Investment-Path Effects on Linked Product Guarantees

Why Investment Timing Matters

A linked contract’s account value is built up over time, not determined by an average return alone. The order and size of returns affect compounding, and contract features can make intermediate account values matter directly. A guarantee may lock in an anniversary value; a charge may be deducted after each year’s investment return; or a lapse rule may end coverage when the account falls below a threshold.

The first distinction is between a contract whose benefit depends only on the terminal account value and one whose benefit also depends on the path. If there are no intermediate-dependent features, two return paths with the same product of annual accumulation factors lead to the same terminal account value. A guarantee based only on that terminal value then produces the same payment. But equal arithmetic average returns do not ensure equal terminal account values, and path-dependent features can produce different outcomes even when terminal values match.

Definition: An investment path is the ordered sequence of investment returns and account values during a projection. A product feature is path-dependent when its payment, account value, or continuation status depends on one or more intermediate values or on the order in which returns occur, rather than only on the terminal account value.

For the hypothetical annual models below, $A_t$ denotes the account value at the end of year $t$, after any stated year-end charge. Unless specified otherwise, returns are effective annual returns, cash flows occur at the stated year-ends, and there are no deaths, withdrawals, or lapses. These are scenario calculations, not estimates of actual product experience.

Projecting the Account and the Guarantee

With annual return $r_t$ and a fixed monetary charge $c_t$ deducted at the end of the year, the account recursion is

A_t=A_{t-1}(1+r_t)-c_t.

The order in this equation matters: the opening account first earns the year’s return, then the charge is deducted. A different contract could specify a charge at the start of the year, a percentage-of-account charge, or another deduction basis; its recursion must reflect that contract wording.

For a maturity guarantee $G$ that tops up the account value to at least $G$, the payment is $\max(A_n,G)$. The insurer’s top-up is $\max(G-A_n,0)$. For a ratchet guarantee that locks the highest anniversary value, the payment instead depends on the maximum of the account values observed over the term.

Key distinction: A terminal-value guarantee responds to $A_n$. A ratchet guarantee, anniversary trigger, or lapse rule may respond to intermediate account values as well. Do not use a terminal-value-only calculation when the contract specifies an intermediate-value feature.

Worked Example: Equal Average Returns, Different Guarantee Costs

A hypothetical two-year linked contract starts with an account of $100. There are no charges or lapses. At maturity, the payment is the greater of the account value and $108. Compare annual returns of $+30\%$ then $-20\%$ with returns of $+5\%$ then $+5\%$. The effective annual discount rate for the insurer’s top-up is $5\%$.

Method. Project each account by multiplying by each year’s accumulation factor, then apply the maturity floor. The arithmetic average return in both scenarios is $5\%$, because $(30\%-20\%)/2=5\%$ and $(5\%+5\%)/2=5\%$.

For the first path,

A_1=100(1.30)=130,\qquad A_2=130(0.80)=104.

The maturity payment is $\max(104,108)=108$, so the insurer’s top-up is $108-104=4$. Its value at issue is

\frac{4}{(1.05)^2}=\frac{4}{1.1025}=3.6281\ldots\approx \$3.63.

For the second path,

A_1=100(1.05)=105,\qquad A_2=105(1.05)=110.25.

The maturity payment is $\max(110.25,108)=110.25$, so the top-up is zero. Thus the same arithmetic average return produces different terminal accounts and different guarantee costs. The first path’s loss after a strong first year illustrates why averaging the annual returns and applying that average as if it were earned each year is incorrect.

Check. The two actual accumulation factors are $1.30(0.80)=1.04$ and $1.05(1.05)=1.1025$. Applied to $100, these give $104 and $110.25 as projected. The guarantee payments are at least $108 in both cases.

Why a Ratchet Makes the Order Matter

A return permutation leaves the product of accumulation factors unchanged: multiplication is commutative. Without charges or other intermediate-dependent features, the same returns in a different order therefore give the same terminal account. But a ratchet can preserve an earlier high account value as a guaranteed benefit, so the sequence can still change the amount payable.

