Deriving the Geometric-Gradient Annuity Value

Payment Patterns and Timing

A geometric-gradient annuity has payments that change by the same proportion from one payment to the next. This pattern appears in hypothetical settings such as a contract with annual benefit increases or a planned series of escalating deposits. The important valuation question is not only how quickly payments grow, but also when each payment is made.

We first consider an annuity-immediate: the first payment is made at the end of period 1, and the final payment is made at the end of period $n$. Let $P$ be the first payment, $g$ the effective growth rate per payment period, and $i$ the effective interest rate per payment period. Both rates are measured over the same period. The payment at the end of period $k$ is $P(1+g)^{k-1}$.

Definition: A finite geometric-gradient annuity-immediate with first payment $P$, growth rate $g$, and $n$ payments has payments $P,\ P(1+g),\ P(1+g)^2,\ldots,P(1+g)^{n-1}$ at times $1,2,\ldots,n$, respectively. Its present value is measured immediately before time 1, at time 0.

Assume $n$ is a positive integer, $i>-1$, and $g>-1$. The condition $i>-1$ ensures a positive accumulation factor $1+i$; $g>-1$ ensures the successive payment factors $1+g$ are positive. The usual formula also requires $i\ne g$. When $i=g$, the annuity still has a well-defined value, but the formula involving division by $i-g$ cannot be used directly.

Deriving the Present Value

Discount each payment to time 0 using $v=1/(1+i)$. The first payment is discounted one period, the second two periods, and the $k$th payment $k$ periods. Thus each successive present-value term is multiplied by the same ratio, $(1+g)/(1+i)$.

\begin{aligned} \text{PV} &=Pv+P(1+g)v^2+P(1+g)^2v^3+\cdots+P(1+g)^{n-1}v^n\\ &=Pv\left[1+(1+g)v+\big((1+g)v\big)^2+\cdots+\big((1+g)v\big)^{n-1}\right]. \end{aligned}

For a finite geometric series $1+r+\cdots+r^{n-1}$, multiplying by $1-r$ gives $1-r^n$. Therefore, when $r\ne1$, the sum is $(1-r^n)/(1-r)$. Here $r=(1+g)v=(1+g)/(1+i)$. If $i\ne g$, then $r\ne1$, and simplifying gives the standard result.

\begin{aligned} \text{PV} &=Pv\frac{1-\left((1+g)v\right)^n}{1-(1+g)v}\\ &=P\frac{1-\left(\frac{1+g}{1+i}\right)^n}{i-g}, \qquad i\ne g. \end{aligned}

The simplification uses $1-(1+g)v=(i-g)/(1+i)$, while $v=1/(1+i)$. These factors of $1+i$ cancel. This derivation also shows why the denominator is $i-g$, rather than $i+g$: the growth factor is compared with the discount factor through the ratio $(1+g)/(1+i)$.

Theorem: For $n$ payments at times $1,\ldots,n$, with first payment $P$, effective interest rate $i$, and effective payment-growth rate $g$, the present value is $$ \text{PV}= \begin{cases} \displaystyle P\frac{1-\left(\frac{1+g}{1+i}\right)^n}{i-g}, & i\ne g,\\[8pt] \displaystyle \frac{nP}{1+i}, & i=g. \end{cases} $$ The interest and growth rates must use the same period as the payment interval.

To see the equal-rate case, set $g=i$ in the discounted cash-flow series itself, not in the expression with denominator $i-g$. Then $(1+g)v=1$, so all $n$ discounted terms are $Pv=P/(1+i)$. Their sum is $nP/(1+i)$. Substitution into the unequal-rate formula would produce division by zero; it does not mean the value is infinite or zero.

Worked Example: An Increasing Four-Payment Stream

A hypothetical fund will pay $1{,}000$ dollars at the end of year 1. Each later annual payment is 4% larger than the preceding one. If the effective annual interest rate is 8%, find the present value of the four payments. Use the annuity-immediate timing stated in the problem.

Method. Identify $P=1{,}000$, $g=0.04$, $i=0.08$, and $n=4$. The rates are both annual effective rates, so they match the annual payment period. Since $i\ne g$, use the closed-form expression:

\begin{aligned} \text{PV} &=1000\frac{1-\left(\frac{1.04}{1.08}\right)^4}{0.08-0.04}\\ &=1000\frac{1-\left(\frac{26}{27}\right)^4}{0.04}\\ &=1000\frac{1-\frac{456976}{531441}}{0.04}\\ &=1000\frac{\frac{74465}{531441}}{0.04}\\ &=3502.9758\ldots \end{aligned}

The payments are $1{,}000$, $1{,}040$, $1{,}081.60$, and $1{,}124.864$ dollars. As a check, their discounted values are approximately $925.9259$, $891.6324$, $858.6089$, and $826.8086$ dollars. These sum to $3502.9758\ldots$, agreeing with the formula. Rounded to the nearest cent, the present value is $\$3{,}502.98$. The timing check is that the first payment was discounted one year, not zero years.

Worked Example: Growth Rate Equal to the Interest Rate

A hypothetical agreement pays $750$ dollars at the end of year 1 and increases each annual payment by 5%. There are six payments, and the effective annual interest rate is also 5%. Find the present value.

