acts 150 - formula (test 3)
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8/4/2019 ACTS 150 - Formula (Test 3)
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Chapter 3: Survival Distributions and Life Tables
Distribution function of X:
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 of T(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 = 1s(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
dttqx = tpx (x + t) = fT(x)(t)
d
dttpx = tpx (x + t)
d
dtTx = lx
d
dtLx = dx
ddt
ex = (x)ex 1
Mean and variance of T 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
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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 age x to x + n: nLx
nLx = Tx Tx+n =n1k=0
Lx+k
=
n0
lx+t dt
Average lifetime after x: ex
ex =Txlx
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)=
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 by dx: a(x)
a(x) =
10
t tpx (x + t) dt
10
tpx (x + t) dt
=Lx lx+1lx lx+1
Recursion formulas:
E[K] = ex = px(1 + ex+1)E[T] = ex = px(1 + ex+1) + qx a(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
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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 A1x: m
n-year pure endowment: A 1x: n
E[Z] = A 1x:n = vn npx nEx
V ar[Z] = 2A 1x: n (A1
x: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|nAx
m|nAx = mEx A1
x+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|
Ax2Ax =
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 A1
x+1: n1
m|nAx = vpx (m1)|nAx+1
Ax: 1 = v
Ax: 2 = vqx + v2
px
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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 = nA
1x: n
(IA)1x: n + (DA)1x: n = (n + 1)A
1x: 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 + i2
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
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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 npx ax+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 + vpx ax+12ax = 1 + v
2px2ax+1
ax:n = 1 + vpx ax+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
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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
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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) =Ax
1 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) =Axax
= 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: na
x: 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 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 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:n
Px:n + d
ax:n =1
P(Ax: n ) +
ax:n =1
Px:n + d
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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: n sx: n tkx
kVx = 1 ax+k
ax
kVx =Px+k PxPx+k + d
kVx =Ax+k Ax
1 Ax
h-payment reserves:
ht V = Ax+t:nt hP(Ax:n )ax+t: ht
hk Vx:n = Ax+k:nk hPx:n ax+k: hk
hk V(Ax:n ) = Ax+k:nk h P(Ax:n )ax+k: hk
hkV
1(m)
x:n = Ax+k:nk hP(m)x:n a
(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)2assuming 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 )2assuming 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 =nnVx nV
1x: n )
= Ax+n 0 = Ax+n
( nPx Px)sx: n =nnVx nVx
= Pxax+n
(Px:n Px)sx: n = nVx:n nVx
= 1 nVx
( mPx: n mPx)sx: m =mmVx: 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)
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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
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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 time h 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 +
Exam M - Life Contingencies - LGD c 10
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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 and T(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 tpxy dt (exy)2
V ar[T(xy)] = 2
0
t tpxy dt (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
Exam M - Life Contingencies - LGD c 11
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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 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]
Exam M - Life Contingencies - LGD c 12
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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 surviving t years
despite all decrements
= 1 tq()
x
= e
tR
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 of J: fJ(j) = q(j)x
=
0
fT,J(t, j)dt
Marginal PDF of T: 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 age a at time t = 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 cause j only
tp(j)x = exp
t
0
(j)x (s)ds
= 1 tq
(j)x
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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
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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 premium
kAS 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 withdrawal
kCV cash amount due to the policy holder as a withdrawal benefit
bk death benefit due at time t = 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+h h+1CV
nhE()x+h
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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 =pxqx
= 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 =qxy
qxy + i
axy =1 + i
qxy + i
exy =pxyqxy
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De Moivres Law
Chapter 3
s(x) = 1x
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
2nqx
ex:n = ex: n +n
2nqx
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
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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 =sqx
1 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 ) =hk 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
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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 event n1.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, , Nk are 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
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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 functions mX(t):
mX(t) = E[etX ]
Sum of random variables Sk = X1+ +Xk:
mSk(t) =kj=1
mXj(t)
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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
) and fY(y) = 1
fX( y
)
Raising to a power:
Y = X1
> 0 : FY(y) = FX(y)
fY(y) = y1fX(y
) < 0 : FY(y) = 1 FX(y
)fY(y) = y
1fX(y)
> 0: transformed distribution = 1: inverse distribution
< 0: inverse transformed
Exponentiation:
Y = eX FY(y) = FX(ln(y))fY(y) =
1y
fX(ln(y))
Mixing:The random variable X depends upon a para-meter , itself a random variable . For a givenvalue = , the individual pdf is fX|(x|).
