Exam M Notes F05

8/11/2019 Exam M Notes F05

1/26

Chapter 3: Survival Distributions and Life Tables

Distribution function ofX:

FX(x) = Pr(X x)

Survival function s(x):

s(x) = 1 FX(x)

Probability of death between age x andage y:

Pr(x < X z) = FX(z) FX(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) = FX(z) FX(x)

1 FX(x)

= s(x) s(z)

s(x)

Notations:

tqx = Pr[T(x) t]

= prob. (x) dies within t years

= distribution function ofT(x)

tpx = Pr[T(x)> t]

= prob. (x) attains age x + t

= 1 tqx

t|uqx = Pr[t < T(x) t + u]

= t+uqx tqx

= tpx t+upx

= tpx uqx+t

Relations with survival functions:

tpx = s(x + t)

s(x)

tqx = 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]

= kpx k+1px

= kpx qx+k= k|qx

Force of mortality(x):

(x) = fX(x)

1 FX(x)

= s(x)

s(x)

Relations between survival functions andforce of mortality:

s(x) = exp

x

0

(y)dy

npx = expx+n

x

(y)dyDerivatives:

d

dt tqx = tpx (x + t) =fT(x)(t)

d

dt tpx = tpx (x + t)

d

dtTx = lx

d

dtLx = dx

ddt

ex = (x)ex 1

Mean and variance ofT and K:

E[T(x)] complete expectation of life

ex =

0

tpx dt

E[K(x)] curtate expectation of life

ex

=

k=1

kp

x

V ar[T(x)] = 2

0

t tpx dt e2x

V ar[K(x)] =

k=1

(2k 1) kpx e2x

Total lifetime after age x: Tx

Tx=

0

lx+t dt

Exam M - Life Contingencies - LGD c 1


2/26

Total lifetime between age x and x + 1: Lx

Lx = Tx Tx+1

=

1

0 lx+t dt=1

0 lx tpx dtTotal lifetime from agex to x + n: nLx

nLx = Tx Tx+n =n1k=0

Lx+k

=

n0

lx+t dt

Average lifetime after x: ex

ex = Tx

lx

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

ex: 1 = Lx

lx

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

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

s(x) =

1

2

Central death rate: mx

mx = lx lx+1

Lx

nmx = lx lx+n

nLx

Fraction of year lived between age x andage x + 1 bydx: a(x)

a(x) =

10

t tpx (x + t) dt

10

tpx (x + t) dt

=Lx lx+1

lx lx+1

Recursion formulas:

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

ex = ex:n + npx ex+n

ex = ex:n + npx ex+n

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

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



3/26

Chapter 4: Life Insurance

Whole life insurance: Ax

E[Z] = Ax =

0

vt tpx x(t)dt

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

n-year term insurance: A1x:n

E[Z] = A1x: n =

n0

vt tpx x(t)dt

V ar[Z] = 2

A1

x:n (A1

x:n)2

m-year deferred whole life: m| Ax

E[Z] = m| Ax=

m

vt tpx x(t)dt

m|Ax = Ax A

1x: m

n-year pure endowment: A 1x: n

E[Z] = A 1x:n =vn npx nEx

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

2 =v2n npx nqx

n-year endowment insurance: Ax: n

E[Z] = Ax: n =

n

0

vt tpx x(t)dt + vn

npx

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

Ax: n = A1x: n +

A 1x: n

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

m|nAx = mEx A

1x+m: n

= A1x: m+n

A1x:m

= m|Ax m+n|Ax

Discrete whole life: Ax

E[Z] = Ax=

k=0

vk+1 kpx qx+k

=

k=0

vk+1 k|qx

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

Discrete n-year term: A1x: n

E[Z] = A1x: n =n1k=0

vk+1 kpx qx+k

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

2

Discrete n-year endowment: Ax:n

E[Z] = Ax:n =n1k=0

vk+1 kpx qx+k+ vn npx

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

Ax:n = A1x:n +

A 1x: n

Recursion and other relations:

Ax =

A

1

x: n + n|Ax

2Ax = 2A1x:n + n|

2Ax

n|Ax = nEx Ax+n

Ax = A1x: n + nEx

Ax+n

Ax = vqx+ vpxAx+12Ax = v

2qx+ v2px

2Ax

A1x: n = vqx+ vpx A 1

x+1: n1

m|nAx = vpx (m1)|nAx+1

Ax: 1 = v

Ax: 2 = vqx+ v2

px



4/26

Varying benefit insurances:

