Contribution Refunds on Withdrawal

Why Withdrawal Refunds Need Two Calculations

A contribution refund can look simple: accumulate the member’s contributions with credited interest, then pay the resulting amount when the member withdraws. In a pension plan, however, the benefit may depend on completed service, and withdrawal competes with other exits. The valuation therefore has two distinct parts: determine what is payable at each withdrawal time, then determine the probability and present value of that payment.

This tutorial treats hypothetical plans with specified contribution amounts, interest credits, service rules, and exit probabilities. The formulas are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. The focus is a cash refund triggered by withdrawal; the benefit payable on another cause of exit must be determined separately under that cause’s terms.

Definition: A contribution refund on withdrawal is a payment made when a member withdraws from a plan, usually based on the member’s own contributions accumulated with credited interest. A plan may restrict eligibility by service or may pay only a specified percentage of the accumulated amount. A refund is not automatically payable on every exit: the plan’s stated trigger and service conditions determine whether it is due.

Calculate the Refund Before Valuing It

Let $C_k$ be a member contribution made at the beginning of year $k$, and let $j$ be the effective annual rate credited to the account. If the account starts at $B_0=0$, no withdrawals occur before the year in question, and each contribution is credited interest for that year, the account balance at the end of year $k$ is

B_k=(B_{k-1}+C_k)(1+j).

This timing matters. A contribution at the beginning of the year earns a full year of interest; a contribution at the end of the year does not. A refund may also exclude employer contributions, apply a service-based percentage, or be limited to the member’s own account. Represent these terms explicitly rather than assuming the balance is automatically payable.

Definition: If $s_k$ is completed service at withdrawal and $\alpha(s_k)$ is the fraction of the account payable under the plan’s service rule, the refund at the end of year $k$ is $R_k=\alpha(s_k)B_k$. If the member is ineligible, set $\alpha(s_k)=0$. If the rule pays the full account after a service threshold, $\alpha(s_k)$ is zero below that threshold and one at or above it.

The balance recursion is appropriate only when the account receives the stated annual effective credit, contributions occur at the stated times, and there are no intervening account adjustments. If contributions are made at year-end, or interest is credited on a different basis, change the accumulation accordingly. The service fraction is applied to the balance specified by the plan; it should not be applied to employer contributions if the rule refunds only member contributions.

Worked Example: Accumulating Contributions Under a Service Rule

A hypothetical plan receives a member contribution of $900 at the beginning of each year and credits 5% effective annual interest. The member withdraws at the end of year 3. The plan returns the full member account after at least one year of service and makes no refund below that threshold. Find the withdrawal refund and its expected present value if the unconditional probabilities of withdrawal at the ends of years 1, 2, and 3 are 0.10, 0.096, and 0.09, respectively. Use a 4% effective annual discount rate.

Method. First accumulate the account with contributions at the beginning of each year. The service condition is met at each of the three stated withdrawal times, so the refund fraction is one.

At the end of year 1,

$$ B_1=(0+900)(1.05)=945. $$

At the end of year 2, the new contribution is added before the year’s interest is credited:

B_2=(945+900)(1.05)=1{,}937.25.

At the end of year 3,

B_3=(1{,}937.25+900)(1.05)=2{,}979.1125.

Thus the refund on a year-3 withdrawal is $2,979.1125, or $2,979.11 to the nearest cent. For the expected present value, use the unconditional probability of withdrawal in each year and discount each end-of-year payment at 4%:

\begin{aligned} \operatorname{EPV} &=\frac{0.10(945)}{1.04} +\frac{0.096(1{,}937.25)}{1.04^2} +\frac{0.09(2{,}979.1125)}{1.04^3}\\ &=90.8653846\ldots+171.9452681\ldots+238.3578\ldots\\ &=501.1685\ldots \end{aligned}

The expected present value is $501.17, rounded to the nearest cent. The method avoids treating the year-2 balance as if it were $1,932.25: multiplying $1{,}845$ by $1.05$ gives $1,937.25. Using that corrected balance also gives the correct year-3 account and present value.

