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:
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using MortalityTables, Plots
vbt2001MN = MortalityTables.table("2001 VBT Select and Ultimate - Male Nonsmoker, ANB")
vbt2001MN.select[35]
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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
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vbt2001MN.select[35][59]
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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
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vbt2001MN.select[35][60]
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0.00776
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vbt2001MN.select[25][60]
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0.00776
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vbt2001MN.ultimate[60]
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0.00776
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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")
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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
\(_{n}q_x = q_x + p_x*q_{(x+1)} + p_x*p_{(x+1)}*q_{(x+2)} + ...\)
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: \(_np_x = p_x*p_{(x+1)}*p_{(x+2)}*...*p_{(x+n)}\)
\[1 = _np_x + _nq_x \rightarrow _nq_x = 1 - _np_x\]