This follows directly from the ratchet definition. Let $H_n$ be the largest account value observed at the initial date or an anniversary, and let the maturity payment be $\max(A_n,H_n)$. The account $A_n$ can be identical across two paths while $H_n$ differs. The resulting payment can therefore differ even though the unguaranteed terminal account does not.

Worked Example: Same Terminal Account, Different Ratchet Benefits

A hypothetical contract starts with $100, has no charges or lapses, and pays at year 2 the greater of the year-2 account and the highest account value at the initial date or either anniversary. Compare returns of $+30\%$ then $-20\%$ with returns of $-20\%$ then $+30\%$. The effective annual discount rate is $5\%$.

Method. Project both account paths, record the maximum account value on each path, and discount the resulting year-2 payment. For the $+30\%,-20\%$ path,

A_1=100(1.30)=130,\qquad A_2=130(0.80)=104,\qquad H_2=\max(100,130,104)=130.

The maturity payment is $\max(104,130)=130$. Its value at the end of year 1, one year before payment, is $130/1.05=123.8095\ldots$, or $123.81.

For the $-20\%,+30\%$ path,

A_1=100(0.80)=80,\qquad A_2=80(1.30)=104,\qquad H_2=\max(100,80,104)=104.

The maturity payment is $\max(104,104)=104$. Its value at the end of year 1 is $104/1.05=99.0476\ldots$, or $99.05. In both scenarios the terminal account is $104, but the ratchet benefit differs by $26 and the year-1 present values differ by $24.76.

Check. Both return products equal $1.30(0.80)=0.80(1.30)=1.04$, confirming the shared terminal account of $104. The first path reaches $130 at an anniversary, whereas the second never exceeds $104. The calculation avoids the mistake of valuing a ratchet as though it were only a terminal-value floor.

Charges, Lapses, and the Timing of Payments

Fixed monetary charges can make account values depend on the order of returns, even if two paths have the same compounded return before charges. In the recursion $A_t=A_{t-1}(1+r_t)-c_t$, a charge deducted early reduces the amount exposed to later investment returns. A charge deducted later does not reduce the account during earlier years. Percentage charges applied in the same multiplicative way each year may behave differently; the contract’s specified basis and timing must be modeled rather than assumed.

Lapse introduces a second timing effect. If a policy lapses at an anniversary, a surrender payment may be made then and future premiums, charges, and guarantees may cease. A projection that continues the account through later returns after a specified lapse trigger includes cash flows that should not occur.

Worked Example: Fixed Charges and a Lapse Trigger

A hypothetical linked contract starts with $100. A fixed $4 charge is deducted at each year-end after investment return. At the end of year 1, the contract lapses if the account is below $90; on lapse, the account value is paid immediately with no additional surrender charge. If the contract remains in force, its year-2 payment is the greater of the account and $100. Compare returns of $+20\%$ then $-10\%$ with returns of $-10\%$ then $+20\%$. Use a $5\%$ effective annual discount rate.

Method. Apply each annual return before the $4 charge, then apply the lapse test at the end of year 1. For the positive-first path,

A_1=100(1.20)-4=116,\qquad A_2=116(0.90)-4=100.40.

Because $116$ is not below $90, the contract stays in force. The year-2 payment is $\max(100.40,100)=100.40$. Its value at issue is $100.40/(1.05)^2=91.0657\ldots$, or $91.07.

For the negative-first path,

$$ A_1=100(0.90)-4=86. $$

Since $86<90$, the contract lapses at the end of year 1 and pays $86 then. The year-2 return and guarantee do not apply. The value at issue of the lapse payment is $86/1.05=81.9047\ldots$, or $81.90.

Without the lapse rule, continuing the second path would give $86(1.20)-4=99.20$ at year 2, but that is not the contract outcome because the policy has already lapsed. Both paths have the same return product, $1.20(0.90)=0.90(1.20)=1.08$. The different results arise from the fixed charges’ timing and, decisively, the lapse trigger.

