soa exam mlc - formulas (5)

17
Chapter 3: Survival Distributions and Life Tables Distribution function of X: Fx(:r) = Pr(X S; :1;) Survival function B(.1:): Probability of death between age :r and age y: Pr(.r < X S; z) F.J( (z) - Fx (:1:) - B(Z) Probability of death between age and age y given survival to age :r:: Pr(:1; < X S; zlX > Notations: tlJx PriT(.r) tl prob. (3:) dies within t years distribution function of T(a:) tPx Pr[T(:c) > t] attains age ;1; + t Pr[t < Tel') t + 'Ill t+ul]x - t(jx t])a' t+u])x tPx' u(]x-t-t Relations with survival functions: Curtate future lifetime (K(:r) greatest integer in T(x)): Pr[K(.l') k] Pr[k T(:r) < k + 1] k]Jx k+lPx kP", . qx+k klJx Exam rv[ Life C;onting;;;ncicH - LGD@ Force of mortality flea:): /1(:1: ) 8' (x) sex) Relations between survival functions and force of mortality: exp ( -I "(Y)d ll ) x+n ) nPx exp - ! p.(y)ely ( Derivatives: d dt t(jx d dt tPx tPx . It (:r: + t) d -T dt··'" d -L dt x d -1 Mean and variance of T and ](: E[T(:r)] Vo.7:T(:r:) ] Vo:r[K(.r}] complete expectation of life =./ tP:B elt o curtate expectation of life ex) ,J ,·2 . t . tPx u,t - ex 2./ 00 o 00 2)2k -1) kP:r e". 2 k=l Total lifetime after age .r: Ta ex; T-r: ./ lx+t dt o 1

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Page 1: SOA Exam MLC - Formulas (5)

Chapter 3: Survival Distributions and Life Tables

Distribution function of X:

Fx(:r) = Pr(X S; :1;)

Survival function B(.1:):

Probability of death between age :r and age y:

Pr(.r < X S; z) F.J( (z) - Fx (:1:)

- B(Z)

Probability of death between age and age y given survival to age :r::

Pr(:1; < X S; zlX >

Notations:

tlJx PriT(.r) tl prob. (3:) dies within t years

distribution function of T(a:)

tPx Pr[T(:c) > t] attains age ;1; + t

Pr[t < Tel') t + 'Ill t+ul]x - t(jx

t])a' t+u])x

tPx' u(]x-t-t

Relations with survival functions:

Curtate future lifetime (K(:r) greatest integer in T(x)):

Pr[K(.l') k] Pr[k T(:r) < k + 1]

k]Jx k+lPx

kP", . qx+k

klJx

Exam rv[ Life C;onting;;;ncicH - LGD@

Force of mortality flea:):

/1(:1:)

8' (x)

sex)

Relations between survival functions and force of mortality:

exp ( -I"(Y)dll)

x+n )

nPx exp - ! p.(y)ely (

Derivatives:

d dt t(jx

d dt tPx tPx . It (:r: + t)

d-Tdt··'" d-Ldt x

d -1

Mean and variance of T and ](:

E[T(:r)]

Vo.7:T(:r:) ]

Vo:r[K(.r}]

complete expectation of life

=./ tP:B elt o

curtate expectation of life ex)

,J ,·2 . t . tPx u,t - ex2./00

o 00

2)2k -1) kP:r e".2

k=l

Total lifetime after age .r: Ta

ex;

T-r: ./ lx+t dt

o

1

Page 2: SOA Exam MLC - Formulas (5)
Page 3: SOA Exam MLC - Formulas (5)
Page 4: SOA Exam MLC - Formulas (5)

1

~, Varying benefit insul'ances:

(IA)x = ./It + IJlIt, tPx !/'x(t)dt

0 11

(L4)~:m ./It + IJI/ ' t1Jxltx(t)dt

0 (X)

(IA)", ./ t ' l,t , tP:r p'o,(t)dt

0 11

(IA);"fll ./ t ' 7,t , tPx Itx (t)dt

0 11

CD"4.);':fll ./(n ItJ ' tPx fJ,x(t)dt

0 T1

(DA);':fll ./(n - t)vt , tPdl'x(t)dt

0

(IA)x Ax + VP:L,(1A)x+l

lIqx + 1)1'rr' + (DA)~:fll nvqx + vpx(DA)x~l:n_ll

(IA);:fll + (15A);:fll

(IA)~:fll + (D)l)~:fll = (n + l)A;:fll

(IA )~:fll + (DA)~:fll (n + l)A~;m

Accumulated cost of insurance:

