( )x

( )x = q ( )x

1 t q ( )x

UDDASDT

t q ( j )x = t q ( j )x

q (1)x = q (1)

x 1 12

q (2)x

q (2)x = q (2)x 1 12

q (1)x

q (1)x = q (1)x 1 12

q (2)x 12

q (3)x + 13

q (2)x q (3)x



19/26

Chapter 1: The Poisson Process

Poisson process with rate :

P r [N (s + t) N (s) = k] = e t(t )k

k!E [N (t)] = tV ar[N (t)] = t

Interarrival time distribution:The waiting time between events. Let T n de-note the time since occurence of the event n 1.Then the T n are independent random variablesfollowing an exponential distribution with mean1/ .

P r [T i t] = 1 e t

f T (t) = e t

E [T ] = 1

V ar[T ] = 12

Waiting time distribution:Let S n be the time of the n-occurence of theevent, i.e. S n =

ni=1 T i .

S n has a gamma distribution with parameters nand = 1 /

S n GammaRV[ = n, = 1]

P r [S n t] =

j = ne t

(t ) j

j !

f S n (t) = et (t )n 1

(n i)!

E [S n ] = n

V ar[S n ] =

n2

Sum of Poisson processes:If N 1, , N k are independent Poisson processeswith rates 1, , k then, N = N 1 + + N kis a Poisson process with rate = 1 + + k .

Special events in a Poisson process:Let N be a Poisson process with rate .Some events i are special with a probabilityP r [event is special] = i and N i counts the spe-cial events of kind i. Then, N i is a Poissonprocess with rate i = i and the N i are inde-pendent of one another.If the probability i (t) changes with time, then

E [ N i (t)] = t

0

(s)ds

Non-homogeneous Poisson process:

(t) intensity functionm(t) mean value function

=t

0

(y)dy

P r [N (t) = k] = e m ( t )[m(t)]k

k!

Compound Poisson process:

X (t) =N (t )

i=1

Y i N (t) PoissonRV w/ rate

E [X (t)] = t E [Y ]V ar[X (t)] = t E [Y 2]

Two competing Poisson processes:Probability that n events in the Poisson process ( N 1,1) occur before m events in the Poisson process(N 2,2):

P r [S 1n < S 2m ] =n + m 1

k= n

n + m 1k

11 + 2

k 21 + 2

n + m 1 k

P r [S 1n < S 21 ] = 1

1 + 2

n

P r [S 11 < S 2

1] =

11 + 2

Exam M - Loss Models - LGD c 1


20/26

Chapter 2&3: Random Variables

k-th raw moment:

k = E [X k ]

k-th central moment:

k = E [(X )k ]

Variance:

V ar[x] = 2 = E [X 2] E [X ]2

= 2 = 2 2

Standard deviation:

= V ar[X ]Coefficient of variation:

/

Skewness:

1 = E [(X )3]

3 =

33

Kurtosis:

1 = E [(X )4]

4 =

44

Left truncated and shifted variable (akaexcess loss variable):

Y P = X d|X > d

Mean excess loss function:

eX (d) e(d) = E [Y P ]= E [X d|X > d ]

=

d S (x)dx1 F (d)

=

E [X ] E [X d]1 F (d)

Higher moments of the excess loss vari-able:

ekx (d) = E [(X d)k |X > d ]

=

d (x d)k f (x)dx1 F (d)

= x>d(x d)k p(x)

1 F (d)

Left censored and shifted variable:

Y L = ( X d)+

= 0 X < d

X d X d

Moments of the left censored and shiftedvariable:

E [(X d)k+ ] =

d

(x d)k f (x)dx

=x>d

(x d)k p(x)

= ek

(d)[1 F (d)]

Limited loss:

Y = ( X u) = X X < uu X u

Limited expected value:

E [X u] = 0

F (x)dx +u

0

S (x)dx

=u

0

[1 F (x)]dx if X is always positive

Moments of the limited loss variable:

E [(X u)k ] =u

xk f (x)dx + uk [1 F (u)]

=x u

xk p(x) + uk [1 F (u)]

Moment generating functions mX (t):

mX (t) = E [etX ]

Sum of random variables S k = X 1 + + X k :

mS k (t) =k

j =1

mX j (t)



21/26

Chapter 4: Classifying and Creating distributions

k-point mixture ( a i = 1 ):

