Annuity Factors from a One-Period Recurrence

Why One-Period Recurrences Help

An annuity-immediate has payments at the end of each payment period. Its present value or accumulated value can be found by applying a factor, but the factors themselves can also be built one period at a time. This is useful when a term is extended by one payment, when a factor is needed without a memorized expression, or when checking whether a calculation has used the correct payment timing.

The key is to separate one payment from the rest of the stream and move the remaining value by exactly one period. For present value, discounting moves a value back one period. For accumulated value, accumulation moves a value forward one period. Those opposite directions produce different recurrences.

Definition: Assume a level annuity-immediate pays one unit at the end of each of $n$ periods. Let $i$ be the effective interest rate per payment period, constant over all periods, and let $v=1/(1+i)$. The present-value factor is $a_{\overline{n}|}$, and the accumulated-value factor at the time of the final payment is $s_{\overline{n}|}$. Set both factors equal to 0 when $n=0$. The recurrences below apply for integer $n\geq 1$ and $i>-1$; in ordinary Exam FM settings, rates are typically nonnegative.

Present Value: Separate the First Payment

For $n$ unit payments at times $1,2,\ldots,n$, the first payment has present value $v$. The remaining $n-1$ payments form a level annuity-immediate whose value one period before its first payment is $a_{\overline{n-1}|}$. That value is at time 1 for the original stream, so it must be discounted one period to time 0.

a_{\overline{n}|}=v+v a_{\overline{n-1}|} =v\left(1+a_{\overline{n-1}|}\right), \qquad a_{\overline{0}|}=0.

The initial value matters: it says there are no payments, and hence no value, when $n=0$. The recurrence then builds the factor for each additional payment period. For example, $a_{\overline{1}|}=v(1+0)=v$, as it must be for a single payment one period from now.

There is another useful present-value recurrence. Keep the first $n-1$ payments and add the final payment at time $n$. The first $n-1$ payments have value $a_{\overline{n-1}|}$ at time 0, and the additional unit payment has value $v^n$ at time 0.

a_{\overline{n}|}=a_{\overline{n-1}|}+v^n, \qquad a_{\overline{0}|}=0.

These are two ways to organize the same cash flows. The first recurrence separates the first payment and discounts the rest of the stream by one period. The second keeps the earlier payments and adds the final payment. Neither changes the timing of any payment.

Worked Example: Building a Present-Value Factor Recursively

A hypothetical fund will receive four annual payments of $250, each at year-end. The effective annual interest rate is 25%. Find the present value at time 0 using the one-period recurrence, and verify the result from the individual discounted payments.

Method. Use the present-value recurrence, with one annual period as the unit of time. The discount factor is $v=1/(1.25)=0.8$. Avoid treating the first payment as if it were made immediately; it is paid at the end of year 1.

Start with $a_{\overline{0}|}=0$. Apply $a_{\overline{n}|}=0.8(1+a_{\overline{n-1}|})$ in order:

\begin{aligned} a_{\overline{1}|}&=0.8(1+0)=0.8,\\ a_{\overline{2}|}&=0.8(1+0.8)=1.44,\\ a_{\overline{3}|}&=0.8(1+1.44)=1.952,\\ a_{\overline{4}|}&=0.8(1+1.952)=2.3616. \end{aligned}

The value of the four $250 payments is $250a_{\overline{4}|}=250(2.3616)=\$590.40$.

Check. Discount the payments separately: $250(0.8)+250(0.8^2)+250(0.8^3)+250(0.8^4)=200+160+128+102.40=\$590.40$. The recurrence and direct valuation agree. No rounding was needed.

Accumulated Value: Separate the Final Payment

For accumulated value at time $n$, the final unit payment is made at time $n$, so it contributes 1. The preceding $n-1$ payments have accumulated to $s_{\overline{n-1}|}$ at time $n-1$. One more period of interest carries that amount to time $n$. Therefore, the factor satisfies a different recurrence:

s_{\overline{n}|}=1+(1+i)s_{\overline{n-1}|}, \qquad s_{\overline{0}|}=0.

The multiplier is $1+i$ because the earlier deposits are being carried forward one period. The added 1 is the final payment itself, which has no time to earn interest before the valuation time. In particular, $s_{\overline{1}|}=1+(1+i)0=1$.

Worked Example: Building an Accumulated-Value Factor Recursively

A hypothetical account receives four annual deposits of $100 at each year-end. The effective annual interest rate is 10%. Find the account value immediately after the fourth deposit, using the recurrence, and check it by accumulating each deposit separately.

Method. Use the accumulated-value recurrence because the requested value is at time 4. Avoid discounting the deposits: each earlier deposit must earn interest forward to the valuation date.

Starting with $s_{\overline{0}|}=0$, apply $s_{\overline{n}|}=1+1.1s_{\overline{n-1}|}$:

\begin{aligned} s_{\overline{1}|}&=1+1.1(0)=1,\\ s_{\overline{2}|}&=1+1.1(1)=2.1,\\ s_{\overline{3}|}&=1+1.1(2.1)=3.31,\\ s_{\overline{4}|}&=1+1.1(3.31)=4.641. \end{aligned}

Thus, the account value immediately after the fourth deposit is $100s_{\overline{4}|}=100(4.641)=\$464.10$.

Check. The deposit at time 1 earns interest for three years, the deposit at time 2 earns interest for two years, and so on. The accumulated total is $100(1.1^3)+100(1.1^2)+100(1.1)+100=133.10+121+110+100=\$464.10$. The recurrence is confirmed, with no rounding.

Why the Recurrences Hold

Each recurrence follows from valuing the same set of payments in two pieces. The one-period movement is essential: a value at time 1 is not yet a present value at time 0, and a value at time $n-1$ is not yet an accumulated value at time $n$.