(IA)x =

0

t + 1vt tpx x(t)dt

(IA)1x: n =

n0

t + 1vt tpx x(t)dt

(IA)x =

0

t vt tpx x(t)dt

(IA)1x: n =

n0

t vt tpx x(t)dt

(DA)1

x: n =

n

0

(n t)vt

tpx x(t)dt

(DA)1x: n =

n0

(n t)vt tpx x(t)dt

(IA)x = Ax+ vpx(IA)x+1

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

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

x+1:n1

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

A1x: n(IA)1x: n + (D

A)1x: n = (n + 1)A1x: n

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

1x: n

Accumulated cost of insurance:

nkx=A1x:n

nEx

Share of the survivor:

accumulation factor = 1nEx

=(1 + i)n

npx

Interest theory reminder

a n = 1 vn

i

a n = 1 vn

= i

a n

a = 1

, a =

1

i, a =

1

d

(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 + i

i2

Doubling the constant force of interest

1 + i (1 + i)2

v v2

i 2i + i2

d 2d d2

i 2i + i

2

2

Limit of interest rate i= 0:

Axi=0 1

A1x: ni=0 nqx

n|Axi=0 npx

Ax: ni=0 1

m|nAxi=0

m|nqx

(IA)xi=0 1 + ex

(IA)xi=0 ex



5/26

Chapter 5: Life Annuities

Whole life annuity: ax

ax = E[a T] =

0

a t tpx (x + t)dt

=

0

vt tpx dt=

0

tExdt

V ar[aT

] =2Ax (Ax)2

2

n-year temporary annuity: ax: n

ax:n =

n

0

vt

tpx dt=

n

0

tExdt

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

2

2

n-year deferred annuity: n|ax

n|ax =

n

vt tpx dt=

n

tExdt

n|ax = vn npxax+n = nEx ax+n

V ar[Y] = 2

v2n npx(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 + nEx ax+n

Most important identity

1 = 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 + vpxax+1

ax = ax: n + n|ax

ax = 1 + vpxax+12ax = 1 + v

2px2ax+1

ax:n = 1 + vpxax+1: n1

a(m)x:n = a

(m)x nEx a

(m)x+n

ax = vpx+ vpxax+1

ax:n = vpx+ vpx ax+1: n1(Ia)x = 1 + vpx[(Ia)x+1+ ax+1]

= ax+ vpx(Ia)x+1

Whole life annuity due: ax

ax = E[a K+1] =

k=0

vk kpx

V ar[aK+1] =

2Ax (Ax)2

d2

n-yr temporary annuity due: ax:n

ax: n = E[Y] =n1k=0

vk kpx

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

2

d2

n-yr deferred annuity due: n|ax

n|ax = E[Y] =

k=nvk kpx

= ax ax: n

= nEx ax+n

n-yr certain and life due: ax:n

ax: n = ax+ a n ax: n

= a n +

k=n

vk kpx

= a n +n| ax



6/26

Whole life immediate: ax

ax = E[a K] =

k=1

vk kpx

ax =

1 (1 + i)Ax

i

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

1

m

a(m)

x:n

= a(m)

x: n

1

m(1 nEx)

Accumulation function:

sx:n = ax:n

nEx=

n0

1

ntEx+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: n1

ax: ni=0 ex:n



7/26

Chapter 6: Benefit Premiums

Loss function:Loss = PV of Benefits PV of PremiumsFully continuous equivalence premiums

(whole life and endowment only):

P(Ax) =Axax

P(Ax) = Ax1 Ax

P(Ax) = 1

ax

V ar[L] =

1 +

P

2

2Ax (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=Px

P(Ax) = dAx1 Ax

P(Ax) = 1

ax d

V ar[L] =

1 +

P

d

2 2Ax (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:

hP(Ax) =A

xa

x: h

hP(Ax: n) =

Ax: na

x: h

hPx = Ax

ax: h

hPx: n = Ax: n

ax: h

Pure endowment annual premium P 1x:n :it is the reciprocal of the actuarial accumulated

value sx: n because the share of the survivor whohas depositedP 1x: n at the beginning of each yearfor n years is the contractual $1 pure endow-ment, i.e.