F Y (y) = a1F X 1 (y) + + ak F X k (y)

f Y (y) = a1f X 1 (y) + + ak f X k (y)E [Y ] = a1E [X 1] + + ak E [X k ]

E [Y 2] = a1E [X 21 ] + + ak E [X 2k ]

Tail weight:

existence of moments light tailhazard rate increases light tail

Multiplication by a constant ,y > 0:

Y = X F Y (y) = F X ( y) and f Y (y) = 1

f X ( y)

Raising to a power:

Y = X 1

> 0 : F Y (y) = F X (y )f Y (y) = y 1f X (y )

< 0 : F Y (y) = 1 F X (y )f Y (y) = y 1f X (y )

> 0: transformed distribution = 1: inverse distribution

< 0: inverse transformed

Exponentiation:

Y = eX F Y (y) = F X (ln( y))f Y (y) = 1y f X (ln( y))

Mixing:The random variable X depends upon a para-meter , itself a random variable . For a givenvalue = , the individual pdf is f X | (x |).

f X (x) = f X | (x |)f ()d If X is a Poisson distribution with parame-

ter , and follows a Gamma distributionwith parameters ( , ), then the resultingdistribution is a Negative Binomial distri-bution with parameters ( r = , = )

If X is an Exponential distribution withmean 1/ and is a Gamma distributionwith parameters ( , ), then the resulting

distribution is a 2-parameter Pareto distri-bution with parameters ( = , = 1 )

If X is a Normal distribution (, V ) and follows a Normal distribution (, A),then the resulting distribution is a Normaldistribution (, A + V )

Splicing ( a j > 0, a j = 1 ):

f (x) =a1f 1(x) c0 < x < c 1

... ...

ak f k (x) ck 1 < x < c k

Discrete probability function (pf):

pk = P r [N = k]

Probability generating function (pgf):

P N (z) = E [zN ] = p0 + p1z + p2z2 + P (1) = E [N ]P (1) = E [N (N 1)]

Poisson distribution:

pk = e k

k!P (z) = e (z 1)

E [N ] = V ar[N ] =

Negative Binomial distribution:Number of failures before success r with proba-bility of success p = 1 / (1 + )

pk = k + r 1

k

1 +

k 11 +

r

P (z) = [1 (z 1)] r

E [N ] = r V ar[N ] = r (1 + )

If N 1 and N 2 are independent NegativeBinomial distributions with parameters(r 1, ) and ( r 2, ), then N 1 + N 2 is a Neg-

ative Binomial distribution with parame-ters ( r = r1 + r 2, )



22/26

If N is a Negative Binomial distributionwith parameters ( r, ) and some events jare special with probability j , then N j ,the count of special events j , is a Nega-tive Binomial distribution with parame-ters ( r j = r, j = j )

Geometric distribution:Special case of the negative binomial distribu-tion with r = 1

pk = k

(1 + )k+1

P (z) = [1 (z 1)] 1 = 1

1 (z 1)

E [N ] = V ar[N ] = (1 + )

Binomial distribution:Counting number of successes in m trials givena probability of sucess q

pk = mk q

k (1 q )m k

P (z) = [1 + q (z 1)]m

E [N ] = mq V ar[N ] = mq (1 q )

The (a,b, 0) class:

pk pk 1

= a + bk

a > 0 negative binomial or geometricdistribution

a = 0 Poisson distributiona < 0 Binomial distribution

Compound frequency models:

P (z) = P prim (P sec (z))

Compound frequency and (a,b, 0) class:Let S be a compound distribution of primaryand secondary distributions N and M with pn =P r [N = n] and f n = P r [M = n]. If N is a mem-ber of the ( a,b, 0) class

gk = 11 af 0

k

j =1

a + j b

kf j gk j

g0 = P prim (f 0)

gk =

k

k

j =1

j f j gk j

Mixed Poisson models:If P (z) id the pgf of a Poisson distribution with

rate drawn from a discrete random variable ,then

P (z |) = e (z 1)

P (z) = p(1)e 1 (z 1) + + p( n )e n (z 1)

Exposure modications on frequency:n policies in force. N j claims produced by the j -th policy. N = N 1 + + N n . If the N j are in-dependent and identcally distributed, then theprobability generating function for N is

P N (z) = [ P N 1 (z)]n

If the company exposure grows to n , let N represent the new total number of claims,

P N (z) = [ P N 1 (z)]n = [ P N (z)]nn

Conditional expectations:

E [X | = ] = xf X |(x |)dxE [X ] = E [E [X |]]

V ar[X ] = E [V ar[X |]] + V ar[E [X |]]

The variance of the random variable X is thesum of two parts: the mean of the conditionalvariance plus the variance of the conditionalmean.

