Life Insurance
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# Life Insurance

## Notation Summary

$${p}_{x}$$ - probability of living from age $x$ to age $x+1$

$q_{x}$ - probability of death between ages $x$ and $x+1$

$$_{n}q_{x}$$ - probability of death between ages $x$ and $x+n$ (1 year is implied if $n$ is missing)

$1 = p_{x} + q_{x}$

${A}_{x}$ - Present value of $1 of life insurance for a life age x$\ddot{a}_{x}$- Present value of a$1 life-contingent annuity due

${a}_{x}$ - Present value of a $1 life-contingent immediate annuity ## Introduction to Life Insurance ### Interest Theory With interest rates rising, banks are advertising Certificates of Deposites which earn 2% interest ($i$) for a year. The future value of$c$dollars invested today would be $$FV = c*(1+i)^{n}$$ Likewise, the present value can be calculated by discounting using $$v^{n} = \frac{1}{(1+i)^{n}} = (1+i)^{-n}$$ So the present value of a 1.00 dollar payment in 1 year is worth $$PV = FV*v = 1*(1+i)^{-1} = 0.98039...$$ ### Incorporating probability Suppose that you will only receive this dollar if you survive the year to claim it with probability$p$. If we assume that this 1 year survival$_{1}p$is 99% then that present value decreases to $$PV = FV * v^{1} * _{1}p = 0.98039 * 0.99 = 0.97059...$$ The payment of an amount conditioned upon the survival to time period$n$for a life age$x$is known as a pure endowment and is represented in actuarial notation as $$_{n}E_{x} = _{1}p_{x}v^{n}$$ Because the compliment of surviving one year is the probability of dying within 1 year ($_{1}q$) then $$_{1}p + _{1}q = 1$$ Then the present value of one dollar payable at the end of 1 year | death would be $$_1q*v^1 = (1-_1p)v^{1} = (1-0.99)*0.98039...= 0.0098039...$$ ## Life contingencies The previous example of a 1-year term insurance issued to an individual at age$x$$$_1q_x*v^{1} = A_x^{1} = 0.0098039$$ This is also commonly referred to as Annual Renewable Term (ART) or Yearly Renewable Term (YRT) Once the term length$n$extends beyond the first year, the probability of surviving to the next period needs to be accounted for as shown below for a 2-year term insurance $$A_x^{2} = _1q_x*v^1+_1p_x*_1q_{(x+1)}*v^2$$ The preceeding subscript for mortality and survival may be left off if it is only for one year $$A_x^{2} = q_x*v^1+p_x*q_{(x+1)}*v^2$$ This formula can be extended for any duration$n$$$A_x^{n} = q_x*v^1 + p_x*q_{(x+1)}*v^2 + ...p_{(x+n-1)}*q_{(x+n)}*v^n$$ If this coverage was issued to a 35 year old, then it would be represented as $$A_{35}^{n} = q_{35}*v^1 + p_{35}*q_{(35+1)}*v^2 + ...p_{(35+n-1)}*q_{(35+n)}*v^n$$ It is important to note that in the above example the 2nd year mortality probability was noted as$q_{(35+1)}$rather than$q_{36}$because the mortality of a 36 year old may be different than the mortality of a 35 year old who survived for 1-year. Before insurance is issued, the policyholder generally has to pass certain underwriting requirements (eg medical exams); therefore, it would be expected that the mortality of a 50 year-old$(q_{50})$who just went through underwriting would be less than that of a 50 year old who went through underwriting 20 years ago$(q_{30+20})$. This period of lower mortality is referred to as the select period. Because this reduced mortality deminishes (ie$q_{35} \leq q_{34+1} \leq q_{33+2}$) eventually there may be an ultimate rate used for a given attained-age where the age at issue does not matter; for example, if$q_{0+80} = q_{1+79} = q_{2+78} = … = q_{80}$then that mortality is the ultimate rate. ### Valuation Basic Tables Various mortality tables can be found on https://mort.soa.org/ and can be read across and then down. For example, if using the 2001 Valuation Basic Table (VBT) Select and Ultimate Table - Male Nonsmoker, then mortality for a non-smoking male issued to someone age 35 would follow: 1 2 3 using MortalityTables, Plots vbt2001MN = MortalityTables.table("2001 VBT Select and Ultimate - Male Nonsmoker, ANB") vbt2001MN.select  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 86-element OffsetArray(::Vector{Float64}, 35:120) with eltype Float64 with indices 35:120: 0.00031 0.00041 0.00052 0.00063 0.00073 0.00084 0.00094 0.00104 0.00115 0.0013 0.00149 0.00172 0.00198 ⋮ 0.53905 0.57031 0.60339 0.63838 0.67541 0.71458 0.75603 0.79988 0.84627 0.89536 0.94729 1.0  If issue-age was 35 and duration is now 25 (with attained age between 59 and 60) mortality would be 1 vbt2001MN.select  1 0.00668  Afterward the ultimate rate would apply which is evident because the mortality for a 60 year old issued at age 35 is the same as the mortality for a 60 year old issued at age 25 1 vbt2001MN.select  1 0.00776  1 vbt2001MN.select  1 0.00776  1 vbt2001MN.ultimate  1 0.00776  1 2 3 4 5 6 7 8 9 10 11 vbt2001FN = MortalityTables.table("2001 VBT Select and Ultimate - Female Nonsmoker, ANB") issue_age = 10 mort = [ vbt2001MN.select[issue_age][issue_age:end], vbt2001FN.select[issue_age][issue_age:end], ] plot( mort, label = ["Male Non-smoker" "Female Non-smoker"], title = "Comparison of 2001 VBT S&U \n Mortality for a 10 Year-Old", xlabel="duration")  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 issue_ages = 10:45 durations = 1:30 # compute the relative rates with the element-wise division ("brodcasting" in Julia) function rel_diff(a, b, issue_age,duration) att_age = issue_age + duration - 1 return a[issue_age][att_age] / b[issue_age][att_age] end diff = [rel_diff(vbt2001MN.select,vbt2001FN.select,ia,dur) for ia in issue_ages, dur in durations] contour(durations, issue_ages, diff, xlabel="Duration",ylabel="Issue Ages", title="Relative difference between Male and Female Mortality \n 2001 VBT Non-smoker ANB", fill=true )  The chart illustrates that male mortality is significantly higher than females for approximately ages 16-35 for non-smoking idividuals according to the 2001 VBT. Like previous examples, these rates are for death in only a single year (ie$_1q_x$), but we could also represent mortality as the probability of death in the next$n$years $$_{n}q_x = q_x + p_x*q_{(x+1)} + p_x*p_{(x+1)}*q_{(x+2)} + ...$$ Probability of death in the next n years = probability of dying in the next year + probability of living 1 year and dying the following year + probability of surviving this and the following year and dying in the third + ... Reminder the probability of survival and death are mutually exclusive events; therefore, it is often easier to calculate the probability of surviving n years ($_np_x\$) and calculate the compliment: $$_np_x = p_x*p_{(x+1)}*p_{(x+2)}*...*p_{(x+n)}$$

$1 = _np_x + _nq_x \rightarrow _nq_x = 1 - _np_x$