Recursion

Recursion#

Using annual values provided we can calculate other values at other ages and durations by applying recursion formulas and other actuarial identities.

Chain rule#

\(_{t+n}p_x =\ _np_x \cdot\ _tp_{x+n}\)

survival probability chain rule

\(_{t+n}E_x =\ _nE_x \cdot\ _tE_{x+n}\)

pure endowment chain rule

Expected future lifetime#

For both complete and curtate future lifetime, we can decompose the formula for the life expectancy of \((x)\) into a temporary life expectancy, plus the probability of surviving till the end of that term times the remaining life expectancy.

\(e_{x} = e_{x:\overline{m|}} +\ _mp_x\ e_{x+m}\)

recursion formula for curtate expectation of lifetime

\(\overset{\circ}{e}_{x} = \overset{\circ}{e}_{x:\overline{m|}} +\ _mp_x\ \overset{\circ}{e}_{x+m}\)

recursion formula for complete expectation of lifetime

\(e_{x:\overline{m+n|}} = e_{x:\overline{m|}} +\ _mp_x\ e_{x:\overline{n|}}\)

recursion formula for limited curtate expectation of lifetime

\(\overset{\circ}{e}_{x:\overline{m+n|}} = \overset{\circ}{e}_{x:\overline{m|}} +\ _mp_x\ \overset{\circ}{e}_{x+m:\overline{n|}}\)

recursion formula for limited complete expectation of lifetime

\(e_{x} = p_x(1 + e_{x+1})\)

special case of a one-year recursion formula for curtate expectaion of lifetime

\(\overset{\circ}{e}_{x} = \overset{\circ}{e}_{x:\overline{1|}} + p_x\ \overset{\circ}{e}_{x+1}\)

special case of a one-year recursion formula for complete expectation of lifetime

\(e_{x:\overline{1|}} = p_x\)

shortcut for one-year limited curtate expectaion of lifetime

Life insurance#

\(A_{x} = v ~ q_x + v ~ p_x ~ A_{x+1} ~\Rightarrow~ A_{x+1} = \dfrac{A_{x} - v ~ q_x}{v ~ p_x}\)

whole life insurance recursion

\(A^{1}_{x:\overline{t|}} = v ~ q_x + v ~ p_x ~ A^{1}_{x+1:\overline{t-1|}} \)

term life insurance recursion

\(A^{1}_{x:\overline{1|}} = v ~ q_x \)

shortcut for one-year term life insurance

\(^{2}A^{1}_{x:\overline{1|}} = v^2 ~ q_x\)

shortcut for second moment of one-year term life insurance

\(A_{x:\overline{0|}} =~ _0E_x = 1\)

endowment insurance at end of term is pure endowment

\(A_{x:\overline{1|}} = q_x ~ v + p_x ~ v = v\)

shortcut for one-year endowment insurance

\(^2A_{x:\overline{1|}} = v^2\)

shortcut for second moment of one-year endowment insurance

\(IA^{1}_{x:\overline{t|}} = v ~ q_x + v ~ p_x ~ (A_{x+1} + IA^{1}_{x+1:\overline{t-1|}})\)

increasing insurance recursion

\(DA^{1}_{x:\overline{t|}} = t ~ v ~ q_x + v ~ p_x ~ (DA^{1}_{x+1:\overline{t-1|}})\)

decreasing insurance recursion

Life annuities#

\(\ddot{a}_{x} = 1 + v ~ p_x ~ \ddot{a}_{x+1} ~\Rightarrow ~\ddot{a}_{x+1} = \dfrac{\ddot{a}_{x} - 1}{v ~ p_x}\)

whole life annuity recursion

\(\ddot{a}_{x:\overline{t|}} = 1 + v ~ p_x ~ \ddot{a}_{x+1:\overline{t-1|}} \)

temporary annuity recursion

\(\ddot{a}_{x:\overline{1|}} = 1\)

shortcut for one-year temporary annuity

Methods#

The Recursion class implements methods to apply recursive, shortcut and actuarial formulas, and traces the steps taken to find the solution.

