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Exam ALTAM · Valuation Under Competing Risks · Tutorial 359 of 999

Worked Retirement and Withdrawal Valuation

Learn to convert annual competing-decrement assumptions into retirement and withdrawal cash-flow values and combine them consistently at the valuation date.

Intermediate 10 min read

What You'll Learn

  • Calculate the probability an employee remains active at the start of each year.
  • Convert conditional retirement and withdrawal probabilities into unconditional exit probabilities.
  • Value a retirement pension whose first installment is payable at retirement.
  • Discount withdrawal benefits paid at the end of the exit year.
  • Add cause-specific present values without double-counting or omitting exits.
  • Check the effects of payment timing and conditional probabilities.

Introduction

A pension valuation with retirement and withdrawal decrements requires two related calculations: the probability that a member is still active when each year begins, and the benefit value associated with each type of exit during that year. The year-specific exit probabilities are conditional on being active, so they cannot be used as unconditional probabilities without first allowing for earlier exits.

This tutorial develops a discrete annual model and works through hypothetical pension designs. The focus is on valuing retirement and withdrawal benefits on a consistent time basis, using the competing-decrement methods aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: In year \(k\), let \(q_{r,k}\) be the probability of retirement and \(q_{w,k}\) the probability of withdrawal, each conditional on being active at the start of that year. Assume these are the only exits in the model and that \(q_{r,k}+q_{w,k}\leq 1\). The probability of remaining active through year \(k\), conditional on being active at its start, is \(p_k=1-q_{r,k}-q_{w,k}\).

Let \(A_k\) denote the probability that a member is active at the start of year \(k\), with \(A_1=1\). Then the unconditional probability of retirement during year \(k\) is \(A_kq_{r,k}\), and the unconditional probability of withdrawal during that year is \(A_kq_{w,k}\). The active probability for the following year is \(A_{k+1}=A_kp_k\).

$$ A_k=\prod_{j=1}^{k-1}(1-q_{r,j}-q_{w,j}),\qquad P(\text{retire in year }k)=A_kq_{r,k},\qquad P(\text{withdraw in year }k)=A_kq_{w,k}. $$

For annual effective interest rate \(i\), use \(v=1/(1+i)\). If an exit benefit \(B_k\) is paid at the end of year \(k\), its expected present value contribution is the unconditional probability of that exit multiplied by \(v^kB_k\). A retirement pension may have a different payment pattern. If a fixed pension of \(R\) is payable at retirement and then once each year for a total of \(n\) guaranteed installments, its value at the retirement date is \(R\ddot{a}_{\overline{n}|}\), where \(\ddot{a}_{\overline{n}|}=\sum_{j=0}^{n-1}v^j\). Its time-zero contribution for retirement in year \(k\) is \(A_kq_{r,k}v^kR\ddot{a}_{\overline{n}|}\).

These formulas assume annual decrement probabilities apply over the stated years and that exit benefits are paid at year-end. The retirement annuity formula also assumes its installments are guaranteed. A pension contingent on survival after retirement requires an appropriate post-retirement survival model; it is not valued as a guaranteed annuity.

Why the Cause-Specific Formula Works

The member can receive a retirement benefit in year \(k\) only if the member is active at the start of that year and then retires during it. Those events have probability \(A_kq_{r,k}\). Once retirement in year \(k\) has occurred, the associated pension value is measured at time \(k\), so discounting that value to time zero multiplies it by \(v^k\). The same reasoning applies to a withdrawal benefit, using its own conditional probability and benefit amount.

The annual events “retire in year \(k\)” and “withdraw in year \(k\)” are mutually exclusive. Across years, the possible exit years are also mutually exclusive. Therefore, adding all cause-year contributions gives the expected present value without overlap.

$$ \operatorname{EPV} =\sum_{k=1}^{m} A_kv^k \left(q_{r,k}R\ddot{a}_{\overline{n}|}+q_{w,k}W_k\right), $$

Here \(m\) is the last year in which active exits are modeled, and \(W_k\) is the withdrawal benefit payable at the end of year \(k\). This expression assumes the same fixed retirement pension applies to retirement in each modeled year. If retirement benefits vary by year, replace \(R\) with the relevant year-specific pension amount.

Worked Valuation

Worked Example: A Three-Year Plan With Level Decrements

A hypothetical plan has a three-year valuation period. At the start of each year, an active member has a 10% probability of retirement and a 10% probability of withdrawal; otherwise the member remains active. The effective annual interest rate is 5%. Retirement provides two guaranteed annual pension installments of $3,000, with the first paid immediately at retirement. Withdrawal provides a $1,000 lump sum at the end of the withdrawal year. Calculate the combined time-zero EPV.

