Deriving Present Values of Deferred Annuities

Why the Valuation Date Matters

A deferred ordinary annuity has level payments that begin in the future rather than one period from now. Its present value is not found by changing the ordinary-annuity formula itself: first, value the payment stream at the date one period before its first payment, then discount that value back to today. The key is to count the periods from the valuation date to that first payment correctly.

For example, if a first payment is due at time 4, the date one period before it is time 3. The annuity’s value at time 3 is the value of payments made at times 4, 5, and onward. To get the value at time 0, discount the time-3 value by three periods—not four. The derivation below makes that timing explicit.

Definition: A deferred ordinary annuity is a finite series of equal payments, each made at the end of a period, with the first payment postponed beyond the first period. Let $P$ be each payment, $n$ the number of payments, and $k$ the number of full periods from time 0 to the date immediately before the first payment. Thus, payments occur at times $k+1,k+2,\ldots,k+n$. Assume a constant effective interest rate $i$ per payment period, with $i>-1$, and write $v=1/(1+i)$.

Deriving the Deferred-Annuity Formula

At time $k$, the payment stream looks like an $n$-payment ordinary annuity: the first payment is one period away, the next is two periods away, and the last is $n$ periods away. The present value of an ordinary annuity at its valuation date is $P a_{\overline{n}|}$, where $a_{\overline{n}|}=v+v^2+\cdots+v^n$. Therefore the value at time $k$ is $P a_{\overline{n}|}$. Since time $k$ is $k$ periods after time 0, discount that value by $v^k$.

\begin{aligned} \text{Value at time }k &=P(v+v^2+\cdots+v^n)\\ &=P a_{\overline{n}|},\\[4pt] \text{Present value at time }0 &=P a_{\overline{n}|}v^k. \end{aligned}

The result can also be verified by discounting every payment directly. A payment at time $k+j$ is discounted by $v^{k+j}$, for $j=1,\ldots,n$. Factoring $v^k$ out of the sum gives the same expression.

\begin{aligned} PV_0 &=P(v^{k+1}+v^{k+2}+\cdots+v^{k+n})\\ &=Pv^k(v+v^2+\cdots+v^n)\\ &=Pv^k a_{\overline{n}|}. \end{aligned}

This equality is a useful timing check: the deferred-annuity formula must match the sum of the discounted payments at their actual dates. If $i\ne 0$, the ordinary-annuity factor can be written $a_{\overline{n}|}=(1-v^n)/i$, so the present value is $P v^k(1-v^n)/i$. If $i=0$, that quotient is undefined, but the sum has a direct value: each of the $n$ payments is worth $P$, so the value at time $k$ and the present value are both $nP$.

Locate the first payment.
If the first payment is at time $k+1$, the date one period before it is time $k$.

Value the annuity at time $k$.
There are $n$ ordinary-annuity payments from that date, so the value is $P a_{\overline{n}|}$.

Discount to time 0.
Time $k$ is $k$ periods after time 0, so multiply by $v^k$, not $v^{k+1}$.

Check the dates.
Confirm that the direct discounting exponents are $k+1$ through $k+n$.

Worked Valuations

Worked Example: Three Years of Deferral

A hypothetical fund will pay $800 at the end of each year for four years, with the first payment at the end of year 4. Find the value at year 3 and the present value at year 0. Assume an annual effective interest rate of $i=5\%$, unchanged over the valuation period.

Method and timeline. The payment dates are years 4, 5, 6, and 7. The date immediately before the first payment is year 3, so $k=3$, $n=4$, and $P=\$800$. The annual discount factor is $v=1/1.05=0.95238095$.

Value at year 3. At year 3, the four payments are one, two, three, and four years away. Their discount factors are $v=0.95238095$, $v^2=0.90702948$, $v^3=0.86383760$, and $v^4=0.82270247$. Thus,

800(0.95238095+0.90702948+0.86383760+0.82270247) =800(3.54595050) =\$2{,}836.76.

Present value. Discount the year-3 value by three years:

\$2{,}836.7604(0.86383760)=\$2{,}450.50.

Check and common mistake avoided. Direct discounting uses the four actual payment dates: $800(v^4+v^5+v^6+v^7)$. The discount-factor sum is $0.82270247+0.78352617+0.74621540+0.71068133=3.06312537$, giving $800(3.06312537)=\$2{,}450.50$, as required. Do not discount the year-3 value four periods: that would move it to year $-1$, not year 0.

Worked Example: Separating the Deferral from the Payment Count

A hypothetical equipment contract pays $1,000 at the end of each year for three years, beginning at the end of year 3. Calculate the contract’s value at year 2 and its present value. Assume an annual effective rate of $i=10\%$.

Method and timeline. Payments occur at years 3, 4, and 5. Therefore, the valuation date immediately before the first payment is year 2, giving $k=2$, $n=3$, and $P=\$1{,}000$. The discount factor is $v=1/1.10=0.90909091$.