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

P minus P over P problems:The difference in magnitude of level benefit 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 benefits:

( nPx P1x: n)sx:n = Ax+n =

nPx P1x: nP 1x:n

(Px:n nPx)sx:n = 1 Ax+n =Px:n nPx

P 1x: n

(Px: n P1x: n)sx:n = 1 =

Px: n P1x:n

P 1x:n

Miscellaneous identities:

Ax:n = P(Ax:n)

P(Ax: n) +

Ax:n = Px:nPx:n + d

ax:n = 1

P(Ax: n) +

ax:n = 1

Px:n + d



8/26

Chapter 7: Benefit Reserves

Benefit reserve tV:The expected value of the prospective loss at

time t.

Continuous reserve formulas:

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

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

tExPremium 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:

kVx = Ax+k Pkax+k

kVx: n = Px+k: nk Px:n ax+k:nkkVx: n =

1

Px:nP

x+k:nk

A

x+k: nk

tVx: n = Px: nsx: n tkx

kVx = 1 ax+k

ax

kVx = Px+k Px

Px+k+ d

kVx = Ax+k Ax

1 Ax

h-payment reserves:

htV = Ax+t:nt h

P(Ax:n)ax+t: hthk Vx:n = Ax+k:nk hPx:nax+k: hk

hk V(

Ax:n) = Ax+k:nk hP(Ax:n)ax+k: hk

hkV

1(m)

x:n = Ax+k:nk hP(m)x:na

(m)

x+k: hk

Variance of the loss function

V ar[ tL] =1 + P

2 2Ax+t (Ax+t)2

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

(1 Ax)2 assuming EP

V ar[ tL] =

1 +

P

2 2Ax+t: nt (

Ax+t: nt)2

V ar[ tL] =2Ax+t:nt (

Ax+t:nt)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

=A1

x: 4

4Ex

1kx = dx

lx+1=

qxpx

Accumulated differences of premiums:

( nPx P1

x:n)sx: n = n

nVx nV1

x: n)

= Ax+n 0 =Ax+n

( nPx Px)sx: n = n

nVx nVx

= Pxax+n

(Px:n Px)sx: n = nVx:n nVx

= 1 nVx

( mPx: n mPx)sx: m = m

mVx: n mmVx

= Ax+m:nm

Ax+m

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

m+n+pVx = 1

(1 mVx)(1 nVx+m)(1 pVx+m+n)



9/26

Chapter 8: Benefit Reserves

Notations:bj : death benefit payable at the end of year of death for the j -th policy yearj1: benefit premium paid at the beginning of the j -th policy year

bt: death benefit payable at the moment of deatht: annual rate of benefit premiums payable continuously at tBenefit reserve:

hV =

j=0

bh+j+1vj+1

jpx+h qx+h+j

j=0

h+j vj

jpx+h

tV =

0

bt+u vu

upx+t x(t + u)du 0

t+u vu

upx+tdu

Recursion relations:

hV + h = v qx+h bh+1+ v px+h h+1V

(hV + h)(1 + i) = qx+h bh+1+ px+h h+1V

(hV + h)(1 + i) = h+1V + qx+h(bh+1 h+1V)

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

Net amount at Risk for policy year h + 1

Net Amount Risk bh+1 h+1V

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

nV =n1

h=0[P vqx+h(bh+1 h+1V)] (1 + i)

nh

= Ps n n1h=0

vqx+h(bh+1 h+1V)(1 + i)nh

If the death benefit is equal to the benefit reserve for the first n policy years

nV =Ps n

If the death benefit is equal to $1 plus the benefit reserve for the first n policy years

nV =Ps n

n1

h=0 vq

x+h(1 + i)

nh



10/26

If the death benefit is equal to $1 plus the benefit reserve for the first n policy years and qx+h qconstant

nV =Ps n vqs n = (P vq)s n

Reserves at fractional durations:

( hV + h)(1 + i)s = spx+h h+sV + sqx+hv

1s

UDD ( hV + h)(1 + i)s = (1 s qx+h) h+sV + s qx+hv

1s

= h+sV + s qx+h(v1s h+sV)

h+sV = v1s 1sqx+h+s bh+1+ v

1s 1spx+h+s h+1V

UDD h+sV = (1 s)( hV + h) + s(h+1V)

i.e. h+sV = (1 s)( hV) + (s)(h+1V) + (1 s)(h)

unearned premium

Next year losses:

h losses incurred from timeh to h + 1

E[h] = 0

V ar[h] = v2(bh+1 h+1V)

