# In spedygiorgio/lifecontingencies: Financial and Actuarial Mathematics for Life Contingencies

knitr::opts_chunkset(warning=F, message=F, fig.width=6, fig.height=3.5, fig.align='center',fig.show='hold')  # Function PensFund In this brief document we present a general function to valuate actuarial liabilities and normal costs, allowing different methods in pension funding for defined benefit schemes. Each method is defined by the way that contributions are accrued in the active age before the specific risk apply. Following [@winklevoss1993pension] and [@bowers1997actuarial], the Normal Cost and Actuarial Liability can be seen as portions ofPVFNC^{r}_x$, the present value of future benefit at age$x$for the risk in$r$, and that's the way that we treat this in the function's code. The methods that are implemented here are the "Benefit Prorate Method, Constant Dollar", "Benefit Prorate Method, Constant Percent", "Cost Prorate Method, Constant Dollar" and "Cost Prorate Method, Constant Benefit" (the last two also called "entry age methods"). For formula details the reader can go to chapters 5 and 6 on [@winklevoss1993pension]. We define a general function that allows to perform the calculations in a multi-decrement environment, with flexibility to consider different kind of benefit scheme, using functions of the "lifecontingencies" package (Spedicato et. al., 2018). The arguments (following general notation in Gian Paolo Clemente's vignette titled "Pension Funding with lifecontingencies") are: •$x$: the age of the insured at the time of valuation •$y$: the age which the person entered to the job. This age is usually different than the age that the plan began •$r$: age where the risk is evaluated •$acttableAccPeriod$: multi-decrement object lifetable, where one of its decrements is the risk of interest. One of the most common applications is for and obligated retirement at age$r$, and even in this case is needed a multi-decrement table with risks of death and retirement (with certainty for age$r$) •$decrement$: decrement to evaluate •$i$: interest rate (real) •$j$: salary rate increment (real, productivity and seniority included) •$delta$: benefit rate increment (real) •$n$: years that the benefit will be payed (can be 1 for single payments). If it is omitted, will take value$\omega-r$•$avg$: number of past salaries to be considered for average •$acttablePaymPeriod$: lifetable object for payment period •$CostMet$: Method Cost, that could be "BPM_CD" (Benefit Prorate Method, Constant Dollar), "BPM_CP" (Benefit Prorate Method, Constant Percent), "CPM_CD" (Cost Prorate Method, Constant Dollar), "CPM_CB" (Cost Prorate Method, Constant Benefit). The general function is: require("lifecontingencies") PensFund <-function(x, y, r, acttableAccPeriod, decrement, i, j, delta, n, avg, acttablePaymPeriod, CostMet){ if (missing(n)) n <- getOmega(acttableAccPeriod)-x-1 if (x<y) stop("Entry age greater than actual age") if (missing(decrement)) stop("Which is the contingency of benefit?") if (class(acttableAccPeriod) != "mdt") stop("Error! Needed Mdt") if (!