For present value, isolate the first payment.
A unit paid at time 1 is worth $v$ at time 0. The remaining $n-1$ payments have value $a_{\overline{n-1}|}$ at time 1, so their value at time 0 is $v a_{\overline{n-1}|}$. Adding these pieces gives $a_{\overline{n}|}=v(1+a_{\overline{n-1}|})$.

For accumulated value, isolate the final payment.
The final unit payment is worth 1 at time $n$. The earlier $n-1$ payments have value $s_{\overline{n-1}|}$ at time $n-1$, which grows to $(1+i)s_{\overline{n-1}|}$ at time $n$. Adding these pieces gives $s_{\overline{n}|}=1+(1+i)s_{\overline{n-1}|}$.

Use the payment amount only after finding the factor.
For a level payment $R$, multiply the appropriate unit-payment factor by $R$. The factors describe timing and interest; $R$ supplies the payment size.

The same payment stream can be valued at different dates, but the recurrences themselves answer different questions. Present value moves payments back to time 0; accumulated value moves payments forward to the end of the term. Under the stated constant-rate assumptions, the relationship between the two factors is

s_{\overline{n}|}=(1+i)^n a_{\overline{n}|}.

To see why, take each payment at time $k$, where $1\leq k\leq n$. Its present value is $v^k$; multiplying by $(1+i)^n$ gives $(1+i)^{n-k}$, its accumulation to time $n$. Applying this to every payment converts the present-value sum into the accumulated-value sum. This identity can be used as a check, but it does not make the two recurrences interchangeable.

Worked Example: Extending a Present-Value Factor by One Payment

A hypothetical contract pays $600 at each year-end for five years. The effective annual interest rate is 20%. Find its present value by extending the four-payment factor with the final payment, then confirm the factor using the first-payment recurrence.

Method. Use $a_{\overline{n}|}=a_{\overline{n-1}|}+v^n$ to add the year-5 payment. This makes the extra payment's timing explicit. The common mistake to avoid is adding the undiscounted $600 to the present value of the first four payments.

Here $v=1/1.2=5/6$. Build the factor through four payments:

\begin{aligned} a_{\overline{1}|}&=\frac{5}{6},\\ a_{\overline{2}|}&=\frac{5}{6}+\frac{25}{36}=\frac{55}{36},\\ a_{\overline{3}|}&=\frac{55}{36}+\frac{125}{216}=\frac{455}{216},\\ a_{\overline{4}|}&=\frac{455}{216}+\frac{625}{1296}=\frac{3355}{1296}. \end{aligned}

The fifth payment adds $v^5=(5/6)^5=3125/7776$. Expressing the four-payment factor with denominator 7776 gives $a_{\overline{4}|}=20130/7776$, so

a_{\overline{5}|}=\frac{20130}{7776}+\frac{3125}{7776} =\frac{23255}{7776} \approx 2.990612.

The first-payment recurrence gives the same factor: $v(1+a_{\overline{4}|})=(5/6)(1+3355/1296)=(5/6)(4651/1296)=23255/7776$. Therefore, the present value is $600(2.9906121399\ldots)=\$1,794.37$, rounded to the nearest cent.

Check. The final payment contributes $600v^5$, not $600, at time 0. The two recurrences agree on the factor, and the final value has been rounded only once.

Common Trap and Practical Checks

A frequent error is using the accumulated-value recurrence to calculate a present-value factor. For example, at $i=10\%$, using $a_{\overline{0}|}=0$, the incorrect calculation $a_{\overline{1}|}=1+1.1(0)=1$, followed by $a_{\overline{2}|}=1+1.1(1)=2.1$, gives 2.1. This is not the present value of two year-end unit payments.

The correction is to discount, not accumulate. Here $v=1/1.1$, so $a_{\overline{1}|}=v=0.9090909\ldots$ and $a_{\overline{2}|}=v(1+a_{\overline{1}|})=1.735537\ldots$. The factor 2.1 is instead $s_{\overline{2}|}$, the accumulated value at time 2. Check both the valuation date and the direction of movement before choosing a recurrence.

The recurrences require payments at regular period-ends and a consistent effective rate per payment period. If payments occur at the beginning of periods, the timing is different; if the interest rate varies by period, a single constant-rate factor recurrence is not appropriate as written. For an annual payment stream, $i$ must be the effective annual rate. For another payment frequency, use the effective rate for that period.

These one-period steps are especially useful for understanding how an annuity factor changes when a payment is added. For present value, adding payment $n$ adds $v^n$ to the factor. For accumulated value, adding the final payment changes the factor from $s_{\overline{n-1}|}$ to $1+(1+i)s_{\overline{n-1}|}$, because the earlier balance first earns one more period of interest and the new payment is then included.

Check Your Understanding

For each problem, identify the valuation date and use the recurrence appropriate to that date. Assume payments occur at the end of each stated period.

At an effective annual rate of 5%, a fund receives three annual payments of $200. Find the present value at time 0. (Answer: $544.65)
An account earns an effective annual rate of 8% and receives three annual deposits of $150. Find its value immediately after the third deposit. (Answer: $486.96)
At an effective annual rate of 25%, find the present value of four annual payments of $300. Show how the value of the fourth payment extends the three-payment present value. (Answer: $585.60)
At an effective annual rate of 20%, an account receives four annual deposits of $250. Find the value immediately after the fourth deposit using the accumulated-value recurrence. (Answer: $1,342.00)
At an effective annual rate of 10%, a stream pays $100 at each of three year-ends. Find both its present value at time 0 and its accumulated value immediately after the third payment, and verify that the accumulated value is $(1.1)^3$ times the present value. (Answer: Present value $248.69; accumulated value $331.00)

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