Caveats:

Not all possible recursion rules and actuarial equations have (yet) been implemented in the present version of the package.
You may set the recursion depth to a larger limit than the default of 3 (with the keyword argument depth when initializing a Recursion class object).
But generally, the current implementation may be fragile if the solution is not available within a relatively shallow search.

Notes:

If a colab or jupyter notebook is auto-detected, the steps are displayed in latex format; else as raw text.
These display options can be changed by calling the blog_options static method

from actuarialmath import Recursion, ConstantForce, Contract
import describe
describe.methods(Recursion)

class Recursion - Solve by appling recursive, shortcut and actuarial formulas repeatedly

    Args:
      depth : maximum depth of recursions (default is 3)
      verbose : whether to echo recursion steps (True, default)

    Notes:
      7 types of function values can be loaded for recursion computations:

      - 'q' : (deferred) probability (x) dies in t years
      - 'p' : probability (x) survives t years
      - 'e' : (temporary) expected future lifetime, or moments
      - 'A' : deferred, term, endowment or whole life insurance, or moments
      - 'IA' : decreasing life insurance of t years
      - 'DA' : increasing life insurance of t years
      - 'a' : deferred, temporary or whole life annuity of t years, or moments

    Methods:
    --------

    set_q(val, x, s, t, u):
      Set mortality rate u|t_q_[x+s] to given value

    set_p(val, x, s, t):
      Set survival probability t_p_[x+s] to given value

    set_e(val, x, s, t, curtate, moment):
      Set expected future lifetime e_[x+s]:t to given value

    set_E(val, x, s, t, endowment, moment):
      Set pure endowment t_E_[x+s] to given value

    set_A(val, x, s, t, u, b, moment, endowment, discrete):
      Set insurance u|_A_[x+s]:t to given value

    set_IA(val, x, s, t, b, discrete):
      Set increasing insurance IA_[x+s]:t to given value

    set_DA(val, x, s, t, b, discrete):
      Set decreasing insurance DA_[x+s]:t to given value

    set_a(val, x, s, t, u, b, variance, discrete):
      Set annuity u|_a_[x+s]:t to given value

    blog_options(latex, notebook):
      Static method to change display options for tracing the recursion steps

Examples#

The given values of functions are set by calling the respective “setter” methods set_q, set_p, set_e, set_E, set_A, set_IA, set_DA, or set_a. Then derived function values, at other ages or durations, can be computed – the successful recursion steps taken are displayed if the verbose flag is initiated to True (which is the default value).

AMLCR2 Exercise 2.6

Given \(P_x =0.99\) , \(P_{x+1} = 0.985\), \(_3P_{x+1} = 0.95\), \(q_{x+3} = 0.02\),

Calculate (a) \(P_{x+3}\), (b) \(_2P_x\), (c) \(_2P_{x+1}\), (d) \(_3P_x\), (e) \(_{1|2}q_x\).

from actuarialmath.recursion import Recursion
x = 0
life = Recursion(depth=3).set_interest(i=0.06)\
                         .set_p(0.99, x=x)\
                         .set_p(0.985, x=x+1)\
                         .set_p(0.95, x=x+1, t=3)\
                         .set_q(0.02, x=x+3)

print(life.p_x(x=x+3), 0.98)
print(life.p_x(x=x, t=2),  0.97515)
print(life.p_x(x=x+1, t=2), 0.96939)
print(life.p_x(x=x, t=3), 0.95969)
print(life.q_x(x=x, t=2, u=1),  0.03031)

\[\begin{split}\begin{array}{llll} ~\texttt{Survival}~p_{x+3}~:=\\ ~~~p_{x+3} = 1 - ~q_{x+3}& \quad \texttt{complement of mortality}\end{array}\end{split}\]