(decrement %in% getDecrements(acttableAccPeriod))) stop("Error! Not recognized decrement type") if (missing(acttableAccPeriod)) stop("Error! Need a lifetable or actuarialtable") if (missing(x) | missing(y)) stop("Error! Need age and seniority!") if (x > getOmega(acttableAccPeriod)) { out = 0 return(out)} if (missing(CostMet)) CostMet = 'BPM_CD' if (missing(avg)) stop("Error! Need avg period of computing benefit") if (missing(j)) stop("Error! Need average salary increase rate") if (any(x < 0, y < 0, avg < 0, delta < 0)) stop("Error! Negative parameters") ##########survive all casuses acttableAccPeriodTotal = pxt(acttableAccPeriod, x = 0:(getOmega(acttableAccPeriod)), t=1) acttableAccPeriodTotal = probs2lifetable(acttableAccPeriodTotal, type = "px") if (missing(acttablePaymPeriod)) acttablePaymPeriod = acttableAccPeriodTotal ########## Set benefit # fix age valuation x0=x B = ifelse(avg==0, 1, (1+j)^(r-x0) * annuity(j, avg, m=0, k=1, type = "immediate") / avg ) q.r = qxt(object=acttableAccPeriod, x=r, t=1, decrement=decrement) Br = q.r * axn(acttablePaymPeriod, r, i = (1+i)/(1+delta)-1, n=n, k=1)*B ########## Actuarial liability in age x, fot the age r (r>=x) ALyNC.x.r = function(x, r){ ## PVFB PVFB.x = Exn(acttableAccPeriodTotal, x, r-x, i=i) * Br ## Normal Costs and liabilities under Actuarial Cost Methods if(CostMet == 'BPM_CD'){ K <- (x-y) / (r-y) k <- 1 / (r-y)} if(CostMet == 'BPM_CP'){ K <- (1+j)^(x-x0) * annuity(j, x-y+1, m=0, k=1, type = "immediate") / ( (1+j)^(r-x0) * annuity(j, r-y+1, m=0, k=1, type = "immediate") ) k <- (1+j)^(x-x0) / ( annuity(j, r-y, m=0, k=1, type = "immediate") * (1+j)^(r-x0))} if(CostMet == 'CPM_CD'){ K <- axn(acttableAccPeriodTotal, y, i = i, n=x-y, k=1) / axn(acttableAccPeriodTotal, y, i = i, n=r-y, k=1) k <- Exn(acttableAccPeriodTotal, y, x-y, i=i) / axn(acttableAccPeriodTotal, y, i = i, n = r-y, k = 1) } if(CostMet == 'CPM_CB'){ K <- axn(acttableAccPeriodTotal, y, i = (1+i)/(1+j)-1, n=x-y, k=1) / axn(acttableAccPeriodTotal, y, i = (1+i)/(1+j)-1, n=r-y, k=1) k <- Exn(acttableAccPeriodTotal, y, x-y, i=i) * (1+j)^(x-y) / axn(acttableAccPeriodTotal, y, i = (1+i)/(1+j)-1, n=r-y, k=1) } ## Results outAL.x.r <- PVFB.x * K outNC.x.r <- PVFB.x * k outPVFB.x <- PVFB.x out.x.r <- c(outAL.x.r, outNC.x.r, outPVFB.x) return(out.x.r) } ######## AL y NC for the rest of ages ages_risk = x:r outAL.x = sapply(ages_risk, function(x) ALyNC.x.r(x, r)[1]) outNC.x = sapply(ages_risk, function(x) ALyNC.x.r(x, r)[2]) outPVFB.x = sapply(ages_risk, function(x) ALyNC.x.r(x, r)[3]) outNC.x[length(ages_risk)] = NA out.x <- data.frame(x = ages_risk, AL = outAL.x, NC = outNC.x, PVFB = outPVFB.x) return(out.x) }  ## First (main) example A first example of application is of a person aged 30 (entered to job at age 20), with a benefit in case of disability, that is the average of the last 5 salaries before get disabled. For calculate that we'll use the multi-decrement table 'SoAISTdata' for the accrued