Method. First calculate active probabilities and unconditional exit probabilities. Then value the retirement pension at the retirement date and discount each cause-year benefit to time zero.

The probability of remaining active each year is \(1-0.10-0.10=0.80\). Thus the probabilities active at the start of years 1, 2, and 3 are \(1\), \(0.80\), and \(0.80^2=0.64\).

Year \(k\)Active at start \(A_k\)Retirement probability \(A_kq_{r,k}\)Withdrawal probability \(A_kq_{w,k}\)
11.000.100.10
20.800.080.08
30.640.0640.064

At \(i=0.05\), \(v=1/1.05=20/21\), \(v^2=400/441\), and \(v^3=8000/9261\). The value at retirement of the two-installment pension is

$$ B_r=3000\ddot{a}_{\overline{2}|} =3000(1+v) =3000\left(1+\frac{20}{21}\right) =\$5{,}857.142857\ldots. $$

The discounted probability weight for retirement is \(0.10v+0.08v^2+0.064v^3 =0.0952380952+0.0725623583+0.0552856064 =0.2230860599\). Therefore the retirement EPV is \(0.2230860599(5{,}857.142857\ldots)=\$1{,}306.646922\), or $1,306.65 to the nearest cent.

The withdrawal benefit is paid at the end of the exit year, so its EPV is \(1000(0.10v+0.08v^2+0.064v^3)\). The three contributions are $95.238095, $72.562358, and $55.285606, totaling $223.086060, or $223.09 to the nearest cent.

Adding the cause totals gives an EPV of \(\$1{,}306.646922+\$223.086060=\$1{,}529.732982\), or $1,529.73 to the nearest cent. The retirement and withdrawal probability masses in each year sum to \(0.20\), \(0.16\), and \(0.128\), respectively, matching 20% of the active probability at each year’s start.

Common mistake avoided. Using 10% as the unconditional year-2 retirement probability would ignore the 20% who had already exited in year 1. The correct year-2 probability is \(0.80(0.10)=0.08\).

Worked Example: Changing Decrement Rates and a Three-Payment Pension

A hypothetical plan is valued for two years at an effective annual interest rate of 4%. In year 1, conditional on being active, retirement and withdrawal probabilities are 12% and 8%. In year 2 they are 15% and 10%. Retirement provides three guaranteed annual installments of $2,400, the first payable immediately at retirement. A withdrawal benefit of $700 is paid at the end of year 1 and $1,100 at the end of year 2. Find the total EPV.

Active and exit probabilities. At the start of year 1, \(A_1=1\). The probability of remaining active through year 1 is \(1-0.12-0.08=0.80\), so \(A_2=0.80\). The unconditional cause probabilities are therefore 0.12 retirement and 0.08 withdrawal in year 1, and \(0.80(0.15)=0.12\) retirement and \(0.80(0.10)=0.08\) withdrawal in year 2.

Here \(v=1/1.04=25/26\) and \(v^2=625/676\). The pension value at retirement is \(2400(1+v+v^2)=2400(2.8860946746)=\$6{,}926.627219\). The discounted retirement probability weight is \(0.12v+0.12v^2=0.12(0.9615384615+0.9245562130)=0.2263313609\). Thus the retirement EPV is \(6{,}926.627219(0.2263313609)=\$1{,}567.712965\).

The withdrawal EPV is \(0.08(700)v+0.08(1{,}100)v^2\). The year-1 contribution is \(56(25/26)=\$53.846154\). The year-2 contribution is \(88(625/676)=\$81.360947\). Their total is $135.207101.

The combined EPV is \(\$1{,}567.712965+\$135.207101=\$1{,}702.920066\), or $1,702.92 to the nearest cent. The cause probabilities for year 2 sum to 0.20, the total decrement probability conditional on being active in that year.

Common mistake avoided. The year-2 retirement probability is not 15% of the original group. It is 15% of the 80% still active, or 12% of the original group.

Worked Example: Service-Linked Withdrawal Benefits

Consider a hypothetical three-year plan with an effective annual interest rate of 5%. The conditional retirement and withdrawal probabilities are 5% and 15% in year 1, 10% and 20% in year 2, and 20% and 10% in year 3. A member who retires receives two guaranteed annual pension installments of $1,500, with the first payable immediately at retirement. A member who withdraws receives $600 for each completed year of service, paid at the end of the withdrawal year. Calculate the total EPV.