0.98 0.98

\[\begin{split}\begin{array}{llll} ~\texttt{Survival}~_{2}p_{x}~:=\\ ~~~_{2}p_{x} = ~_{3}p_{x} / ~p_{x+2}& \quad \texttt{survival chain rule}\\ ~~~~~p_{x+2} = ~_{2}p_{x+1} / ~p_{x+1}& \quad \texttt{survival chain rule}\\ ~~~~~~~_{2}p_{x+1} = ~_{3}p_{x+1} / ~p_{x+3}& \quad \texttt{survival chain rule}\\ ~~~~~_{3}p_{x} = ~_{4}p_{x} / ~p_{x+3}& \quad \texttt{survival chain rule}\\ ~~~~~~~_{4}p_{x} = ~_{3}p_{x+1} * ~p_{x}& \quad \texttt{survival chain rule}\\ ~~~~~~~~~p_{x+3} = 1 - ~q_{x+3}& \quad \texttt{complement of mortality}\end{array}\end{split}\]

0.9751500000000001 0.97515

\[\begin{split}\begin{array}{llll} ~\texttt{Survival}~_{2}p_{x+1}~:=\\ ~~~_{2}p_{x+1} = ~_{3}p_{x} / ~p_{x}& \quad \texttt{survival chain rule}\\ ~~~~~_{3}p_{x} = ~_{4}p_{x} / ~p_{x+3}& \quad \texttt{survival chain rule}\\ ~~~~~~~_{4}p_{x} = ~_{3}p_{x+1} * ~p_{x}& \quad \texttt{survival chain rule}\\ ~~~~~~~~~p_{x+3} = 1 - ~q_{x+3}& \quad \texttt{complement of mortality}\end{array}\end{split}\]

0.9693877551020409 0.96939

\[\begin{split}\begin{array}{llll} ~\texttt{Survival}~_{3}p_{x}~:=\\ ~~~_{3}p_{x} = ~_{4}p_{x} / ~p_{x+3}& \quad \texttt{survival chain rule}\\ ~~~~~_{4}p_{x} = ~_{3}p_{x+1} * ~p_{x}& \quad \texttt{survival chain rule}\\ ~~~~~~~~~p_{x+3} = 1 - ~q_{x+3}& \quad \texttt{complement of mortality}\end{array}\end{split}\]

0.9596938775510204 0.95969

\[\begin{split}\begin{array}{llll} ~\texttt{Mortality}~_{{1|2}}q_{x}~:=\\ ~~~_{{1|2}}q_{x} = ~p_{x} - ~_{3}p_{x}& \quad \texttt{complement survival}\\ ~~~~~_{3}p_{x} = ~_{4}p_{x} / ~p_{x+3}& \quad \texttt{survival chain rule}\\ ~~~~~~~_{4}p_{x} = ~_{3}p_{x+1} * ~p_{x}& \quad \texttt{survival chain rule}\\ ~~~~~~~~~p_{x+3} = 1 - ~q_{x+3}& \quad \texttt{complement of mortality}\end{array}\end{split}\]

0.030306122448979567 0.03031

SOA Question 6.40

For a special fully discrete whole life insurance, you are given:

The death benefit is \(1000(1.03)^k\) for death in policy year k, for \(k = 1, 2, 3...\)
\(q_x = 0.05\)
\(i = 0.06\)
\(\ddot{a}_{x+1} = 7.00\)
The annual net premium for this insurance at issue age x is 110

Calculate the annual net premium for this insurance at issue age \(x + 1\).

print("SOA Question 6.40: (C) 116 ")
x = 0
life = Recursion().set_interest(i=0.06).set_a(7, x=x+1).set_q(0.05, x=x)
a = life.whole_life_annuity(x)
A = 110 * a / 1000
print(a, A)
life = Recursion().set_interest(i=0.06).set_A(A, x=x).set_q(0.05, x=x)
A1 = life.whole_life_insurance(x+1)
P = life.gross_premium(A=A1 / 1.03, a=7) * 1000
print(P)