2px+h qx+h

The Hattendorf theorem

V ar[ hL] = V ar[h] + v2px+h V ar[ h+1L]

= v2(bh+1 h+1V)2px+h qx+h+ v

2px+h V ar[ h+1L]

V ar[ hL] = v2(bh+1 h+1V)

2px+h qx+h

+v4(bh+2 h+2V)2px+h px+h+1 qx+h+1

+v6(bh+3 h+3V)2px+h px+h+1 px+h+2 qx+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]

tpxy = sT(x)T(y)(t, t)= P r[T(x)> t andT(y)> t]

Joint life status:

FT(t) = P r[min(T(x), T(y)) t]

= tqxy

= 1 tpxy

Independant livestpxy = tpx tpy

tqxy = tqx+ tqy tqx tqy

Complete expectation of the joint-life sta-tus:

exy =

0

tpxydt

PDF joint-life status:

fT(xy)(t) = tpxy xy(t)

xy(t) =fT(xy)(t)

1 FT(xy)(t)=

fT(xy)(t)

tpxy

Independant lives

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

fT(xy)(t) = tpx tpy[(x + t) + (y+ t)]

Curtate joint-life functions:

kpxy = kpx kpy [IL]

kqxy = kqx+ kqy kqx kqy [IL]

P r[K=k] = kpxy k+1pxy

= kpxy qx+k:y+k

= kpxy qx+k:y+k = k|qxy

qx+t:y+t = qx+k+ qy+k qx+k qy+k [IL]

exy = E[K(xy)] =

1

kpxy

Last survivor status T(xy):

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

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

FT(xy)+ FT(xy) = FT(x)+ FT(y)

tpxy+ tpxy = tpx+ tpy

Axy+ Axy = Ax+ Ay

axy+ axy = ax+ ay

exy+ exy = ex+ ey

exy+ exy = ex+ ey

n|qxy = n|qx+ n|qy n|qxy

Complete expectation of the last-survivorstatus:

exy =

0

tpxydt

exy =1

kpxy

Variances:

V ar[T(u)] = 2

0

t tpudt (eu)2

V ar[T(xy)] = 2

0

t tpxydt (exy)2

V ar[T(xy)] = 2

0

t tpxydt (exy)2

Notes:

For joint-life status, work with ps:

npxy = npx npy

For last-survivor status, work with qs:

nqxy = nqx nqy

Exactly one status:

np[1]xy = npxy npxy

= npx+ npy 2 npx npy

= nqx+ nqy 2 nqx nqya[1]xy = ax+ ay 2axy



12/26

Common shock model:

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

= sT(x)(t) et

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

= sT(y)(t) et

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

= sT(x)(t) sT(y)(t) et

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

Insurance functions:

Au =

k=0

vk+1 kpu qu+k

=

k=0

vk+1P r[K=k]

Axy =

k=0

vk+1 kpxy qx+k:y+k

Axy =

k=0

vk+1( kpx qx+k+ kpy qy+k

kpxy qx+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 ax

Axy = 1 axy

Axy

= 1 axy

Premiums:

Px = 1

ax d

Pxy = 1

axy d

Pxy = 1

axy d

Annuity functions:

au =

0

vt tpudt

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 ofT(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

Chapter 10 & 11: Multiple Decrement Models

Notations:

tq(j)x = probability of decrement in the next

t years due to cause j

tq()x = probability of decrement in the next

t years due to all causes

=m

j=1

tq(j)x

(j)x = the force of decrement due only

to decrement j

()x = the force of decrement due to all

causes simultaneously

=m

j=1

(j)x

tp()x = probability of survivingt years

despite all decrements

= 1 tq()

x

= e

t

0

()x (s)ds

Derivative:

d

dt tp()x =

d

dt tq()x = tp

()x

()x

Integral forms of tqx :

tq(j)x =

t0

sp()x

(j)x (s)ds

tq()

x

=

t

0

sp()

x

()

x

(s)ds

Probability density functions:

Joint PDF: fT,J(t, j) = tp()x

(j)x (t)

Marginal PDF ofJ: fJ(j) = q(j)x

=

0

fT,J(t, j)dt

Marginal PDF ofT: fT(t) = tp()x

()x (t)

=m

j=1

fT,J(t, j)

Conditional PDF: fJ|T(j|t) = (

j)x (t)

()x (t)

Survivorship group:

Group ofl()a people at some agea at timet = 0.