Share of the survivor:

accumulation factor 1

Interest theory reminder

1 vn

am

l,n1 i n'fll 80fll

5 1 1

-/5 ' 00Cl

i d - nvn

(Ia)fll

1 (IO)OCl 52

(n + l)Ofll (Ia)fll + (Da)m

-5 i'IJ 1 1 +i id 12

Doubling the constant force of interest 5

1 +i (1 + i)2-4

1)2v

-4 2i + i 2

d2d --+ 2d ­i 2i + i 2

5 -+

25

Limit of interest rate i = 0:

Ao, 1

A~:fll nqx

n!Ax 11}1X

Ax:fll

mlnqx

(JA)x 1 +e:r:

(IA)x eo,

Exam l'vl - Life Cont.ingenciel$ - LC;D'V 4

Page 5: SOA Exam MLC - Formulas (5)

Chapter 5: Life Annuities

Whole life annuity: ax

J00

Elan] at!· t]Jx + t)dt

o 00

Jvt'tPxdt J,x,)

tExdt

o o

1lor [an]

n-year temporary annuity:

tJn

. = Jn

v tllx dt o 0

1l oriY]

n-year deferred annuity:

rAJ OC.J . tPx dt JtE~,dt1,t

n n

Vor[Y] 2aX!n)

n-yr certain and life annuity:

+ na,x +

Most important identity

1 ba'T + )Ix

1 )Ix

1 ba'x:111

1 - (2b)

d 1 Ax:111

d 1 (lii J;:111 +

Recursion relations

+ + nl

(Iii)x

1 +vpx

1 + v 2Px

Whole life annuity due: 0,,;

00

E[ii K+lll L 11k. kPx '..=0

Yor[oK+lll

n-yr temporary annuity due:

'11-1

E[Y] = Lvk. k]lx

k=O

n-yr deferred annuity due:

ex)

E[Y] = L . kP" k=n

n-yr certain and life due: ii'x:111

k0111 + L v . kllx

k=n

+n,O'T

Exam f,/l - Life COlltingencieh 5

Page 6: SOA Exam MLC - Formulas (5)

Whole life immediate: ax

= L . ~'P2: k=1

1

m-thly annuities

Vo.r[Y]

1

rn. .. (m) 1(ra) -(I­o'x:nl ax:-:m

'm

Accumulation function:

11

=/-1 o

Limit of interest rate i 0:

;=0 ax ---+ ex

ii,x 1 + c2:

IIx ex

o.x:11I cx:rrl ;=0 ---+ 1 +

ex:rrl

6

Page 7: SOA Exam MLC - Formulas (5)

Chapter 6: Benefit Premiums

Loss function: Loss PV of Benefit,s - PV of Premiums Fully continuous equivalence premiums (whole life and endowment only):

P(Ax) ii",

(L4xP(A",) 1 1

P(A:r) =- -6 (l,;r

.. 2]\/ar[L] (A,,:)(1 + ~r[ Var[L]

Var[L]

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

P(A,,:) Px

dAa:P(Ax)

1- Ax

P(Ax) d ax( 1

prVadL] 1 + d [ (A,,:) 2]

2Ax (Ax?\/ar[L]

(dii.".)2 <Ax - (Ax)2

\/ar[L] = (1- A."Y

Semicontinuous equivalence premiums:

m-thly equivalence premiums:

p(m) #

h-payment insurance premiums:

A,,;

°x:h\

Pure endowment annual premium PJ::~: it is the reciprocal of the actuarial accumulated value because the share of the survivor who has deposited P:r:4 at the beginning of each year for n years is the contractual $1 pure endow­ment, i.e.

(1)

P minus P over P problems: The difference in magnitude of level benefit pre­miums is solely attributable t.o the investment feature of the contract. Hence, comparisons of the policy values of survivors at age :/: + n lllay he done by ana.lyzing future benefits:

P l'"( n Px - x:nl)8x :m

lVIiscellaneous identities:

P(Ax :nl )

P(Ax:m) +6

1

+d

Exa.m tv! LIfe Contin)1;en-C'ies - LGD(':;: 7

Page 8: SOA Exam MLC - Formulas (5)

lkx = lX+1 P.T

ACCUilmlated differences of premiums:

/~. Chapter 7: Benefit Reserves

Benefit reserve tV: The expected value of the prospect.ive loss at time t.