period and 'soa08' for the payment period, both available in "lifecontingencies" package. data(SoAISTdata) data(soa08) MultiDecrTable <- new("mdt", name="testMDT", table=SoAISTdata)  The specific name of this risk in the table is 'inability'. getDecrements(MultiDecrTable)  Now we calculate the liability and normal cost with the four methods, and$i = .04$,$j = .06$,$delta = .03$,$avg = 5$. In this case for the age risk 53 ($r = 53$). x1=30 y1=20 r1=53 decrement1 = "inability" BPM_CD = PensFund(x1, y1, r1, acttableAccPeriod = MultiDecrTable, decrement = decrement1, i = .04, j = .06, delta = .03, avg = 5, acttablePaymPeriod = soa08, CostMet = 'BPM_CD') BPM_CP = PensFund(x1, y1, r1, acttableAccPeriod = MultiDecrTable, decrement = decrement1, i = .04, j = .06, delta = .03, avg = 5, acttablePaymPeriod = soa08, CostMet = 'BPM_CP') CPM_CD = PensFund(x1, y1, r1, acttableAccPeriod = MultiDecrTable, decrement = decrement1, i = .04, j = .06, delta = .03, avg = 5, acttablePaymPeriod = soa08, CostMet = 'CPM_CD') CPM_CP = PensFund(x1, y1, r1, acttableAccPeriod = MultiDecrTable, decrement = decrement1, i = .04, j = .06, delta = .03, avg = 5, acttablePaymPeriod = soa08, CostMet = 'CPM_CB')  The way that each method amortize the benefit cost in active years can be seen in the next graph, showing the liability accumulated relationship between at any age: "CPM_CD"$\geqslant$"CPM_CB"$\geqslant$"BPM_CD"$\geqslant$"BPM_CP". par(mfrow=c(1,2)) plot(x1:r1, BPM_CD$AL, t="o", ylab="ALx", xlab="Age", cex = .7)
points(x1:r1, BPM_CP$AL, t="o", col=2, cex = .7) points(x1:r1, CPM_CD$AL, t="o", col=3, cex = .7)
points(x1:r1, CPM_CP$AL, t="o", col=4, cex = .7) title ("Actuarial liability. Example 1", cex.main=.9) legend("topleft", legend = c("BPM_CD", "BPM_CP", "CPM_CD", "CPM_CB"), lty = rep(1,4), col = 1:4, cex=.6, bty = "n") plot(x1:r1, BPM_CP$NC, t="o", ylab="NCx", xlab="Age", cex = .7)
points(x1:r1, BPM_CD$NC, t="o", col=2, cex = .7) points(x1:r1, CPM_CD$NC, t="o", col=3, cex = .7)
points(x1:r1, CPM_CP$NC, t="o", col=4, cex = .7) title ("Normal Cost. Example 1", cex.main=.9) legend("topleft", legend = c("BPM_CP", "BPM_CD", "CPM_CD", "CPM_CB"), lty = rep(1,4), col = 1:4, cex=.6, bty = "n")  We can check the results with actuarial equivalences, specifically the formulas$AL^{r}_x = PVFB^{r}_x-PVFNC^{r}_x$(6.3a) and$AL^{r}_x = AVPNC^{r}_x$(6.4a) in [@winklevoss1993pension], which means that the actuarial liability at any time can be expressed as the difference between the present value of future benefit and the present value of future normal costs, and as the actuarial value of past normal costs (retrospective view). # AL = PVBF - PVFNC # first get survivial for all causes acttableAccPeriodTotal = pxt(MultiDecrTable, x= 0:(getOmega(MultiDecrTable)), t=1) acttableAccPeriodTotal = probs2lifetable(acttableAccPeriodTotal, type = "px") # are equal? round(BPM_CD$AL[1],10) ==  round(BPM_CD$PVFB[1] - sum(sapply(x1:(r1-1), function(x) Exn(acttableAccPeriodTotal, x1, x-x1, i=.04) * BPM_CD$NC[x-x1+1])),10)