Step 1: Active probabilities and exits. The probability of remaining active is 0.80 in year 1 and 0.70 in year 2. Hence \(A_1=1\), \(A_2=0.80\), and \(A_3=0.80(0.70)=0.56\). The unconditional retirement probabilities by year are \(0.05\), \(0.80(0.10)=0.08\), and \(0.56(0.20)=0.112\). The withdrawal probabilities are \(0.15\), \(0.80(0.20)=0.16\), and \(0.56(0.10)=0.056\).

Step 2: Retirement value. With \(v=20/21\), the two-installment pension is worth \(B_r=1500(1+v)=\$2{,}928.571429\) at retirement. Also \(v^2=400/441\) and \(v^3=8000/9261\). The discounted retirement probability weight is

$$ 0.05v+0.08v^2+0.112v^3 =0.0476190476+0.0725623583+0.0967498111 =0.2169312170. $$

The retirement EPV is \(0.2169312170(2{,}928.571429)=\$635.298564\).

Step 3: Withdrawal value. The withdrawal benefits are $600, $1,200, and $1,800 at the ends of years 1, 2, and 3. Their EPV contributions are \(0.15(600)v=\$85.714286\), \(0.16(1{,}200)v^2=\$174.149660\), and \(0.056(1{,}800)v^3=\$87.074830\). The withdrawal EPV is therefore $346.938776.

Step 4: Combine and check. The total is \(\$635.298564+\$346.938776=\$982.237340\), or $982.24 to the nearest cent. In year 3, the active probability is 0.56 and total conditional exit probability is 0.30, so the total unconditional year-3 exit probability is \(0.56(0.30)=0.168\), equal to \(0.112+0.056\).

Common mistake avoided. Applying the year-3 rates directly to the original group would give 20% retirement and 10% withdrawal. The correct probabilities are 11.2% and 5.6%, because only 56% of the original group remains active at the start of year 3.

Timing and Valuation Checks

The most important timing decision is whether the first retirement installment is paid at the retirement date or one year later. The examples use an annuity-due value, \(\ddot{a}_{\overline{n}|}\), because the first installment is paid immediately upon retirement. If the first installment is one year after retirement and there are \(n\) installments, the value at retirement is instead \(R a_{\overline{n}|}=R\sum_{j=1}^{n}v^j\). The entire pension value is then discounted from the retirement date to time zero.

A reliable valuation keeps three questions separate:

1
Who is exposed to the decrement?
Use the active probability \(A_k\) at the start of the year.
2
Which exit occurs?
Multiply \(A_k\) by the conditional probability for that cause, not by the total decrement probability or by a rate from another year.
3
When is the benefit paid?
Discount a year-end lump sum by \(v^k\); value a retirement pension at retirement using its correct payment pattern, then discount that value by \(v^k\).

These checks matter because a valid total exit probability does not by itself give the correct benefit value: retirement and withdrawal can have different amounts and payment patterns. Calculate their EPVs separately, then add them. Do not apply retirement pension payments to withdrawals or treat a withdrawal refund as payable at retirement.

Key takeaway: For each year, multiply the probability active at its start by the conditional probability of the specific exit cause. Discount the resulting cause-specific benefit using its actual payment timing, and sum across causes and years.

Check Your Understanding

For each hypothetical case, identify the annual active probabilities and use the stated payment timing. Round money to the nearest cent.

  1. A plan has year-1 retirement and withdrawal probabilities of 10% and 20%, and year-2 probabilities of 20% and 10%, conditional on being active. Find the year-2 active probability and the unconditional year-2 probability of each exit cause. (Answer: \(A_2=0.70\); retirement \(=0.14\), withdrawal \(=0.07\).)
  2. At 5% interest, year-1 conditional retirement and withdrawal probabilities are 10% and 10%; in year 2 they are 20% and 10%. Retirement pays a $4,000 lump sum at the end of the exit year, and withdrawal pays $1,000 at the end of that year. Find the total EPV over two years. (Answer: $1,129.25.)
  3. A two-year plan has zero interest. In year 1 the conditional retirement and withdrawal probabilities are 10% and 20%; in year 2 they are 25% and 15%. Retirement pays three guaranteed installments of $1,000, beginning immediately at retirement. Withdrawal pays $400 at the end of year 1 or $600 at the end of year 2. Find the combined EPV. (Answer: $968.00.)
  4. At 5% interest, year-1 conditional retirement and withdrawal probabilities are 8% and 12%; year-2 probabilities are 15% and 10%. Retirement pays two guaranteed installments of $2,500 beginning immediately at retirement. Withdrawal pays $700 at the end of year 1 and $1,000 at the end of year 2. Find the combined EPV. (Answer: $1,055.70.)
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