Why Premium Timing Matters
A premium credited to a policy account does not earn the same interest regardless of when it is paid. A beginning-of-year premium can earn interest for the full policy year; a middle-of-year premium earns interest for only part of that year; and an end-of-year premium earns none during that year. Even if the total premiums are identical, their timing can produce different projected account values.
This tutorial considers a hypothetical account with a stated effective annual credited rate. To isolate timing, assume there are no premium loads, policy charges, withdrawals, benefit payments, or other cash flows. Each premium shown is the amount actually credited to the account. These are simplifying assumptions for the calculations, not a description of every product.
The fractional-year assumption is important: a middle-of-year premium receives half a year of accumulation at the equivalent effective rate, not a full year and not automatically half of the annual interest amount. For example, the half-year factor is \((1+i)^{1/2}\), which is generally not \(1+i/2\).
One-Year Account Projection
Write \(A_{\text{end}}\) for the value at the end of the policy year. If premiums \(P_0\), \(P_{1/2}\), and \(P_1\) are paid at the beginning, middle, and end of the year, respectively, then
The starting account value and beginning premium are together invested for a full year. The middle premium is invested for half a year. The end premium is included in the year-end account value, but does not earn interest in that year. If no premium is paid at one of these times, set its amount to zero.
Worked Example: Comparing Three Premium Dates
A hypothetical account starts the year with $1,000 and earns a 8% effective annual credited rate. A $600 premium is credited either at the beginning, middle, or end of the year. Find the year-end account value in each case, and compare the interest earned on the premium. Assume the whole premium is credited and there are no other transactions.
Method. Apply the accumulation factor for the time the premium is in the account. The initial $1,000 earns a full year of interest in every case:
Beginning of year. The premium earns a full year of interest:
The premium itself grows from $600 to \(600(1.08)=648\), so it earns $48 of interest. The account value is $1,728.
Middle of year. The half-year accumulation factor is \((1.08)^{1/2}=1.0392304845\). Therefore,
The premium earns \(623.5382907-600=23.5382907\) dollars of interest. The account value is $1,703.54.
End of year. The premium receives no interest during the year:
The account value is $1,680, and the premium earns no interest in that policy year. Rounded to the nearest cent, the three values are $1,728.00, $1,703.54, and $1,680.00. The ordering is sensible: under a positive credited rate, the earlier premium dates produce higher values. The common mistake this comparison avoids is applying a full-year factor to every premium, regardless of its payment date.
Projecting Several Premiums
For multiple payments in one year, accumulate each payment from its own date to year-end, then add the accumulated amounts to the starting account value. The same timing principle can be applied successively across policy years. An end-of-year premium enters the account at that year-end; if it remains there, it is part of the starting balance for the next year and earns interest during that next year.
Worked Example: Different Premium Dates Across Two Years
A hypothetical account starts with $1,200. The effective annual credited rate is 4%. In year 1, premiums of $300, $200, and $100 are paid at the beginning, middle, and end of the year, respectively. In year 2, $150 is paid at the beginning and $250 at the middle; there is no end-of-year-2 premium. Find the account value at the end of year 2. Assume all stated premiums are credited in full and there are no other transactions.
Method. First project to the end of year 1. The middle-year accumulation factor is \((1.04)^{1/2}=1.0198039027\). Thus,
The beginning balance plus beginning premium grows to \(1500(1.04)=1560\). The middle premium grows to \(200(1.0198039027)=203.9607805\). Adding the end premium gives
This is the balance entering the year-2 projection. For year 2, add the $150 beginning premium before applying the full-year factor, then add the accumulated middle premium:
Rounded to the nearest cent, the end-of-year-2 account value is $2,349.47. The check is that the $100 paid at the end of year 1 is included in \(A_1\), and so is included in the amount that earns interest during year 2. Omitting it from the year-2 starting balance would understate the projection.
Why the Timing Factors Hold
The factors follow from the assumed accumulation rule. Under an effective annual rate \(i\), an amount invested for one full year grows by \(1+i\). More generally, the model assumes that an amount invested for \(s\) years grows by \((1+i)^s\). A premium paid \(t\) years after the start of a policy year remains invested for \(1-t\) years before year-end, so it accumulates by \((1+i)^{1-t}\).
Measure \(t\) in years from the start of the policy year. For the beginning, middle, and end, \(t=0\), \(t=1/2\), and \(t=1\).
The time from payment to year-end is \(1-t\) years.