Exposure modications on frequency for common distributions:N Parameters for N Parameters for N

Poisson = n

n Negative Binomial r, r = nn r, =



23/26

Chapter 5: Frequency and Severity with Coverage Modications

Notations:

X amount of loss

Y L amount paid per lossY P amount paid per payment

Per loss variable:

f Y L (0) = F X (d)f Y L (y) = f x (y + d)F Y L (y) = F X (y + d)

Per payment variable:

f Y P (y) = f

x(y + d)

1 F X (d)

F Y L (y) = F X (y + d) F X (d)

1 F X (d)

Relations per loss / per payment variable:

Y P = Y L |Y L > 0

E [Y P ] = E [Y L ]P r [Y L > 0]

E [(Y P )k ] = E [(Y L )k ]

P r [Y L

> 0]Simple (or ordinary) deductible d:

X = X Y L = ( X d)+Y P = X d|X > d

E [Y L ] = E [(X d)+ ]= E [X ] E [X d]

E [Y P ] = E [X d|X > d ]

= E [X ] E [X d]

1 F X (d)

Franchise deductible d:

X = X Y L = 0 if X < d, X if X > dY P = X |X > d,

Y P franchise = Y P

ordinary + d

E [Y L ] = E [X ] E [X d] + d[1 F X (d)]E [Y P ] = E [X |X > d ]

= E [X ] E [X d]1 F X (d)

+ d

Effect of deductible for various distribu-tions:

X : exp[mean ] Y P

: exp[mean ]X : uniform[0, ] Y P : uniform[0, d]X : 2Pareto[ , ] Y P : 2Pareto[ = , = +

Loss elimination ratio:

LER X (d) = E [X d]

E [X ]

relations:

c(X d) = ( cX ) (cd)

c(X d)+ = ( cX cd)+A = ( A b) + ( A b)+

Ination on expected cost per loss:

E [Y L ] = (1 + r ) E (X ) E X d1 + r

Ination on expected cost per payment:

E [Y P ] =(1 + r ) E (X ) E X d1+ r

1 F X d1+ r

u-coverage limit:

Y L = X uY P = X u |X > 0

E [Y L ] = E [X u]

=u

0

[1 F x (x)]dx

u-coverage limit with ination:

E [X u] (1 + r )E X u1 + r

u-coverage limit and d ordinary de-ductible:

Y L = ( X u) (X d)Y P = ( X u) (X d)|X > 0

= ( X u) d|X > dE [Y L ] = E [X u] E [X d]

E [Y P ] = E [X u] E [X d]1 F X (d)



24/26

u-coverage limit and d ordinary deductiblewith ination:

E [Y L ] = (1 + r ) E X u1 + r

E X d1 + r

E [Y P ] = (1 + r )E X u1+ r E X d1+ r

1 F X d1+ r

Coinsurance factor :First calculate Y L and Y P with deductible andlimits. Then apply

Y L = Y L

Y P = Y P

Probability of payment for a loss with de-ductible d:

v = 1 F X (d)

Unmodied frequency:The number of claims N (frequency) does notchange, only the probabilities associated withthe severity are modied to include the possi-bility of zero payment.

Bernoulli random variable:

N = 0 @ 1 p1 @ pE [N ] = p

V ar[N ] = (1 p) p

Deductible, limit, ination and coinsurance:

d = d1 + r

u = u1 + r

E [Y L ] = (1 + r ) {E [X u ] E [X d ]}

E [Y P ] = (1 + r )E [X u ] E [X d ]

1 F X (d )E [Y L 2] = 2(1 + r )2 E (X u )2 E (X d )2 2d E [X u ] + 2d E [X d ]

Modied frequency:N (the modied frequency) has a compound distribution: The primary distribution is N , and the secondarydistribution is the Bernoulli random variable I . The probability generating function is

P N (z) = P N [1 + v(z 1)]