Check. The first-year accounts are $116 and $86 after charges, so only the second path breaches the lapse threshold. The resulting cash flows occur at different times: $100.40 at year 2 versus $86 at year 1. The calculation avoids carrying a lapsed contract forward or discounting both payments from the same date.

A Reliable Comparison Workflow

For an integrated scenario comparison, keep the account projection, contractual benefit, and valuation date distinct. The following sequence prevents common timing and guarantee errors.

State the contract mechanics.
Identify when returns are credited, how and when charges are deducted, the guarantee basis, lapse conditions, and payment dates.

Project each path in order.
Calculate account values year by year; do not replace the sequence with its arithmetic average.

Apply path-dependent rules as they occur.
Record ratchets, triggers, and lapses at the specified dates. Once a lapse ends the contract, do not project later contractual benefits.

Determine the actual cash flows.
Apply terminal floors, top-ups, or surrender payments only after identifying which benefits remain payable.

Value from a common date.
Discount each cash flow using the stated effective rate and its actual payment time, then compare the resulting values.

For a fixed scenario with a single payment $B_t$ at time $t$, discounted at an effective annual rate $i$, its value at time 0 is $B_t/(1+i)^t$. This present-value calculation describes the scenario’s discounted payment; it is not by itself an expected liability across uncertain scenarios. If probabilities are supplied, scenario values must be combined using those probabilities and the valuation assumptions specified for the problem.

Common Exam Traps

Substituting an arithmetic average for compounding. A sequence of returns accumulates through the product of the annual factors, not through applying the arithmetic mean once per year.
Assuming order never matters. A return permutation preserves the product of factors only when intermediate features and timing effects do not alter the projection. Ratchets, fixed charges, and lapse rules can make order consequential.
Applying the wrong guarantee definition. A floor on the final account is not interchangeable with a highest-anniversary-value ratchet.
Projecting after lapse. Apply the lapse test at its specified date before projecting future contractual cash flows.
Discounting from the wrong date. A year-1 surrender value is discounted one year; a year-2 maturity benefit is discounted two years to time 0.

The central diagnostic is to ask what information the benefit formula uses. If it uses only the terminal account, equal terminal accounts imply equal benefits under that formula. If it uses a maximum, threshold crossing, charge history, or continuation status, the entire ordered path may be needed. This distinction is useful when comparing linked guarantees, explaining changes in projected reserves, and checking whether a scenario result is consistent with the contract mechanics.

Check Your Understanding

For each problem, project the account or benefit using the specified timing before comparing paths.

A hypothetical contract begins with $200, has no charges, and guarantees at least $200 at maturity after two years. Returns are $+10\%$ and $-10\%$. Find the terminal account and guarantee top-up. (Answer: terminal account $198; top-up $2.)
A ratchet contract starts with $80 and pays the greater of the terminal account and the highest account value at the initial date or either anniversary. Compare returns $+25\%,-20\%$ with returns $-20\%,+25\%$. (Answer: both terminal accounts are $80; payments are $100 and $80, respectively.)
A contract starts with $150 and deducts $3 at each year-end after investment return. Compare returns $-12\%,+15\%$ and $+15\%,-12\%$, with no lapse or guarantee. (Answer: terminal accounts are $145.35 and $146.16, respectively.)
A contract starts with $100 and deducts $2 after each year’s return. It lapses at the end of year 1 if its account is below $95; on lapse it pays the account value without an extra charge. Otherwise, at year 2 it pays the greater of account value and $100. Compare returns $-8\%,+20\%$ with $+20\%,-8\%$. (Answer: the first path lapses and pays $90 at year 1; the second pays $106.56 at year 2.)
A two-year contract has an account value of $102 at maturity and a guaranteed maturity payment of at least $105. There are no other benefits. At an effective annual discount rate of $4\%$, find the insurer’s time-0 value of the guarantee top-up. (Answer: $3/(1.04)^2=\$2.77$, rounded to the nearest cent.)

What You'll Learn