Multiply the premium by \((1+i)^{1-t}\), then add all accumulated payments and the accumulated starting balance.
For example, when \(t=1/2\), the premium is invested for \(1-1/2=1/2\) year and its factor is \((1+i)^{1/2}\). This derivation also indicates the limits of the formula: it depends on a constant effective annual rate with the stated fractional-year accumulation convention. If a contract specifies a different crediting schedule or a different convention for partial years, use that contract’s rule instead.
Solving for an Unknown Premium and Checking Feasibility
The one-year equation can be rearranged to solve for an unknown premium. For a middle-of-year premium \(Q\), with all other cash flows known,
where \(A_{\text{known}}\) is the year-end value from the starting account and all known premiums. The algebra produces a signed amount. If premiums must be nonnegative, first check whether the target is at least \(A_{\text{known}}\). When \(A_{\text{target}}<A_{\text{known}}\), the equation requires a negative amount: a positive premium cannot reduce the account value.
Worked Example: Finding a Feasible Middle-of-Year Premium
A hypothetical account starts the year with $1,600, earns 5% effective annually, receives $200 at the beginning of the year and $100 at the end, and is to have $2,100 at year-end. Find the nonnegative middle-of-year premium required. Assume no other transactions and that the full premium is credited.
Method. First calculate the year-end value without the unknown premium:
The middle premium accumulates by \((1.05)^{1/2}=1.0246950766\). The target requires an additional \(2100-1990=110\) dollars at year-end, so
The required premium is $107.35 to the nearest cent. Substitution checks the result: the unrounded premium accumulates to $110, and \(1990+110=2100\). The common mistake avoided here is dividing by the annual factor \(1.05\); the premium is paid halfway through the year, so its correct factor is the half-year factor.
Worked Example: A Target That No Nonnegative Premium Can Reach
A hypothetical account has $2,000 at the start of year 1 and earns 5% effective annually. A $300 premium is paid at the end of year 1. What nonnegative premium paid halfway through year 2 would produce an account value of $2,500 at the end of year 2? Assume there are no other transactions and the full premiums are credited.
Method. Project the known amounts to the end of year 2 before considering the unknown premium. The initial account grows for two years, while the end-of-year-1 premium grows for one year:
The known account value already exceeds the $2,500 target by $20. A nonnegative middle-of-year-2 premium would add a nonnegative amount at year-end, so no such premium can reach the target. If a signed amount were allowed, the algebraic solution would be
Thus, the signed solution is approximately \(-$19.52\), which represents a withdrawal rather than a premium. The check is immediate: it reduces the known $2,520 by $20 at year-end. The exam trap is to solve mechanically and report a positive premium without checking the sign of the required amount or comparing the target with the known account value.
Common Timing Errors
- Giving every premium a full year of interest. Only a beginning-of-year payment receives the full-year factor under this model. A middle payment receives \((1+i)^{1/2}\), and an end payment receives \(1\).
- Using \(1+i/2\) for a half-year. With fractional-year compound accumulation, the factor is \((1+i)^{1/2}\). These expressions are generally different.
- Excluding an end-of-year premium from the year-end account. It earns no interest during that year, but it is still credited to the account at year-end.
- Ignoring feasibility when solving backward. If the known projected balance is already above the target, a nonnegative premium cannot solve the equation.
- Changing the time unit without changing the rate convention. The formulas here use years and an effective annual rate. A different crediting schedule requires its specified accumulation rule.
Check Your Understanding
For each hypothetical account, assume the stated effective annual rate applies consistently over fractional years, all premiums are credited in full, and there are no other transactions. Round dollar answers to the nearest cent.
- An account starts with $500, earns 6% effective annually, and receives a $100 premium halfway through the year. Find the year-end value. (Answer: $632.96)
- An account starts with $900, earns 4% effective annually, and receives $250 at the beginning and $80 at the end of the year. Find the year-end value. (Answer: $1,276.00)
- An account starts with $1,000 and earns 4% effective annually. During the year it receives $200 at the beginning, $100 halfway through, and $50 at the end. Find the year-end value. (Answer: $1,399.98)
- An account starts with $1,000, earns 4% effective annually, and receives $50 at year-end. What middle-of-year premium is needed to reach $1,200? (Answer: $107.86)
- An account starts year 1 with $1,000 and earns 10% effective annually. It receives $100 at the beginning of year 1, $50 at the end of year 1, and $100 halfway through year 2. Find the value at the end of year 2. (Answer: $1,490.88)