SOA Question 6.40: (C) 116

\[\begin{split}\begin{array}{llll} ~\texttt{Whole Life Annuity}~\ddot{a}_{{x}}~:=\\ ~~~\ddot{a}_{{x}} = 1 + ~E_{x} * ~\ddot{a}_{{x+1}}& \quad \texttt{backward recursion}\\ ~~~~~E_{x} = ~p_{x} * v& \quad \texttt{pure endowment}\\ ~~~~~~~p_{x} = 1 - ~q_{x}& \quad \texttt{complement of mortality}\end{array}\end{split}\]

7.2735849056603765 0.8000943396226414

\[\begin{split}\begin{array}{llll} ~\texttt{Whole Life Insurance}~A_{{x+1}}~:=\\ ~~~A_{{x+1}} = [ ~A_{{x}} / v - ~q_{x} * b ] / ~p_{x}& \quad \texttt{forward recursion}\\ ~~~~~~~~~p_{x} = 1 - ~q_{x}& \quad \texttt{complement of mortality}\end{array}\end{split}\]

116.51945397474269

SOA Question 6.10 : (D) 0.91

For a fully discrete 3-year term insurance of 1000 on (x), you are given:

\(p_x\) = 0.975
i = 0.06
The actuarial present value of the death benefit is 152.85
The annual net premium is 56.05

Calculate \(p_{x+2}\).

print("SOA Question 6.10: (D) 0.91")
x = 0
life = Recursion(depth=5).set_interest(i=0.06)\
                         .set_p(0.975, x=x)\
                         .set_a(152.85/56.05, x=x, t=3)\
                         .set_A(152.85, x=x, t=3, b=1000)
p = life.p_x(x=x+2)
print(p)

SOA Question 6.10: (D) 0.91

\[\begin{split}\begin{array}{llll} ~\texttt{Survival}~p_{x+2}~:=\\ ~~~p_{x+2} = ~E_{x+2} /v& \quad \texttt{one-year pure endowment}\\ ~~~~~E_{x+2} = ~A_{{x+2:\overline{1|}}} - ~A^1_{{x+2:\overline{1|}}}& \quad \texttt{endowment insurance minus term}\\ ~~~~~~~A^1_{{x+2:\overline{1|}}} = [ ~A^1_{{x+1:\overline{2|}}} / v - ~q_{x+1} * b ] / ~p_{x+1}& \quad \texttt{forward recursion}\\ ~~~~~~~~~p_{x+1} = [ ~\ddot{a}_{{x+1:\overline{2|}}} - 1 ] / [ v * ~\ddot{a}_{{x+2:\overline{1|}}} ]& \quad \texttt{annuity recursion}\\ ~~~~~~~~~~~\ddot{a}_{{x+1:\overline{2|}}} = [ ~\ddot{a}_{{x:\overline{3|}}} - 1 ] / ~E_{x}& \quad \texttt{forward recursion}\\ ~~~~~~~~~A^1_{{x+1:\overline{2|}}} = [ ~A^1_{{x:\overline{3|}}} / v - ~q_{x} * b ] / ~p_{x}& \quad \texttt{forward recursion}\\ ~~~~~~~~~~~~~E_{x} = ~p_{x} * v& \quad \texttt{pure endowment}\end{array}\end{split}\]

0.9097382950525702

SOA Question 6.48

For a special fully discrete 5-year deferred 3-year term insurance of 100,000 on (x) you are given:

There are two premium payments, each equal to P . The first is paid at the beginning of the first year and the second is paid at the end of the 5-year deferral period
\(p_x = 0.95\)
\(q_{x + 5} = 0.02\)
\(q_{x + 6} = 0.03\)
\(q_{x + 7} = 0.04\)
\(i = 0.06\)

Calculate P using the equivalence principle.

print("SOA Question 6.48:  (A) 3195")
life = Recursion(depth=3).set_interest(i=0.06)
x = 0
life.set_p(0.95, x=x, t=5)
life.set_q(0.02, x=x+5)
life.set_q(0.03, x=x+6)
life.set_q(0.04, x=x+7)
a = 1 + life.E_x(x, t=5)
A = life.deferred_insurance(x, u=5, t=3)
P = life.gross_premium(A=A, a=a, benefit=100000)
print(P)