Continuous reserve formulas:

Prospective: t V(Ax) Ax+t - P(Ax)a.x+t

Retrospective:

Premium diff.:

Paid-np Ins.:

tti(Ax) = P(AT).'lT;t]--

Annuity res.: tV(A;t.) = 1

Discrete reserve formulas:

h-payment reserves:

~V

hr7' A )Ie' ~ ·'"ix:nl

Variance of the loss function

Vad tL] assuming EP

ass1lming EPVar[t L]

Cost of insurance: funding of the accumu­lated costs of the death claims incurred between age ;1: and x + t by the living at t, e.g.

4E x

qT

~Vx - n~::nl)

Ax-;-n 0 = A~'+n

.~Vx·- nVT

- ~Vr

AXTm:n-~ - Ax+m

Relation between various terminal re­serves (whole life/endowment only):

= 1­

(I - m~~)(l- nV.,,+m)(l-

Exam ?vI - Contil1j2;€llcieb - LGD© 8

Page 9: SOA Exam MLC - Formulas (5)

Chapter 8: Benefit Reserves

Notations: br death benefit payable at the end of year of death for the j-th policy year 71J~l: benefit premium paid at the beginning of the j-th policy year bt : death benefit payable at the moment of death 7ft: annual rate of benefit premiums payable continuously at t Benefit reserve:

00 00

hI! = Lbh+j j]Jx+h qx+h-tj - L j=O j=O

U li V ul)x+tfJ~,(t + n)dl1 L 7ft+u V U P2'+tdv

o

Recursion relations:

hI! + 7fh l' f]x--;-h . bh-'-l + 11 Px+h . h+1 If

(" ~7 + 7fh)(l + i) qx+h . + 11x+h . h+1 If

(hI! + (l+i) h+ll! + - h+1 V)

Terminology: "policy year h+ 1" the policy year from time t = h to time t = h + 1 "h V + == initial benefit reserve for policy year h + 1

terminal benefit reserve for polky year h terminal benefit reserve for policy year h + 1

Net amount at Risk for policy year h + 1

='let Amount Risk

\Vhen the death benefit is defined as a function of the reserve: For each preminm P, the cost of providing the ensuing year's death benefit, based on the net amount at risk at age .T + h, is : - h-ti V). The leftover, P - vqx,h(/)h+l - h+IV) is the source of reserve creation. Accullmlated to age :r + 'TI, we have:

71-1

- htl it)] (1 +L h=O

n~1

- L1Hlx+h(bhl-l - 11-'-1 V)(l + h=O

• If the death benefit is equaJ to the benefit reserve for the first 17 policy yean,

• If the death benefit is equal to plus the benefit reserve for the fiTst n policy years

71-1

nV=V~m- L + h=O

Exam Life Contingencies - LGD(:) 9R

Page 10: SOA Exam MLC - Formulas (5)

• If the death benefit is equal to $1 pIue; the benefit reserve for the first n policy years ane qxlh == q

COllfltant

n V =

Reserves at fractional durations:

(h1/ + 7Th)(l + sPx+h' h~:.sV + UDD '* (hll + 7Th)(l +

V +8' his1/)

h+.sV Vl~8 . I-sqx+h+s . bh+ 1 + . l-sPx+h+s .

UDD '* h+8V (1-8)("V+7Th)+"("+lVr)

i.E. (1 .'\)(h1/)+ V)+ (1-.5)(7Th) '-..r---"

unearned premium

Next year losses:

Ah losses incurred from time h to h + 1

E[Ah] o V01'[Ah l

The Hattendorf theorem

~"

-. Exam Ivi - Life ContingEl1cies LGDZ: 10

Page 11: SOA Exam MLC - Formulas (5)

Chapter 9: Multiple Life Functions

Joint survival function:

(,~, t) Pr[T(.1:) > 8&T(y) > *1 (t,t)

Pr[T(:r.) > t and T(y) > tj

Joint life status:

FT(t) = Prlmin(T(:r),T(y)).s; t]

Independant lives

tjJ2' . tPy

t!J" + tqy tq," . tqll

Complete expectation of the joint-life sta­tus:

= J(Xl

t1ixy dt

o

PDF joint-life status:

(t)

Independant lives

+ t) + I-L(Y + t)

(t) t])x . tJ)ylf.L(:r. + t) + f.1(Y + t)]

Curtate joint-life functions:

/,])xy /,P2' . kPy [IL] k(jxy kqx + k(jy - kqx . Ic(jy [IL]

Pr[K = k] k])xy - k+1Pxy

kPxy . qx+k,y+"

kPxy' =

IJx+k + 00

E[K(;ry)] 2.:: kP,ry 1

Exam Life Com.inl'?>~ncies - LGDCS

Last survivor status T(xy):

T(J::Y) + T(.TY)

T(:ry) .T(XIJ)

+ h(xy)

Fy(xy) + tP:1'1I + tP.TY

Axy+

i'i.xy + + +

Variances:

FarfT(l1)]

Va·r[T(.I:Y)]

\/ar[T(;ry)]

Notes:

T(:r.) +T(y)

T(:r.) . T(y)

+ FT(x) + tPx + tl)y

== Ax + -,,4y

+ ex + ey

Complete expectation of the last-survivor status:

Jtl'Tydt

o

2 J00

t· tl)udt

o 00

2 Jt . tPxy dt ­

o

2 Jt· t'Pxy dt ­

o

For joint-life Htat1lH, work with p's:

nPxll = n])2' ' nPy

For last-survivor status, work with q's:

"Exactly one" status:

nPxy - nPxy

nPx + nPy 2 11 px' nPy

n!]x + nqy - 211.I]x . n(jy

+ o.y - 20xy

11

Page 12: SOA Exam MLC - Formulas (5)

Common shock model:

(t) (t) . 8 z (t)

ST*(x)(t) .

(t) (t) .

(t) . C- At

(t) (t)·· 'T*(y) (t) . 8.,(t)

8Y*(X)(t) . (t) . C-At

J1xy(t) = j1(;r + t) + J1(Y + t) + A

Insurance functions:

A" = L k=Q

. "p." . qu+k

L k=O 00

Pl'[E( k]

. kPxy'

00

k=O

Variance of insurance functions:

Vor[Z] - (An)2

Vor[Z] 2Axy (A2•y)2

,VY(xy )] (A~: i1xy)( Ay ­

Covariance of T(:ry) and T(x!7):

Call [T(:ry), T(.TY)] Call [T(;r) ,T(y)] + {E fT(;r)]

C01l T(y)] + (ex (ey [IL]

Insurances:

1 - Ofj,~.

1 - Oo.xy

1

Premiums:

d

1 _ d

1 -d

Annuity functions:

00

)' v t . tPu dt

o

Reversionary annuities: A reversioanry annuity is payable during the ex­istence of one status n only if another status v has failed. E.g. an annuity of 1 per year payable continnollsly to (y) after the death of (x).

E [T(:J:Y)]}' {(E [T(y)] E :Tery)]}

Exa,m 1..,{ Life Contingencie~ - LCD@ 12

Page 13: SOA Exam MLC - Formulas (5)

Chapter 10 & 11: Multiple Decrement Models

Notations:

probability of decrement in the next

t yearH due to caUHe .7

= probability of decrement in the next

t yearH due to all caUHes m

L j=l

the force of decrement dHe only

to decrement j

Il~T) the force of decrement due to all

causeH simultaneously rn

L j=l

probability of surviving t yearH

despite all decrementH

1 t

-/ (B)ds e (l

Derivative:

_!i () dt

Integral forms of tqx :

t

/ S p~;T) , p,~) o

Exam IvI - Life Contingencie~ LGD© 13

Probability density functions:

.Joint PDF: hAt, j) (t)

Marginal PDF of J: fA.j) = 00 q~j) iX'

= / f:r,J(tLi)dt o

Marginal PDF of T: f:r(t) (t) 171

= LhAt,

Conditional PDF: hlrUlt) = ---,.---

Survivorship group: Group of l~T) people at some age a at time t o. Each member of the group has a joint pdf for time until decrement and cause of decrement.

T.~a+n

/ tP~~), fL!!) (t)dt :r--a

L m

j=1

171

Ll~) j cce 1

Associated single decrement:

probability of decrement from caUHe j only

e~p [-I"~) (,')d'1 1 - tq~(j)

Page 14: SOA Exam MLC - Formulas (5)

Basic relationships:

II rn

PI(;)

t x ;=1

UDD for multiple decrements:

t· t. q~T)

Decrements uniformly distributed in the associated single decrement table:

1 .1(2) \= q~(l) (1 -q I2 x )

~ Qx1(1))(1

2

l(1) ( 1 1(2) 1 ...1. ~ql(2) . ql(3))

qx \ 1 - "2 Qx . - 2 . 3 x x

Actuarial present values

Irh<;tead of summing the benefit8 for each pos­sible cause of death, it is often easier to write the benefit as one benefit given regardless of the cause of death and add/8ubtract other henefit8 according to the ca,lse of death.