The frequency distribution is modied to include only non-zero claim amounts. Each claim amount prob-

ability is modied by dividing it by th eprobability of a non-zero claim. For common distributions, thefrequency distribution changes as follows (the number of positive payments is nothing else than a specialevent with probability = v)

N Parameters for N Parameters for N Poisson = vBinomial m, q m = m, q = vq Negative Binomial r, r = r, = v



25/26

Chapter 6: Aggregate Loss Models

The aggregate loss compound randomvariable:

S =

N

j =1X j

E [S ] = E [N ] E [X ]V ar[S ] = E [N ] V ar[X ] + V ar[N ] E [X ]2

S (E [S ], Va r [S ]) if E [N ] is large

Probability generating function:

P S (z) = P N [P X (z)]

Notations:

f k = P r [X = k] pk = P r [N = k]gk = P r [S = k]

Compound probabilities:

g0 = P N (f 0) if P r [X = 0] = f 0 = 0gk = p0 P r [sum of X s = k]

+ p1 P r [sum of one X = k]+ p2 P r [sum of two X s = k]

+ If N is in the (a,b, 0) class, then

g0 = P N (f 0)

gk = 11 af 0

k

j =1a +

j bk

f j gk j

n-fold convolution of the pdf of X :

f (n )X (k) = P r [sum of n X s = k]

f (n )X (k) =k

j =0f (n 1)X (k j )f X ( j )

f (1)X (k) = f X (k) = f k

f (0)X (k) = 1 , k = 00 , otherwise

n-fold convolution of the cdf of X :

F (n )X (k) =k

j =0

F (n 1)X (k j )f X ( j )

F (0)X (k) = 0 , k < 01 , k 0

Pdf of the compound distribution:

f S (k) =

n =0 pn f

(n )X (k)

Cdf of the compound distribution:

F S (k) =

n =0 pn F (n )X (k)

Mean of the compound distribution:

E [S ] =

k=0

k f S (k) =

k=0

[1 F S (k)]

Net stop-loss premium:

E [(S d)+

] =

d(x d)f

S (x)dx

E [(S d)+ ] =

d+1

(x d)f S (x)

E [(S d)+ ] = E [S ] d d 1

x =0(x d)f S (x)

E [(S d)+ ] =

d

[1 F S (d)]

= E [S ]

d 1

x =0[1 F S (d)]

Linear interpolation a < d < b :

E [(S d)+ ] = b db a

E [(S a)+ ]+d ab a

E [(S b)+ ]

Recursion relations for stop-loss insurance:

E [(S d 1)+ ] = E [(S d)+ ] [1 F S (d)]E [(S s j )] = E [(S s j +1 )] + ( s j +1 s j )P r [S s j +1 ]

where the s j s are the possible values of S , S : s0 = 0 < s 1 < s 2 < andP r [S s j +1 ] = 1 P r [S = s0 or S = s1 or or S = s j ]



26/26

Last Chapter: Multi-State Transition Models

Notations:

Q( i,j )n = P r [M n +1 = j |M n = i]

k Q( i,j )n = P r [M n + k = j |M n = i]P ( i )n = Q(

i,i )n

k P ( i )n = P r [M n +1 = = M n + k = i|M n = i]

Theorem:

k Qn = Qn Qn +1 Qn + k 1k Qn = Qk for a homogeneous Markov chain

Inequality:

k P ( i )n = P (i)

n P ( i )n +1 P

( i )n + k 1 kQ

( i,i )n

Cash ows while in states:

lC ( i ) = cash ow at time l if subject in state i at time l

k vn = present value of 1 paid k years after time t = nAP V s@n (C ( i ) ) = actuarial present value at time t = n of all future payments to be

made while in state i , given that the subject is in state s at t = n

AP V s@n (C ( i ) ) =

k=0k Q(s,i )n n + k C (

i ) kvn

Cash ows upon transitions:

l+1 C ( i,j ) = cash ow at time l + 1 if subject in state i at time l and state j at time l + 1

lQ(s,i )n Q( i,j )n + l = probability of being in state i at time l and in state j at time l + 1

AP V s@n (C ( i,j ) ) = actuarial present value at time t = n of a cash ow to be paid upontransition from state i to state j

AP V s@n (C ( i,j ) ) =

k=0k Q(s,i )n Q

( i,j )n + k n + k+1 C

( i,j ) k+1 vn

chapter 3 survival distributions and life tables exam m notes f05

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