Value at year 2. The three payments are one, two, and three years after year 2:

\begin{aligned} V_2 &=\$1{,}000(0.90909091+0.82644628+0.75131480)\\ &=\$1{,}000(2.48685199)\\ &=\$2{,}486.85. \end{aligned}

Present value. Discount from year 2 to year 0, a two-period interval:

PV_0=\$2{,}486.85199(0.82644628)=\$2{,}055.25.

Check and common mistake avoided. Directly discounting the payments gives $\$1{,}000(v^3+v^4+v^5)$. The factors are $0.75131480$, $0.68301346$, and $0.62092132$; their sum is $2.05524958$, giving $\$2{,}055.25$. The deferral is two periods, even though the first payment is at time 3: the annuity value at time 2 must be discounted two periods, not three.

Worked Example: First Payment at Year 5

A hypothetical benefit arrangement pays $600 at the end of each year for three years, starting at the end of year 5. Find its present value at year 0 and its value one period before the first payment. Assume a constant annual effective interest rate of $i=4\%$.

Method and timeline. The payment dates are years 5, 6, and 7, so the date one period before the first payment is year 4. Hence $k=4$, $n=3$, and $P=\$600$. The annual discount factor is $v=1/1.04=0.96153846$.

Value at year 4. The payments are one, two, and three years away from that date:

\begin{aligned} V_4 &=\$600(0.96153846+0.92455621+0.88899636)\\ &=\$600(2.77509103)\\ &=\$1{,}665.05. \end{aligned}

Present value. Discount the value at year 4 by four periods. Using the unrounded value $V_4=\$1{,}665.05462$, and rounding only the final dollar amounts to the nearest cent:

PV_0=\$1{,}665.05462(0.85480419)=\$1{,}423.30.

Check and common mistake avoided. The payments occur at years 5, 6, and 7, so direct discounting is $\$600(v^5+v^6+v^7)$. The factors are $0.82192711$, $0.79031453$, and $0.75991878$; their sum is $2.37216042$, giving $\$1{,}423.30$. Discounting the year-4 value by five periods would incorrectly treat it as a value at year 5.

Conditions, Interpretation, and Common Traps

The formula $P v^k a_{\overline{n}|}$ assumes that the payments are level, occur at regular period ends, and use the same period as the stated effective interest rate. If payments are annual but the rate is a monthly rate, the rate and period units must first be made consistent. The formula also assumes that $k$ counts periods to the valuation date just before the first payment—not the number of periods to the first payment itself.

The equivalent direct-sum form is especially useful for checking an interpretation. For a first payment at time $m$, set $k=m-1$. The payment dates are then $m,m+1,\ldots,m+n-1$, and the present-value exponents must match those dates exactly. A mismatch between the exponent list and the payment timeline signals a timing error.

First payment date	Value date immediately before it	Periods to discount that value to time 0
Time 1	Time 0	0
Time 3	Time 2	2
Time 5	Time 4	4

A frequent exam trap is to count the first-payment date as the deferral period. For example, a first payment at time 5 means the annuity value is at time 4; discounting that value to time 0 takes four periods. A second trap is to value the annuity at time $k$ using payments at times $k$ through $k+n-1$. That would be an annuity due at time $k$, not the ordinary annuity in this definition; the correct dates are $k+1$ through $k+n$.

Key takeaway: For $n$ level end-of-period payments beginning at time $k+1$, value the payments at time $k$ as $P a_{\overline{n}|}$, then discount $k$ periods. The resulting present value is $P v^k a_{\overline{n}|}$, and it must equal the sum of the payments discounted from their actual dates.

Check Your Understanding

For each problem, identify the valuation date immediately before the first payment. Then apply the formula or verify it by discounting the payments at their stated dates. Assume the stated rate is effective per year.

A fund pays $500 at the end of each year for two years, first paying at year 3. At $i=6\%$, find its value at year 2 and its present value. (Answer: Value at year 2 is $916.70; present value is $815.86.)
A contract pays $900 at the end of each year for three years, first paying at year 4. At $i=5\%$, find its value at year 3 and its present value. (Answer: Value at year 3 is $2,450.92; present value is $2,117.20.)
A no-interest agreement pays $275 at the end of each year for four years, first paying at year 6. Find the value immediately before the first payment and the present value. (Answer: Value at year 5 is $1,100.00; present value is $1,100.00.)
A benefit pays $400 at the end of each year for four years, first paying at year 6. At $i=8\%$, find its value at year 5 and its present value, rounding to cents. (Answer: Value at year 5 is $1,324.85; present value is $901.67.)
An analyst values four annual payments of $750, first paid at year 4, at $i=5\%$. The analyst correctly finds the value at year 3, then discounts that value by four periods. Is the method correct? Give the correct present value. (Answer: No; discount the time-3 value by three periods. The present value is $2,297.34.)

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