Premiums:

p(T) x

EXtn11 rvI Life Contingl;'ncies - LGDe~ 14

Page 15: SOA Exam MLC - Formulas (5)

Chapter 15: lVlodels Including Expenses

Notations:

G expense loaded ( or gross) premium

b face amount of the policy

G /b per unit gross premium

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

Asset shares notations:

G level ann11al contract premimn

kAS asset share assigned to the policy at time t = k

Ck fraction of premium paid for expenses at k cG is the expel1xe premium)

Ck expenseH paid per policy at time t = k

probability of decrement by death

probability of decrement by withdrawal

"CiI cash amount due to the policy holder as a withdrawal benefit

b" death benefit due at time t = k

Recursion formula:

(1 + i)

Direct formula:

f G(l Ch) - eh h+1 Cl1

h=O

EXalTI l\j - Life COTIringencies - LGD@ 15

Page 16: SOA Exam MLC - Formulas (5)

Constant Force of Mortality

• Chapter 3

S CJ:)

Z" nPJ:

1 E[T] E'X]

nPx) 1

1/or[T] Var[.1:] =

p.

ln2 . ]Yledian[T] = Mechan,X p.

ex Px = E[K] qar

• Chapter 4

/-1. + (j f."

f.."+26

Ax (1 nEx)

q+i

• Chapter 5

1

p.+ 1

ax

• Chapter 6

llqx = P;':11i P(A~,) p

For fully discrete whole wi EP,

For fully continuous whole life, w IEP.

1/m'[Loss] =

• Chapter 7

t V(A)x O. t 2: 0

O. k=O,l,2 ....

For discrete whole life, assuming

Vo1'[ "Los.s] p. If = 0, 1,2, ...

For fully continuous whole as:mming EP,

1/a/' [ tLo.~81 1,4x , t 2: 0

• Chapter 9 For two COllstant forces, i. e. acting on (.1:) and Il.F acting on (y), we have:

(hy + i 1 +i

qJ,y + i 1Jxy

exy qJ'Y

Exam fvI Lift Contingencies - LGD<9 16

Page 17: SOA Exam MLC - Formulas (5)

De Moivre's Law

• Chapter 3

8(.1; )

,,1','

t1Jx p(.r + t)

Vor[T]

0(.1;)

• Chapter 4

• Chapter 3

8(.r)

ft·(;r)

"Px

a 2(w-x) I2AX :r 2(w - x)1- ­W ow-xl

AxW -;r W -.rlo-- ex: (w - .1;)

W 07il A~:7il1 W -.r!J.('r) =-­

W -.r (Io)w_xl(IA)x

W -.1: W -.T

(Io)w-xlw-.r-T/ (IA)x w-.rw-.r

qx = !J.('r) = h(.r) , O.:s; t < W -.T (Io)7il(IAX:7il w-.r

"2(lx + Ix+d 1

(Io)7il(IAX:7ilw-;r: W -.1:-2- = E[T] = Median[T]

W - .r _ ~ = E[K] 2 2 • Chapter 5

(w-.r)2 1-Ax d heNo useful formulas: use ii. x -d- an. t 12 chapter 4 formulas.

(w _.r)2 -1

12 • Chapter 9 qx 2dx

1 - ~qx lx + l1:+1 W-;]:1 --(= MDML with p. = 2/(w - .1:))E[8]. =-2 3

n 2(w - .1:)nnPx + '2 "qx

3 T/

y-x]Jx Cyy + y-1,qx cye1':7il + '2 "l]x

For two lives with different w's, simply translate a'w_xl one of the age by the difference in w's. E.g. w-:r:

a'7il w-.r Age 30, w = 100 {o} Age 15, w = 85

Modified De Moivre's Law

V or[T]

(1- ~r • Chapter 9

(w-.1:)C c10 -----:;- ex: (w - .r)

W -.1:

c 2c+ 1 w -.r 2c

ex o.

wIth j.t = - ­w -.r

w -.r

w - .1: = E[T]

(w-.T-n)C

c+1

Exam rvI - Life Contingenciet> - LGD© 17