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    3 Methods To Differentiate Term Insurance

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    han
    Method 1 discusses determining the partial derivative of A1:n with respect to x by using an integral formula from the textbook. Method 2 derives a differential equation for the partial derivative of A1:n with respect to x using the Leibniz rule and integrating by parts. Method 3 changes the time variables to age variables and again uses the Leibniz rule to obtain a differential equation relating the partial derivative of A1:n to δ, μx, and A1:n. The solution is that the partial derivative equals δA1:n - μx(1 - A1:n).

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    3 Methods To Differentiate Term Insurance

    Uploaded by

    han
    1 views2 pages
    Method 1 discusses determining the partial derivative of A1:n with respect to x by using an integral formula from the textbook. Method 2 derives a differential equation for the partial derivative of A1:n with respect to x using the Leibniz rule and integrating by parts. Method 3 changes the time variables to age variables and again uses the Leibniz rule to obtain a differential equation relating the partial derivative of A1:n to δ, μx, and A1:n. The solution is that the partial derivative equals δA1:n - μx(1 - A1:n).

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    3 methods to differentiate term insurance

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    Method 1 discusses determining the partial derivative of A1:n with respect to x by using an integral formula from the textbook. Method 2 derives a differential equation for the partial derivative of A1:n with respect to x using the Leibniz rule and integrating by parts. Method 3 changes the time variables to age variables and again uses the Leibniz rule to obtain a differential equation relating the partial derivative of A1:n to δ, μx, and A1:n. The solution is that the partial derivative equals δA1:n - μx(1 - A1:n).

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    1 views2 pages

    3 Methods To Differentiate Term Insurance

    Uploaded by

    han
    Method 1 discusses determining the partial derivative of A1:n with respect to x by using an integral formula from the textbook. Method 2 derives a differential equation for the partial derivative of A1:n with respect to x using the Leibniz rule and integrating by parts. Method 3 changes the time variables to age variables and again uses the Leibniz rule to obtain a differential equation relating the partial derivative of A1:n to δ, μx, and A1:n. The solution is that the partial derivative equals δA1:n - μx(1 - A1:n).

    Copyright:

    © All Rights Reserved
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    Download as pdf or txt
    of 2

    ∂ 1

    Three methods to determine A


    ∂x x:n
    Method 1: Exercise 4.21.c on page 129 of AM (HW Set 3 #3)

    Method 2: Applying the Leibniz rule (AM p. 701 first result) to the formula
    n t
    =
    Ax1:n ∫0 v t px µ x +t dt
    and noting the two limits of the integral are constant, we have
    ∂ 1 n ∂
    = Ax:n ∫ vt ( t px µ x +t )dt .
    ∂x 0 ∂x
    By the product rule,
    ∂ ∂ ∂ ∂
    ( t px µ x +t )= ( t px )µ x +t + t px µ x +t = t px (µ x − µ x +t )µ x +t + t px µ x +t .
    ∂x ∂x ∂x ∂t
    ∂ ∂ ∂ ∂
    (That µ x +t = µ x +t follows from g ( x + t )= g ′( x + t )= g ( x + t ) which is true
    ∂x ∂t ∂x ∂t
    because of the chain rule.) Hence, we have
    ∂ 1 n n ∂
    Ax:n = µ x Ax1:n − ∫ vt t px (µ x +t ) 2dt + ∫ vt t px µ x +t dt .
    ∂x 0 0 ∂t
    The last integral “invites” you to integrate by parts,
    n ∂ n
    ∫0 t Ex ∂t µ x+t d=t ∫0 t Ex dµ x+t
    t =n
    ( t Exµ x+t ) t =0 − ∫0 µ x +t d ( t E x )
    n
    =

    ( n Exµ x+n − µ x ) − ∫0 ( )
    n
    = µ x + t v t d t p x + t p x dv t .

    ( n Exµ x+n − µ x ) + ∫0 vt t px (µ x+t )2dt + δAx1:n .


    n
    =
    Hence, we have derived the following differential equation for Ax1:n :
    ∂ 1
    Ax:n = µ x Ax1:n − ∫  + ( n Ex µ x + n − µ x ) + ∫ + δAx1:n 
    ∂x
    = ( n Ex µ x + n − µ x ) + (µ x + δ) Ax1:n , x ≥ 0.

    t t

    Method 3: Because vt t px = e ∫0 e ∫0
    − ( δ+µ x + s )ds n − ( δ+µ x + s )ds
    ,=
    we have Ax1:n ∫0 µ x + t dt .
    By changing the time variables t and s to age variables
    y = x+t and z = x+s, we have
    z= y

    e ∫z = x
    y= x + n − ( δ+µ z )dz
    =Ax1:n ∫y = x µ y dy .
    By the Leibniz rule,
    ∂  − ∫z = x ( δ+µ z )dz 
    z= y
    ∂ 1 y= x + n
    = A
    ∂x x:n ∫y = x 
    ∂x 
    e µ y  dy

    =z x= +n
    ∂ ∂
    z x

    + e ∫ ( x + n) − e ∫
    − ( δ+µ z )dz − ( δ+µ z )dz
    z=x z=x
    µ x+n µx x.
    ∂x ∂x
    Now, by the chain rule and Leibniz rule,

    ( )
    =z y=z y
    ∂ − ∫ z x= ∂
    e ∫z x
    ( δ+µ z )dz − ( δ+µ z )dz z= y
    =
    e= × − ∫ (δ + µ z )dz
    ∂x ∂x z=x
    z= y z= y
     ∂ 
    = e ∫z = x × −  0 + 0 − ( δ + µ x ) x  = ( δ + µ x ) e ∫z = x
    − ( δ+µ z )dz − ( δ+µ z )dz
    .
     ∂x 
    ∂ 1
    Hence, Ax:n = (δ + µ x ) Ax1:n + n Ex µ x + n − µ x .
    ∂x

    ∂
    Question: A =?
    ∂x x:n
    Solution (By using the answer above): Because A=
    x:n
    Ax1:n + n Ex and because

    ( )
    z=
    x+n z=
    x+n
    ∂= ∂ − ∫z x= ∂
    = e ∫z x
    ( δ+µ z )dz − ( δ+µ z )dz z= x+ n
    n=
    Ex e × −∫ (δ + µ z )d
    =z n E x ( −µ x + n + µ x )
    ∂x ∂x ∂x z =x
    (the last formula is #10(i) in Midterm Exam 1; it can also be obtained by using
    ∂
    n=px n px (µ x − µ x + n ) ), we have
    ∂x
    ∂
    Ax:n = (δ + µ x ) Ax1:n − µ x + n Ex µ x = δAx1:n − µ x (1 − Ax:n ) .
    ∂x

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