Each member of the group has a joint pdf fortime until decrement and cause of decrement.

nd(j)x = l

()a xap

()a nq

(j)x

= l()a

xa+nxa

tp()a

(j)a (t)dt

nd()x =

mj=1

nd(j)x

l()a =m

j=1

l(j)a

q

(j)

x =

d(j)x

l()x

Associated single decrement:

tq(j)x = probability of decrement from causej only

tp(j)x = exp

t

0

(j)x (s)ds

= 1 tq

(j)x



14/26

Basic relationships:

tp()x = exp

t0

(1)x (s) + +

(m)x (s)

ds

tp()x =

mi=1

tp(i)x

tq(j)x tq

(j)x

tp(j)x tp

()x

UDD for multiple decrements:

tq(j)x = t q

(j)x

tq()x = t q

()x

q(j)x = tp()x

(j)x (t)

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

tp()x

= q(j)x

1 t q()x

tp(j)x =

tp

()x

q(j)xqxt

Decrements uniformly distributed in theassociated single decrement table:

tq(j)x = t q

(j)x

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

1

2

q(2)x

q(2)x = q(2)x

1

1

2q(1)x

q(1)x = q(1)x

1

1

2q(2)x

1

2q(3)x +

1

3q(2)x q

(3)x

Actuarial present values

A=m

j=1

0

B(j)x+tv

t tp()x

(j)x (t)dt

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

Premiums:

P()x =

k=0

B()k+1v

k+1 kp()x q

()x+k

k=0

vk kp()x

P(j)x =

k=0

B(j)k+1v

k+1 kp()x q

(j)x+k

k=0

vk kp()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 ofG that is independent ofb.

Asset shares notations:

G level annual contract premium

kAS asset share assigned to the policy at timet = k

ck fraction of premium paid for expenses atk (i.e. cG is the expenxe premium)

ek expenses paid per policy at timet = k

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

q(w)x+k probability of decrement by withdrawal

kCV cash amount due to the policy holder as a withdrawal benefit

bk death benefit due at timet = k

Recursion formula:

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

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

()x+k k+1AS

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

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

Direct formula:

nAS=n1h=0

G(1 ch) eh vq(d)x+hbh+1 vq

(w)x+hh+1CV

nhE()x+h



16/26

Constant Force of Mortality

Chapter 3

(x) = > 0, x

s(x) = ex

lx = l0ex

npx = en = (px)

n

ex = 1

=E[T] =E[X]

ex:n = ex(1 npx)

V ar[T] = V ar[x] = 1

2

mx =

Median[T] =

ln2

= Median[X]

ex = px

qx=E[K]

V ar[K] = px(qx)2

Chapter 4

Ax =

+

2Ax =

+ 2A1x: n =

Ax(1 nEx)

nEx = en(+)

(IA)x =

( + )2

Ax = q

q+ i

2Ax = q

q+ 2i + i2

A1x: n = Ax(1 nEx)

Chapter 5

ax = 1

+

2ax = 1

+ 2

ax = 1 + i

q+ i

2ax = (1 + i)2

q+ 2i + i2

ax:n = ax(1 nEx)ax:n = ax(1 nEx)

(IA)x = Axax =

( + )2

(IA)x = Axax = (1 + i)

( + )(q+ i)

(IA)x = Axax = q(1 + i)

(q+ i)2

(Ia)x = (ax)2 =

1

( + )2

(Ia)x = (ax)2 =

1 + i

q+ i

2 Chapter 6

Px = vqx= P1x:n

P(

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

tV(Ax) = 0, t 0

kV

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

For fully discrete whole life, assuming EP,

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

For fully continuous whole life, assuming EP,

V ar[ tLoss] = 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 = 1

M + F

Axy = qxyqxy+ i

axy = 1 + i

qxy+ i

exy = pxy

qxy



17/26

De Moivres Law

Chapter 3

s(x) = 1 x

lx = l0 x ( x)

qx = (x) = 1

x

n|mqx = m

x

npx = x n

x

tpx (x + t) = qx= (x) =fT(x), 0 t < x

Lx = 1

2(lx+ lx+1)

ex = x

2 =E[T] = Median[T]

ex = x

2

1

2=E[K]

V ar[T] = ( x)2

12

V ar[K] = ( x)2 1

12

mx = qx

1 12qx=

2dxlx+ lx+1

a(x) = E[S] =1

2ex:n = n npx +

n

2 nqx

ex:n = ex: n +n

2 nqx

Chapter 4

Ax =a x x

A1x: n = a n x

2Ax =a2(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 =

1Axd

and thechapter 4 formulas.