Withdrawal Among Competing Exits

A withdrawal rate in a multiple-decrement model is commonly conditional on the member being active at the start of the period. It is not, by itself, the probability of a withdrawal from the original group. For an annual model, let $q_{w,k}$ be the probability of withdrawal during year $k$, conditional on being active at the start of that year. Let $q_{o,k}$ be the corresponding conditional probability of any other exit, and let $p_k=1-q_{w,k}-q_{o,k}$ be the probability of remaining active through the year. Assume these are mutually exclusive exits, with no re-entry, and that a withdrawal refund is paid at the end of the year of withdrawal.

Theorem: Under the annual competing-exit assumptions above, the probability of withdrawal in year $k$, measured from the original active group, is $$ \Pr(\text{withdrawal in year }k) =\left(\prod_{j=1}^{k-1}p_j\right)q_{w,k}. $$ If $R_k$ is the refund paid at the end of year $k$, the expected present value at time 0, at effective annual discount rate $i$, is $$ \operatorname{EPV} =\sum_{k=1}^{n}v^k \left(\prod_{j=1}^{k-1}p_j\right)q_{w,k}R_k, \qquad v=\frac{1}{1+i}. $$

The probability formula follows by requiring the member to remain active through every earlier year, then withdraw during year $k$. Remaining active through years $1,\ldots,k-1$ has probability $\prod_{j=1}^{k-1}p_j$; conditional on that event, withdrawal in year $k$ has probability $q_{w,k}$. Multiplying gives the unconditional probability of that cause and year. The present-value formula then sums each possible payment multiplied by its probability and discount factor. It applies to end-of-year payments under the stated annual model; different payment timing requires a different discount exponent.

If probabilities are already given as unconditional probabilities from time 0, use them directly in the EPV sum. Do not multiply those probabilities by prior survival again. Conversely, if the data give conditional rates, prior active probabilities must be included.

Worked Example: Valuing a Service-Based Refund Schedule

A hypothetical plan pays a refund of $300 for a withdrawal at the end of year 1, $1,200 at the end of year 2, and $1,800 at the end of year 3. These amounts already incorporate the plan’s contribution and service provisions. The unconditional probabilities of withdrawal in those years are 0.20, 0.13, and 0.0845. Find the expected present value at time 0 using a 5% effective annual interest rate.

Method. Since the probabilities are stated unconditionally, multiply each directly by its year-specific refund and discount to time 0. The fact that other exits compete with withdrawal is already reflected in these unconditional probabilities.

\begin{aligned} \operatorname{EPV} &=\frac{0.20(300)}{1.05} +\frac{0.13(1{,}200)}{1.05^2} +\frac{0.0845(1{,}800)}{1.05^3}\\ &=57.142857\ldots+141.496598\ldots+131.389698\ldots\\ &=330.02915\ldots \end{aligned}

The expected present value is $330.03, rounded to the nearest cent. The common mistake is to use the correct first term but miscalculate the second and third discounted terms. For example, $156/1.1025=141.496598\ldots$, not approximately $141.72; each division should be checked before rounding the total.

Worked Example: Conditional Mean Refund Given Withdrawal

A hypothetical plan deposits $500 at the beginning of each year and credits 4% effective annual interest. A full account refund is paid at the end of the year of withdrawal. In year 1, conditional on being active at the start of the year, withdrawal and other-exit probabilities are 0.10 and 0.20. In year 2, the corresponding probabilities are 0.20 and 0.10. Given that withdrawal occurs by the end of year 2, find the expected refund without discounting.

Method. Calculate both account balances, convert the conditional withdrawal rates to probabilities from time 0, and divide the probability-weighted refunds by the probability of withdrawal by year 2.