SOA Question 6.48:  (A) 3195

\[\begin{split}\begin{array}{llll} ~\texttt{Pure Endowment}~_{5}E_{x}~:=\\ ~~~_{5}E_{x} = ~_{5}p_{x} * v^{5}& \quad \texttt{pure endowment}\end{array}\end{split}\]

\[\begin{split}\begin{array}{llll} ~\texttt{Term Insurance}~A^1_{{x+5:\overline{3|}}}~:=\\ ~~~A^1_{{x+5:\overline{3|}}} = ~A_{{x+5:\overline{3|}}} - ~_{3}E_{x+5}& \quad \texttt{endowment insurance - pure}\\ ~~~~~_{3}E_{x+5} = ~E_{x+5} * ~_{2}E_{x+6}& \quad \texttt{pure endowment chain rule}\\ ~~~~~~~_{2}E_{x+6} = ~E_{x+6} * ~E_{x+7}& \quad \texttt{pure endowment chain rule}\\ ~~~~~~~~~E_{x+7} = ~p_{x+7} * v& \quad \texttt{pure endowment}\\ ~~~~~~~~~E_{x+6} = ~p_{x+6} * v& \quad \texttt{pure endowment}\\ ~~~~~~~E_{x+5} = ~p_{x+5} * v& \quad \texttt{pure endowment}\\ ~~~~~~~~~p_{x+7} = 1 - ~q_{x+7}& \quad \texttt{complement of mortality}\\ ~~~~~A^1_{{x+5:\overline{3|}}} = v * [ ~q_{x+5} * b + ~p_{x+5} * ~A^1_{{x+6:\overline{2|}}} ]& \quad \texttt{backward recursion}\\ ~~~~~~~A^1_{{x+6:\overline{2|}}} = v * [ ~q_{x+6} * b + ~p_{x+6} * ~A^1_{{x+7:\overline{1|}}} ]& \quad \texttt{backward recursion}\\ ~~~~~~~~~p_{x+6} = 1 - ~q_{x+6}& \quad \texttt{complement of mortality}\\ ~~~~~~~~~p_{x+5} = 1 - ~q_{x+5}& \quad \texttt{complement of mortality}\end{array}\end{split}\]

3195.118917658744

SOA Question 6.17

An insurance company sells special fully discrete two-year endowment insurance policies to smokers (S) and non-smokers (NS) age x. You are given:

The death benefit is 100,000; the maturity benefit is 30,000
The level annual premium for non-smoker policies is determined by the equivalence principle
The annual premium for smoker policies is twice the non-smoker annual premium
\(\mu^{NS}_{x+t} = 0.1.\quad t > 0\)
\(q^S_{x+k} = 1.5 q_{x+k}^{NS}\), for \(k = 0, 1\)
\(i = 0.08\)

Calculate the expected present value of the loss at issue random variable on a smoker policy.

print("SOA Question 6.17:  (A) -30000")
x = 0
life = ConstantForce(mu=0.1).set_interest(i=0.08)
A = life.endowment_insurance(x, t=2, b=100000, endowment=30000)
a = life.temporary_annuity(x, t=2)
P = life.gross_premium(a=a, A=A)
print(A, a, P)

life1 = Recursion().set_interest(i=0.08)\
                   .set_q(life.q_x(x, t=1) * 1.5, x=x, t=1)\
                   .set_q(life.q_x(x+1, t=1) * 1.5, x=x+1, t=1)
contract = Contract(premium=P * 2, benefit=100000, endowment=30000)
L = life1.gross_policy_value(x, t=0, n=2, contract=contract)
print(L)