# Restrospective AL
# First get the liability at entry age y
BPM_CD_y = PensFund(y1, y1, r1, acttableAccPeriod = MultiDecrTable, decrement = decrement1,i = .04, j = .06, delta = .03, avg = 5, acttablePaymPeriod = soa08, CostMet = 'BPM_CD')
# are equal?
round(BPM_CD_y$AL[BPM_CD_y$x==45],10) == round(sum(sapply(y1:44, function(y) (1.04)^(45-y) / pxt(acttableAccPeriodTotal, y, 45-y) * BPM_CD_y$NC[BPM_CD_y$x==y])),10)


## Second example

Now we'll evaluate a retirement pension, that receives a payment grade which is a lineal function between age 60 and the maximum possible age of retirement at 70, and goes from 50% to 100%. This is not a grading function in which benefit provided at early or late retirement has the same present value as the benefit payable at normal retirement ([@winklevoss1993pension], chapter 9). Instead, here the person is penalyzed for early retirement. The insured has an annual salary of 5000 at age $x$. Also there is a seniority condition of 10 years. The other parameters are the same that the previous example.

x2=50
y2=35
decrement2 = "retirement"
snty_req = 20
s_x = 5000
Cost_Method2 = "BPM_CD"


The calculation now focus on estimate the actuarial values at age $x2$, for all the retirement scheme (all ages between $x2$ and 70).

AL_xr = data.frame(r=0, AL=0, NC=0)
for(r in x2:70) {
if ((r-y2) < snty_req)
AL_xr[r-x2+1,2:3]=c(0,0)
else {
AL_xr[r-x2+1,2:3] = s_x * pay_grade[pay_grade$x==r,2] * PensFund(x2, y2, r, acttableAccPeriod = MultiDecrTable, decrement = decrement2, i = .04, j = .06, delta = .03, avg = 5, acttablePaymPeriod = soa08, CostMet = Cost_Method2)[1,2:3] } AL_xr[r-x2+1,1] = r }  The total liability at age x is r round(sum(AL_xr[,2]),1) and is distributed between this retirement options, influenced by the progressive function of the early retirement pay_grade) and the individual's expected behavior (probability of retirement). par(mar=c(6, 4, 4, 6) + 0.1) par(mfrow=c(1,1)) plot(AL_xr$r, AL_xr$AL, t="o", ylab="$", xlab="Retirement Age")
title ("Actuarial liability in age x=50, for retirement in age r", cex.main=.9)
par(new=TRUE)
plot(AL_xr$r, qxt(object=MultiDecrTable, x=x2:70, t=1, decrement=decrement2), col=2, ylim = c(0,1), t="o", xlab="", ylab="", axes=FALSE) axis(side=4, at=seq(0, 1, .2)) mtext("Probability", side = 4, line = 3) legend('topleft', legend=c('ALx', 'q(retire)'), col=1:2, lty=c(1,1), bty = 'n', cex = .6)  ## Third example Now let's imagine that the same insured has the benefit of receive a 50% of his/her last salary at each age before retirement if he decide to leave the Company. The benefit is in concept of recognition of past contributions to the pension plan. This benefit can be estimated setting$n=1$(no rent, a single payment) and$avg=0$(only last salary). For example, for the risk of withdrawal at age 45: decrement3 = "withdrawal" x3 = 30 y3 = 20 r3 = 45 BPM_CP3 = s_x * .5 * PensFund(x3, y3, r3, acttableAccPeriod = MultiDecrTable, decrement = decrement3, i = .04, j = .06, delta = .03, avg = 0, n=1, acttablePaymPeriod = soa08, CostMet = 'BPM_CP')  par(mfrow=c(1,2)) plot(x3:r3, BPM_CP3$AL, t="o", ylab="ALx", xlab="Age")
title ("Actuarial liability. Example 3", cex.main=.9)
legend("topleft", legend = c("BPM_CD"), lty = 1, col = 1, cex=.6, bty = "n")
plot(x3:r3, BPM_CP3\$NC, t="o", ylab="NCx", xlab="Age")
title ("Normal Cost. Example 3", cex.main=.9)
legend("topleft", legend = c("BPM_CD"),
lty = 1, col = 1, cex=.6, bty = "n")


## Future improvements:

The following improvements are planned:

1. Because a payment can't be done before the contingency occurs, an assumption about occurrence and payment must be done. We will improve the function to consider payments at half of the year. This affects only the financial component, and little bit the final results.
2. Make the benefit rent payable for a designed survivor too, and with a portion of the titular benefit. This also allows to include death as a main benefit risk.

# References

spedygiorgio/lifecontingencies documentation built on March 21, 2021, 5:36 a.m.