Why a Recurrence Is Useful
A compound reversionary bonus is built one declaration at a time. Each declared addition increases the benefit used in the next calculation, so the amount added in a later year can depend on earlier declarations. A recurrence makes that dependency explicit: it specifies how to move from the benefit after one declaration to the benefit after the next.
The recurrence also gives a compact way to check a longer projection. Once the rule for one year is clear, repeated application produces a product of growth factors. This is useful when declaration rates vary by year, or when a calculation must be checked for accidental simple accumulation.
This is a stated calculation convention, not a claim that every product defines its bonus base in this way. The starting base and the basis on which a particular scale is declared must be specified by the product terms. Here, the rate is applied to the current total benefit, which is what makes the accumulation compound.
Deriving the Year-by-Year Recurrence
The declaration adds a proportion \(r_k\) of the benefit already in force immediately before the declaration. Therefore the new total is the old total plus that addition:
The factor \(1+r_k\) is the growth factor for declaration \(k\). It multiplies the whole benefit carried into that declaration, not just the original base. Applying the recurrence successively gives the product form:
The product formula follows by substitution, not by assuming that all the rates are equal. For example, \(B_2=B_1(1+r_2)\); replacing \(B_1\) with \(B_0(1+r_1)\) gives \(B_2=B_0(1+r_1)(1+r_2)\). Repeating this substitution introduces one factor for each declaration.
Identify \(B_0\), the benefit immediately before the first declaration, in the units used by the problem.
For a percentage, write it as a decimal. For a scale stated as dollars per thousand, divide the scale by 1,000 to obtain the corresponding rate.
For year \(k\), calculate \(C_k=r_kB_{k-1}\), then \(B_k=B_{k-1}+C_k\). Carry the unrounded amount to the next year.
Compute \(B_0\prod_{k=1}^{n}(1+r_k)\) independently and compare it with the year-by-year result, allowing only for stated final rounding.
When the same rate \(r\) applies at every declaration, the product reduces to \(B_n=B_0(1+r)^n\). When the rates differ, replacing them with one rate or adding them is generally not equivalent: the correct factors remain separate.
Worked Example: Three Different Annual Rates
A hypothetical participating policy has a benefit base of $80,000 immediately before its first annual declaration. The declared compound reversionary bonus rates for the next three anniversaries are 2%, 3%, and 4%, respectively. Find the total benefit after the third declaration. Assume each rate applies to the total benefit immediately before that year's declaration.
Method. Use the recurrence in declaration order. Convert the rates to \(0.02\), \(0.03\), and \(0.04\), and do not round intermediate amounts.
After the first declaration, the addition is \(0.02(80{,}000)=1{,}600\), so \(B_1=80{,}000+1{,}600=81{,}600\).
For the second declaration, the addition is \(0.03(81{,}600)=2{,}448\), so \(B_2=81{,}600+2{,}448=84{,}048\).
For the third declaration, the addition is \(0.04(84{,}048)=3{,}361.92\), so \(B_3=84{,}048+3{,}361.92=87{,}409.92\).
Product check. The growth-factor calculation is \(80{,}000(1.02)(1.03)(1.04)=80{,}000(1.092624)=87{,}409.92\). It agrees with the recurrence. The total additions are \(1{,}600+2{,}448+3{,}361.92=7{,}409.92\), also equal to \(87{,}409.92-80{,}000\). The answer is $87,409.92.
Reading the Product and the Addition Amounts
The product \(B_0\prod_{k=1}^{n}(1+r_k)\) gives the total benefit after all declarations. It does not, by itself, give the addition made at any one declaration. The year-\(k\) addition is the difference between consecutive totals, or equivalently \(r_kB_{k-1}\). Keeping those quantities distinct prevents a common reporting error: giving a total benefit when asked for the latest addition, or giving the sum of additions when asked for the final benefit.
A rate quoted as a dollar amount per thousand is a proportion of the relevant benefit base. If the scale states \(q_k\) dollars per $1,000, its rate for this recurrence is \(r_k=q_k/1{,}000\). For instance, $20 per $1,000 corresponds to \(0.02\), or 2%. The multiplication can be performed either by converting to a rate or by dividing the current benefit by 1,000 and multiplying by the scale. Both methods must use the same current benefit.
Worked Example: Converting a Per-Thousand Scale
A hypothetical policy has a benefit base of $125,000 before its first declaration. Its compound bonus scale is $18 per $1,000 in year 1, $20 per $1,000 in year 2, and $22 per $1,000 in year 3. Each scale applies to the total benefit immediately before the declaration. Calculate the final total benefit.
Method. Convert the three scale amounts to rates: \(18/1{,}000=0.018\), \(20/1{,}000=0.020\), and \(22/1{,}000=0.022\).
Year 1's addition is \(0.018(125{,}000)=2{,}250\), giving \(B_1=127{,}250\).