Method. Here $P=750$, $g=i=0.05$, and $n=6$. The unequal-rate formula has a zero denominator, so instead evaluate the discounted series. The payment in year $k$ is $750(1.05)^{k-1}$, and its time-0 value is:

\frac{750(1.05)^{k-1}}{(1.05)^k} =\frac{750}{1.05} =714.285714\ldots

Thus every payment contributes the same present value. Summing six terms gives

\text{PV}=6\left(\frac{750}{1.05}\right) =\frac{4500}{1.05} =4285.714285\ldots

The payments themselves are $750$, $787.50$, $826.875$, $868.21875$, $911.6296875$, and $957.211171875$ dollars. Each is discounted by the matching power of 1.05, leaving $750/1.05$ for each term. The present value, rounded to the nearest cent, is $\$4{,}285.71$. The key point is that equal growth and interest make the discounted terms level; they do not make the payments level.

Deferred Payments and Timing

The derived formula applies to payments at times $1,\ldots,n$. If the first payment occurs later, retain the same payment pattern but discount its value back across the deferral. For example, if the payments are made at times $2,\ldots,n+1$, their present value is the value of an $n$-payment annuity-immediate at time 1, multiplied by $v$ to move it to time 0. A deferral changes the valuation date, not the growth ratio between consecutive payments.

Worked Example: A Deferred Increasing Stream

A hypothetical contract makes its first payment of $1{,}000$ dollars at the end of year 2. Each subsequent annual payment is 5% greater, and there are three payments in total. The effective annual interest rate is 10%. Find the present value at time 0.

Method. The payments occur at times 2, 3, and 4 and are $1{,}000$, $1{,}050$, and $1{,}102.50$ dollars. First value the same three-payment pattern at time 1, where it would have payments one, two, and three years later. Here $P=1{,}000$, $g=0.05$, $i=0.10$, and $n=3$:

\begin{aligned} V_1 &=1000\frac{1-\left(\frac{1.05}{1.10}\right)^3}{0.10-0.05}\\ &=1000\frac{1-\left(\frac{21}{22}\right)^3}{0.05}\\ &=1000\frac{1-\frac{9261}{10648}}{0.05}\\ &=2605.1841\ldots \end{aligned}

The value is at time 1, so discount it one year to time 0:

V_0=\frac{V_1}{1.10} =\frac{2605.1841\ldots}{1.10} =2368.3492\ldots

As a direct check, discount the three payments from their actual dates: $1000/(1.10)^2=826.4463$, $1050/(1.10)^3=788.8805$, and $1102.50/(1.10)^4=753.0223$. Their sum is $2368.3491\ldots$, consistent with the deferred-formula calculation up to displayed rounding. The present value rounded to the nearest cent is $\$2{,}368.35$. The common timing error is to use the time-1 value as if it were already the time-0 value.

Common Mistakes and Exam Checks

A geometric-gradient formula is compact, but it is only reliable after the payment schedule and the rate period have been identified. In particular, $P$ is the first payment, not the payment at time 0, and $g$ applies between consecutive payments.

Common mistake: Treating the first payment of an annuity-immediate as if it occurred at time 0. In the four-payment example, that error multiplies the correct value by $1.08$, giving $3502.98(1.08)=3783.22$ dollars (to the nearest cent). The correction is to discount the first payment one full period, as shown by the factor $Pv$ in the series.

Another frequent error is substituting $i=g$ into the unequal-rate formula and stopping at $0/0$. Return to the discounted terms: when the rates match, each term is $P/(1+i)$, so the answer is $nP/(1+i)$. Also check the unit of each rate. If payments are quarterly, an annual effective rate cannot be used as though it were an effective quarterly rate; first express the interest and growth factors over the payment interval.

When $g=0$, the payments are level, and the geometric-gradient series reduces to an ordinary finite annuity-immediate. When $g<0$ but $g>-1$, payments decrease over time and the same derivation applies. In either case, the final check should be a cash-flow ledger: list each payment date, payment amount, and discount factor, then compare the sum with the formula.

Check Your Understanding

For each problem, identify the payment period and timing before applying a formula.

A fund pays $600$ dollars at the end of each of four successive years, with the first payment at year 1 and each payment 3% greater than the last. The effective annual interest rate is 7%. Find the present value at time 0. (Answer: $\$2{,}120.32$)
A six-payment annuity-immediate has first payment $2{,}500$ dollars. Payments grow by 4% per year, and the effective annual interest rate is 4%. Find its present value. (Answer: $\$14{,}423.08$)
A five-payment stream has its first payment of $900$ dollars at the end of year 2; each later payment grows by 2% per year. The effective annual interest rate is 5%. Find its time-0 present value. (Answer: $\$3{,}854.97$)
A five-payment annuity-immediate starts at $1{,}000$ dollars and decreases by 2% per year. The effective annual interest rate is 4%. Find its present value at time 0. (Answer: $\$4{,}284.05$)
A six-payment annuity-immediate grows at the same effective annual rate as the 5% effective annual interest rate. If its present value is $10{,}000$ dollars, find the first payment. (Answer: $\$1{,}750.00$)

What You'll Learn