SOA Question 6.17:  (A) -30000
37251.49857703495 1.8378124241073746 20269.478042694158

\[\begin{split}\begin{array}{llll} ~\texttt{Term Insurance}~A^1_{{x:\overline{2|}}}~:=\\ ~~~A^1_{{x:\overline{2|}}} = ~A_{{x:\overline{2|}}} - ~_{2}E_{x}& \quad \texttt{endowment insurance - pure}\\ ~~~~~_{2}E_{x} = ~_{2}p_{x} * v^{2}& \quad \texttt{pure endowment}\\ ~~~~~~~_{2}p_{x} = ~p_{x+1} * ~p_{x}& \quad \texttt{survival chain rule}\\ ~~~~~A^1_{{x:\overline{2|}}} = v * [ ~q_{x} * b + ~p_{x} * ~A^1_{{x+1:\overline{1|}}} ]& \quad \texttt{backward recursion}\\ ~~~~~~~p_{x+1} = 1 - ~q_{x+1}& \quad \texttt{complement of mortality}\\ ~~~~~~~~~E_{x} = ~p_{x} * v& \quad \texttt{pure endowment}\\ ~~~~~~~~~p_{x} = 1 - ~q_{x}& \quad \texttt{complement of mortality}\end{array}\end{split}\]

\[\begin{split}\begin{array}{llll} ~\texttt{Temporary Annuity}~\ddot{a}_{{x:\overline{2|}}}~:=\\ ~~~\ddot{a}_{{x:\overline{2|}}} = 1 + ~E_{x} * ~\ddot{a}_{{x+1:\overline{1|}}}& \quad \texttt{backward recursion}\\ ~~~~~E_{x} = ~p_{x} * v& \quad \texttt{pure endowment}\\ ~~~~~~~p_{x} = 1 - ~q_{x}& \quad \texttt{complement of mortality}\end{array}\end{split}\]

\[\begin{split}\begin{array}{llll} ~\texttt{Pure Endowment}~_{2}E_{x}~:=\\ ~~~_{2}E_{x} = ~_{2}p_{x} * v^{2}& \quad \texttt{pure endowment}\\ ~~~~~_{2}p_{x} = ~p_{x+1} * ~p_{x}& \quad \texttt{survival chain rule}\\ ~~~~~~~p_{x} = 1 - ~q_{x}& \quad \texttt{complement of mortality}\\ ~~~~~~~~~p_{x+1} = 1 - ~q_{x+1}& \quad \texttt{complement of mortality}\end{array}\end{split}\]

-30107.42633581115

SOA Question 2.5 : (B) 37.1

You are given the following:

\(e_{40:20} = 18\)
\(e_{60} = 25\)
\(_{20}q_{40} = 0.2\)
\(q_{40} = 0.003\)

Calculate \(e_{41}\).

hints:

solve for \(e_{40}\) from limited lifetime formula
compute \(e_{41}\) using backward recursion

print("SOA Question 2.5:  (B) 37.1")
life = Recursion(verbose=True).set_e(25, x=60, curtate=True)\
                              .set_q(0.2, x=40, t=20)\
                              .set_q(0.003, x=40)\
                              .set_e(18, x=40, t=20, curtate=True)
e = life.e_x(41, curtate=True)
print(e)

SOA Question 2.5:  (B) 37.1

\[\begin{split}\begin{array}{llll} ~\texttt{Lifetime}~e_{{x+41}}~:=\\ ~~~e_{{x+41}} = [ ~e_{{x+40}} - ~e_{{x+40:\overline{1|}}} ] / ~p_{x+40}& \quad \texttt{forward recursion}\\ ~~~~~e_{{x+40}} = ~e_{{x+40:\overline{20|}}} + ~_{20}p_{x+40} * ~e_{{x+60}}& \quad \texttt{backward recursion}\\ ~~~~~~~~~_{20}p_{x+40} = 1 - ~_{{20}}q_{x+40}& \quad \texttt{complement of mortality}\\ ~~~~~~~e_{{x+40:\overline{1|}}} = ~p_{x+40}& \quad \texttt{1-year curtate shortcut}\\ ~~~~~~~~~p_{x+40} = 1 - ~q_{x+40}& \quad \texttt{complement of mortality}\end{array}\end{split}\]

37.11434302908726