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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

    Exam M - Life Contingencies - LGD c 1

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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

    Exam M - Life Contingencies - LGD c 2

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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

    Exam M - Life Contingencies - LGD c 3

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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

    Exam M - Life Contingencies - LGD c 4

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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

    Exam M - Life Contingencies - LGD c 5

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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

    Exam M - Life Contingencies - LGD c 6

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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

    Exam M - Life Contingencies - LGD c 7

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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)

    Exam M - Life Contingencies - LGD c 8

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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

    Exam M - Life Contingencies - LGD c 9

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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

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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]

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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, )

    Exam M - Loss Models - LGDc 3

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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

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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 ]

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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