Chapter 9

exx = x

3

( MDML with = 2/( x))

exx = 2( x)

3exy = yxpx eyy + yxqx 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

Modified De Moivres Law

Chapter 3

s(x) =

1 x

clx = l0

x

c ( x)c

(x) = c

x

npx = 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

tqx = t qx , 0 t 1

(x + t) = qx1 tqx, 0 t 1

lx+t = lx t dx , 0 t 1

sqx+t = sqx1 tqx

, 0 s + t 1

V ar[T] = V ar[K] + 1

12

mx = (x +1

2) =

qx

1 12qx

Lx = lx 1

2dx

a(x) = 12

ex: 1 = px+1

2qx

tpx (x + t) = qx , 0 t 1

Chapter 4

Ax = i

Ax

A(m)x = i

i(m)Ax

A1x: n = iA1x: n

(IA)1x: n = i

(IA)1x: n

Ax: n = i

A1x: n + A

1x:n =

i

A1x:n + nEx

n|Ax =

i

n|Ax

2Ax = 2i + i2

22Ax

Chapter 5

a(m)x ax m 1

2m

a(m)x = (m)ax (m)

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

with: (m) = id

i(m)d(m) 1

and (m) = i i(m)

i(m)d(m)

m 1

2m

a(m)x: n = a

(m)x nEx a

(m)x+n

Chapter 6

P(Ax) = i

Px

P(A1x: n) = i

P1x: n

P(Ax: n) = i

P1x: n + P

1x: n

P(m)x = Px

(m) (m)(Px+ d)

P(m)

x: n = Px: n

(m) (m)(P1x:n + d)

nP(m)

x = nPx

(m) (m)(P1x:n + d)

hP(m)(A1x: n) = i

hP1(m)x:n

Chapter 7

hk V

(m)x: n =

hk Vx: n + (m) hP

(m)x: n kV

1

x:h

hk V

(m)(Ax: n) = h

k V(Ax:n) + (m) hP

(m)(Ax:n) kV1

x:

hk V(

Ax: n) = i

hk V

1

x:h + hk V

1

x:h

Chapter 10UDDMDT

tq(j)x = t q

(j)x

q(j)x = (j)x (0)

q()x = ()x (0)

tp(j)x =

tp

()x

q(j)xq

()x

(j)x = q

(j)x

1 t q()x

()x = q

()x

1 t q()x

UDDASDT

tq(j)x = t q

(j)x

q(1)x = q(1)x

1

1

2q(2)x

q(2)x = q(2)x

1

1

2q(1)x

q(1)x = q(1)x

1

1

2q(2)x

1

2q(3)x +

1

3q(2)x q

(3)x



19/26

Chapter 1: The Poisson Process

Poisson process with rate :

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

k!E[N(t)] = t

V ar[N(t)] = t

Interarrival time distribution:The waiting time between events. Let Tn de-note the time since occurence of the eventn1.Then the Tn are independent random variablesfollowing an exponential distribution with mean1/.

P r[Ti t] = 1 et

fT(t) = et

E[T] = 1

V ar[T] = 1

2

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

ni=1 Ti.

Sn has a gamma distribution with parameters nand = 1/

Sn GammaRV[= n, = 1

]

P r[Sn t] =j=n

et(t)j

j!

fSn(t) = et (t)

n1

(n i)!

E[Sn] = n

V ar[Sn] = n

2

Sum of Poisson processes:IfN1, , Nkare independent Poisson processeswith rates 1, , k then, N = N1+ +Nkis 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 Ni counts the spe-cial events of kind i. Then, Ni is a Poissonprocess with rate i = i and the Ni are inde-pendent of one another.If the probability i(t) changes with time, then

E[Ni(t)] =

t0

(s)ds

Non-homogeneous Poisson process:

(t) intensity function

m(t) mean value function

=

t

0

(y)dy

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

k!