The account balances at the two possible withdrawal times are

B_1=500(1.04)=520,\qquad B_2=(520+500)(1.04)=1{,}060.8.

The unconditional probability of withdrawal in year 1 is 0.10. The probability of remaining active through year 1 is $1-0.10-0.20=0.70$, so the unconditional probability of withdrawal in year 2 is $0.70(0.20)=0.14$. Therefore, the probability of withdrawal by the end of year 2 is $0.10+0.14=0.24$. The conditional expected refund is

\begin{aligned} E[R\mid\text{withdrawal by year 2}] &=\frac{0.10(520)+0.14(1{,}060.8)}{0.24}\\ &=\frac{52+148.512}{0.24}\\ &=835.4666\ldots \end{aligned}

The answer is $835.47, rounded to the nearest cent. A common error is to use 0.20 as the unconditional year-2 withdrawal probability. That is the rate conditional on being active at the start of year 2; only 0.70 of the original group reaches that point, so the year-2 probability from time 0 is 0.14.

Service Rules, Timing, and Exam Traps

A service rule changes the amount payable, not the logic of competing-risk probabilities. For example, if a full refund is payable only after two completed years, a withdrawal before that point may have $R_k=0$, while later withdrawals may receive the accumulated account. In the EPV formula, use the actual amount $R_k$ for each year, including zero when the member is ineligible. Do not assign the full account to a member who has not satisfied the stated condition.

The most reliable sequence is to determine the refund schedule before calculating its value:

Fix the payment timing.
Identify when contributions enter the account, when interest is credited, and when a refund is paid.

Calculate the account and apply the plan rule.
Find the account balance at each possible withdrawal time, then apply service eligibility and the payable fraction.

Use the right probability basis.
Convert conditional withdrawal rates to unconditional probabilities if needed, accounting for earlier competing exits.

Discount each possible payment.
Match the exponent to the payment time and sum probability-weighted present values.

Keep three quantities distinct: the account balance, the amount payable under the service rule, and the present value of the contingent payment. They are generally not equal. Also check that annual exit probabilities are nonnegative and sum to no more than one in each year, and that the total probabilities of mutually exclusive withdrawal years are no greater than one.

Key takeaway: A withdrawal refund is valued by pairing the plan’s year-specific payable amount with the probability of withdrawal in that year. Conditional decrement rates require prior active probabilities; unconditional cause-and-time probabilities do not.

Check Your Understanding

For each problem, state the relevant timing and apply the service rule before valuing the refund.

A hypothetical plan deposits $600 at the beginning of each year for three years and credits 2% effective annual interest. A full refund is payable on withdrawal at the end of year 3. Find the refund to the nearest cent. (Answer: $1,872.96)
A member contributes $400 at the beginning of each year. The account earns 3% effective annual interest, and the plan pays 80% of the account on withdrawal after two completed years. Find the refund for withdrawal at the end of year 2, to the nearest cent. (Answer: $669.09)
Conditional on being active at the start of each year, withdrawal and other-exit probabilities are 0.08 and 0.12 in year 1, then 0.15 and 0.10 in year 2. Refunds are $700 and $1,100 at the respective year-ends. At 5% effective annual interest, find the expected present value to the nearest cent. (Answer: $173.06)
A plan deposits $500 at the beginning of each year and credits 4% effective annual interest. A full account refund is paid at the end of either of the first two years. Given year-1 conditional withdrawal and other-exit probabilities of 0.10 and 0.20, and year-2 probabilities of 0.20 and 0.10, find the expected refund without discounting, conditional on withdrawal by the end of year 2. (Answer: $835.47)
Conditional withdrawal and other-exit probabilities are 0.10 and 0.10 in each of years 1, 2, and 3. A service rule pays no refund before two completed years, $900 for a year-2 withdrawal, and $1,300 for a year-3 withdrawal. At 4% effective annual interest, calculate the expected present value from time 0, to the nearest cent. (Answer: $140.53)

What You'll Learn