Year 2's addition is \(0.020(127{,}250)=2{,}545\), giving \(B_2=129{,}795\).
Year 3's addition is \(0.022(129{,}795)=2{,}855.49\), giving \(B_3=132{,}650.49\).
Product check. \(125{,}000(1.018)(1.020)(1.022)=125{,}000(1.06120392)=132{,}650.49\). The additions sum to \(2{,}250+2{,}545+2{,}855.49=7{,}650.49\), which is also \(132{,}650.49-125{,}000\). The final benefit is $132,650.49.
Zero Rates, Changing Rates, and Rounding
A zero declaration rate is handled directly by the recurrence: if \(r_k=0\), then \(B_k=B_{k-1}(1+0)=B_{k-1}\). It contributes a factor of 1 to the product. It does not cancel a previous addition. Similarly, when rates change, each declaration contributes its own growth factor; the benefit entering the next year is the result after the preceding year's rate has been applied.
In a projection involving dollars, retain full precision through the recurrence and round the final amount to the nearest cent unless the question specifies another convention. Rounding each year's addition before carrying it forward can create a small difference, especially over many declarations. If a scale is stated in whole dollars per thousand, first interpret that scale as the stated rate; do not round the benefit base to a whole thousand unless explicitly required.
Worked Example: A Zero Declaration Between Positive Rates
A hypothetical policy has a benefit base of $60,000 before the first of four annual declarations. The rates are 1.5%, 0%, 2.5%, and 1%. Assume the rates apply to the current total benefit as defined above. Find each year's addition and the final benefit.
Method. Apply the recurrence in order, including the zero rate as a declaration with growth factor 1.
For year 1, the addition is \(0.015(60{,}000)=900\), so \(B_1=60{,}900\).
For year 2, the addition is \(0(60{,}900)=0\), so \(B_2=60{,}900\).
For year 3, the addition is \(0.025(60{,}900)=1{,}522.50\), so \(B_3=62{,}422.50\).
For year 4, the addition is \(0.01(62{,}422.50)=624.225\), so \(B_4=63{,}046.725\). Rounded to the nearest cent, the final benefit is $63,046.73.
Product check. \(60{,}000(1.015)(1)(1.025)(1.01)=63{,}046.725\), which rounds to the same amount. The unrounded additions total \(900+0+1{,}522.50+624.225=3{,}046.725\); adding this to $60,000 gives $63,046.725. The check confirms that the zero-rate year leaves the accumulated benefit unchanged.
Common Mistake: Adding the Rates
A frequent error is to add annual rates and apply the sum once to the original base. In the first worked example, this incorrect approach gives \(80{,}000(1+0.02+0.03+0.04)=87{,}200\). It misses the growth on prior bonus additions. The correct result is $87,409.92, because the later rates apply to a benefit that already includes earlier additions.
This distinction can also be stated algebraically. With two rates, compound accumulation gives
whereas simple accumulation adds only \(B_0(r_1+r_2)\) to the original base. The additional term \(B_0r_1r_2\) is the growth on the first addition at the second declaration. For positive rates it is positive, which explains why the simple calculation is lower in this example.
| Question being asked | Calculation under this model |
|---|---|
| Total benefit after declaration \(k\) | \(B_k=B_{k-1}(1+r_k)\) |
| Addition made at declaration \(k\) | \(C_k=r_kB_{k-1}=B_k-B_{k-1}\) |
| Total after \(n\) declarations | \(B_n=B_0\prod_{k=1}^{n}(1+r_k)\) |
A reliable exam check is to write down, before calculating, both the specified base and the amount the question asks for. Then carry each total forward in order, keep additions separate from totals, and compare the result with the product of growth factors. The product form is valid when each year's rate applies to the preceding total as defined here; if a problem defines a different basis, this recurrence must not be used without adjustment.
Check Your Understanding
For each problem, use the stated compound convention, retain full precision during the calculation, and round dollar answers to the nearest cent.
- A hypothetical policy has a $90,000 starting benefit and annual rates of 2%, 1.5%, and 2.5%. Find the total after the third declaration. (Answer: $95,506.43)
- A $70,000 starting benefit receives compound rates of $12, $15, and $18 per $1,000 in successive years. Find the final total benefit. (Answer: $73,196.85)
- A $48,000 base has declaration rates of 0%, 3%, and 3%. Find the final benefit and the total amount added. (Answer: Final benefit $50,923.20; total additions $2,923.20)
- A hypothetical benefit grows from $100,000 through three compound declarations. The first two rates are 2% and 3%, and the total after the third declaration is $109,262.40. Find the third rate. (Answer: 4%)
- A $40,000 base receives a 5% compound rate at each of two declarations. Find the compound total and compare it with the total if both 5% additions were instead calculated only on the original base. (Answer: Compound total $44,100; original-base total $44,000; compound result is $100 higher)