Compound Poisson process:

X(t) =

N(t)i=1

Yi N(t) PoissonRV w/ rate

E[X(t)] = t E[Y]

V ar[X(t)] = t E[Y2]

Two competing Poisson processes:Probability that n events in the Poisson process (N1,1) occur before m events in the Poisson process(N2,2):

P r[S1n< S2m] =

n+m1k=n

n+m1

k

1

1+ 2

k 21+ 2

n+m1k

P r[S1n< S21 ] =

1

1+ 2

n

P r[S1

1< S2

1] =

1

1+ 2

Exam M - Loss Models - LGDc 1


20/26

Chapter 2&3: Random Variables

k-th raw moment:

k =E[Xk]

k-th central moment:

k =E[(X )k]

Variance:

V ar[x] = 2 =E[X2] 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):

YP =X d|X > d

Mean excess loss function:

eX(d) e(d) =E[YP]

= E[X d|X > d]

=

d

S(x)dx

1 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)kf(x)dx

1 F(d)

=

x>d

(x d)kp(x)

1 F(d)

Left censored and shifted variable:

YL = (X d)+

= 0 X < dX d X d

Moments of the left censored and shiftedvariable:

E[(X d)k+] =

d

(x d)kf(x)dx

=x>d

(x d)kp(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

=

u0

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

Moments of the limited loss variable:

E[(X u)k] =

u

xkf(x)dx + uk[1 F(u)]

= xu

xkp(x) + uk[1 F(u)]

Moment generating functionsmX(t):

mX(t) =E[etX]

Sum of random variablesSk =X1+ +Xk:

mSk(t) =kj=1

mXj(t)



21/26

Chapter 4: Classifying and Creating distributions

k-point mixture (

ai = 1):

FY(y) = a1FX1 (y) + + akFXk(y)

fY(y) = a1fX1 (y) + + akfXk(y)E[Y] = a1E[X1] + + akE[Xk]

E[Y2] = a1E[X21 ] + + akE[X

2k ]

Tail weight:

existence of moments light tail

hazard rate increases light tail

Multiplication by a constant ,y >0:

Y =X FY(y) =FX( y

) andfY(y) =1

fX( y

)

Raising to a power:

Y =X1

>0 : FY(y) =FX(y)

fY(y) = y1fX(y

) 0: transformed distribution = 1: inverse distribution

0,

aj = 1):

f(x) =

a1f1(x) c0< x < c1...

...akfk(x) ck1< x < ck

Discrete probability function (pf):

pk = P r[N =k]

Probability generating function (pgf):

PN(z) = E[zN] =p0+p1z+p2z

2 +

P(1) = E[N]

P(1) = E[N(N 1)]

Poisson distribution:

pk = ek

k!

P(z) = e(z1)

E[N] =

V ar[N] =

Negative Binomial distribution:Number of failures before success r with proba-bility of successp = 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 N1 and N2 are independent NegativeBinomial distributions with parameters(r1, ) and (r2, ), thenN1+ N2 is a Neg-

ative Binomial distribution with parame-ters (r= r1+ r2, )



22/26

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

ters (rj =r, j =j)

Geometric distribution:Special case of the negative binomial distribu-tion withr= 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 =

m

k

qk(1 q)mk

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

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

The (a,b, 0) class:

pkpk1

=a +b

k

a >0 negative binomial or geometric

distribution

a= 0 Poisson distribution

a


23/26

Chapter 5: Frequency and Severity with Coverage Modifications

Notations:

X amount of loss

YL amount paid per lossYP amount paid per payment

Per loss variable:

fYL (0) = FX(d)

fYL (y) = fx(y+ d)

FYL (y) = FX(y+ d)

Per payment variable:

fYP(y) = fx

(y+ d)

1 FX(d)

FYL (y) = FX(y+ d) FX(d)

1 FX(d)

Relations per loss / per payment variable:

YP = YL|YL >0

E[YP] = E[YL]

P r[YL >0]

E[(YP)k] = E[(YL)k]

P r[YL

>0]

Simple (or ordinary) deductible d:

X = X

YL = (X d)+

YP = X d|X > d

E[YL] = E[(X d)+]

= E[X] E[X d]

E[YP] = E[X d|X > d]