fX(x) =
fX|(x|)f()d
IfX 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
)
IfX is a Normal distribution (, V) and follows a Normal distribution (, A),then the resulting distribution is a Normal
distribution (, A + V)
Splicing (aj > 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 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 N1 and N2 are independent NegativeBinomial distributions with parameters(r1, ) and (r2, ), then N1 + N2 is a Neg-
ative Binomial distribution with parame-ters (r = r1 + r2, )
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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 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
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 < 0 Binomial distribution
Compound frequency models:
P(z) = Pprim (Psec(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 fn = P r[M = n]. IfN is a mem-ber of the (a,b, 0) class
gk =1
1 af0
kj=1
a +
j b
k
fj gkj
g0 = Pprim(f0)
gk =
k
kj=1
j fj gkj
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(z1)
P(z) = p(1)e1(z1) + +p(n)e
n(z1)
Exposure modifications on frequency:n policies in force. Nj claims produced by the
j-th policy. N = N1 + + Nn. If the Nj are in-dependent and identcally distributed, then theprobability generating function for N is
PN(z) = [PN1 (z)]n
If the company exposure grows to n, let N
represent the new total number of claims,
PN(z) = [PN1 (z)]n = [PN(z)]
n
n
Conditional expectations:
E[X| = ] =
xfX|(x|)dx
E[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 modifications on frequency for common distributions:N Parameters for N Parameters for N
Poisson
=n
n Negative Binomial r, r = n
nr, =
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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 if X > 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)
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u-coverage limit and d ordinary 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
d
1+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 for N
Poisson = vBinomial m, q m = m, q = vqNegative Binomial r, r = r, = v
Exam M - Loss Models - LGDc 6
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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) if P r[X = 0] = f0 = 0
gk = p0 P r[sum of Xs = k]
+ p1 P r[sum of one X = k]
+ p2 P r[sum of two Xs = k]
+
If N 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 of X:
f(n)X (k) = P r[sum of n 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 of X:
F(n)X (k) =
k
j=0
F(n1)X (k j)fX(j)
F(0)X (k) =
0 , k < 01 , k 0
Pdf of the compound distribution:
fS(k) =n=0
pn f(n)X (k)
Cdf of the compound distribution:
FS(k) =
n=0
pn F(n)X (k)
Mean of the compound distribution:
E[S] =k=0
k fS(k) =k=0
[1 FS(k)]
Net stop-loss premium:
E[(S d)+
] =
d
(x d)fS
(x)dx
E[(S d)+] =d+1
(x d)fS(x)
E[(S d)+] = E[S] d d1x=0
(x d)fS(x)
E[(S d)+] =d
[1 FS(d)]
= E[S]
d1x=0
[1 FS(d)]
Linear interpolation a < d < b:
E[(S d)+] =b d
b aE[(S a)+]+
d a
b aE[(S b)+]
Recursion relations for stop-loss insurance:
E[(S d 1)+] = E[(S d)+] [1 FS(d)]
E[(S sj)] = E[(S sj+1)] + (sj+1 sj)P r[S sj+1]
where the sj s are the possible values of S, S : s0 = 0 < s1 < s2 < andP r[S sj+1] = 1 P r[S = s0 or S = s1 or or S = sj ]
Exam M - Loss Models - LGDc 7
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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 paid k 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 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 Vs@n(C(i,j)) = actuarial present value at time t = n of a cash flow to be paid upon
transition from state i 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