= E[X] E[X d]

1 FX(d)

Franchise deductible d:

X = X

YL = 0 if X < d, X ifX > d

YP = X|X > d,

YPfranchise = YPordinary+ d

E[YL] = E[X] E[X d] + d[1 FX(d)]

E[YP] = E[X|X > d]

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

+ d

Effect of deductible for various distribu-tions:

X: exp[mean ] YP

: exp[mean ]X: uniform[0, ] YP : uniform[0, d]

X: 2Pareto[, ] YP : 2Pareto[ =, = +

Loss elimination ratio:

LERX(d) =E[X d]

E[X]

relations:

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

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

Inflation on expected cost per loss:

E[YL] = (1 + r)

E(X) E

X

d

1 + r

Inflation on expected cost per payment:

E[YP] =(1 + r)

E(X) E

X d1+r

1 FX d1+r

u-coverage limit:

YL = X u

YP = X u|X >0

E[YL] = E[X u]

=

u0

[1 Fx(x)]dx

u-coverage limit with inflation:

E[X u] (1 + r)E

X

u

1 + r

u-coverage limit and d ordinary de-ductible:

YL = (X u) (X d)

YP = (X u) (X d)|X >0

= (X u) d|X > d

E[YL] = E[X u] E[X d]

E[YP] = E[X u] E[X d]1 FX(d)



24/26

u-coverage limit anddordinary deductiblewith inflation:

E[YL] = (1 + r)

E

X

u

1 + r

E

X

d

1 + r

E[YP] = (1 + r)E

X u1+r E

X d1+r

1 FX

d1+r

Coinsurance factor:First calculate YL and YP with deductible andlimits. Then apply

YL = YL

YP = YP

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

v= 1 FX(d)

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

Bernoulli random variable:

N =

0 @ 1 p1 @ p

E[N] = p

V ar[N] = (1 p)p

Deductible, limit, inflation and coinsurance:

d = d

1 + r

u = u

1 + r

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

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

1 FX(d)E[YL

2] = 2(1 + r)2

E

(X u)2E

(X d)2

2dE[X u] + 2dE[X d]

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

PN(z) =PN[1 + v(z 1)]

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

ability is modified 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 forN

Poisson =vBinomial m, q m =m, q =vqNegative Binomial r, r =r, =v



25/26

Chapter 6: Aggregate Loss Models

The aggregate loss compound randomvariable:

S =

Nj=1

Xj

E[S] = E[N] E[X]

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

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

Probability generating function:

PS(z) =PN[PX(z)]

Notations:

fk = P r[X=k]

pk = P r[N=k]

gk = P r[S= k]

Compound probabilities:

g0 = PN(f0) ifP r[X= 0] = f0= 0

gk = p0 P r[sum ofXs =k]

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

+ p2 P r[sum of two Xs =k]

+

IfN is in the (a,b, 0) class, then

g0 = PN(f0)

gk = 1

1 af0

kj=1

a +

j b

k

fj gkj

n-fold convolution of the pdf ofX:

f(n)X (k) = P r[sum ofn Xs =k]

f(n)X (k) =

kj=0

f(n1)X (k j)fX(j)

f(1)X (k) = fX(k) =fk

f(0)X (k) =

1 , k= 00 , otherwise

n-fold convolution of the cdf ofX:

F(n)X (k) =

k

j=0

F(n1)X (k j)fX(j)

F(0)X (k) =

0 , k


26/26

Last Chapter: Multi-State Transition Models

Notations:

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

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

(i,i)n

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

Theorem:

kQn = Qn Qn+1 Qn+k1

kQn = Qk for a homogeneous Markov chain

Inequality:

kP(i)n =P

(i)n P

(i)n+1 P

(i)n+k1 kQ

(i,i)n

Cash flows while in states:

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

kvn = present value of 1 paidk years after time t = n

AP Vs@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 Vs@n(C(i)) =

k=0

kQ(s,i)n n+kC(i) kvnCash flows upon transitions:

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

lQ(s,i)n Q

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

AP Vs@n(C(i,j)) = actuarial present value at timet = n of a cash flow to be paid upon

transition from statei to state j

AP Vs@n(C(i,j)) =

k=0

kQ

(s,i)n Q

(i,j)n+k

n+k+1C

(i,j) k+1vn

Documents

Exam M Notes F05