Recursive Valuation of Varying Payment Streams

Why Use Backward Recursion?

A standard annuity factor is efficient when the payments and interest assumptions fit its pattern. But a finite stream may have irregular payment dates, a different effective rate in each interval, or payment amounts determined by a changing rule. In those situations, forcing the cash flows into a standard factor can obscure the timing or use the wrong rate. Backward recursion handles each interval on its own: start with the last payment, then work toward the valuation date.

The method applies to a deterministic finite stream: the payment amounts and dates are known, and each interval has a specified positive accumulation factor. It does not require equal payment spacing or a constant rate. If a payment is uncertain, this deterministic calculation alone does not account for that uncertainty.

Define the Values at the Payment Dates

Let $t_0<t_1<\cdots<t_n$ be the valuation date and the successive payment dates. Let $C_k$ be the payment at $t_k$. Write $q_k$ for the accumulation factor from $t_{k-1}$ to $t_k$. Thus, one dollar invested at $t_{k-1}$ accumulates to $q_k$ dollars at $t_k$. Assume $q_k>0$.

Definition: Let $V_k$ be, at time $t_k$ just after the payment at that time, the value of all payments strictly after $t_k$. Set $V_n=0$. The backward recursion is $$ V_{k-1}=\frac{C_k+V_k}{q_k}, \qquad k=n,n-1,\ldots,1. $$ The value at $t_0$ of payments after $t_0$ is $V_0$. If there is also a payment $C_0$ at $t_0$, the value including that payment is $C_0+V_0$.

When the interval is exactly one year and its effective annual rate is $i_k$, $q_k=1+i_k$. If a single effective annual rate $i$ applies throughout an interval of length $\Delta t$ years, the usual effective-rate model gives $q_k=(1+i)^{\Delta t}$. For any other timing or rate convention, use the accumulation factor specified by that model; do not assume a one-year factor applies to a shorter or longer interval.

The recursion values each interval separately. $V_k$ is first put together with the payment at $t_k$, since both are values at that date. Their combined amount is then discounted from $t_k$ to $t_{k-1}$ by dividing by $q_k$. The last interval is evaluated first because its endpoint is known: after the final payment, there are no later payments, so $V_n=0$.

Why the Recursion Works

The recurrence is a direct consequence of equivalence under the stated accumulation factor. An amount $X$ at $t_{k-1}$ grows to $q_kX$ at $t_k$. To have $C_k+V_k$ at $t_k$, its value at $t_{k-1}$ must therefore be $(C_k+V_k)/q_k$. Applying that argument successively from the last interval back to the first gives the full value.

Expanding the recurrence also shows how it accounts for every payment. Starting from $V_n=0$, the value at $t_0$ is

V_0=\frac{C_1}{q_1}+\frac{C_2}{q_1q_2}+\cdots+\frac{C_n}{q_1q_2\cdots q_n}.

For example, $C_2$ is first discounted from $t_2$ to $t_1$ by $q_2$, then from $t_1$ to $t_0$ by $q_1$. Its total discount factor is $q_1q_2$. The recursion and term-by-term discounting therefore give the same result, but the recursion organizes the calculation around successive dates and makes changing interval assumptions explicit.

A Reliable Calculation Process

Before calculating, put the payment dates in chronological order and identify the factor for each interval. If payment amounts depend on an earlier payment, determine the amounts from their rules first. Then proceed backward.

List dates, payments, and interval factors.
For each $k$, record $t_k$, $C_k$, and the accumulation factor $q_k$ from $t_{k-1}$ to $t_k$. Check that each factor matches its own interval.

Set the terminal value.
At the last payment date, there are no payments strictly after the date, so set $V_n=0$.

Work backward one interval at a time.
For each earlier date, calculate $V_{k-1}=(C_k+V_k)/q_k$. Keep enough digits to avoid accumulating rounding error.

State the value and check its timing.
Identify the date at which the final value applies. Include a payment at the valuation date separately if one is part of the requested total.

Worked Examples

Worked Example: Irregular Dates and Interval-Specific Rates

A fund will pay \$600 at $t_1=0.5$, \$900 at $t_2=1.5$, and \$1,200 at $t_3=3$, where time is measured in years from $t_0=0$. The effective accumulation factors over the three respective intervals are $q_1=1.04$, $q_2=1.06$, and $q_3=1.05$. Find the value at $t_0$. These factors are given for their full intervals; no annual-rate conversion is required.

Method. Start with $V_3=0$ and apply the backward recurrence using the factor for each interval.

From $t_2$ to $t_3$, the value at $t_2$ of the final payment is

V_2=\frac{1{,}200+0}{1.05}=1{,}142.8571429.

From $t_1$ to $t_2$, include the \$900 payment and the value at $t_2$ of the last payment:

V_1=\frac{900+1{,}142.8571429}{1.06} =\frac{2{,}042.8571429}{1.06} =1{,}927.2237197.

Finally, discount the \$600 payment and the later stream from $t_1$ to $t_0$:

V_0=\frac{600+1{,}927.2237197}{1.04} =\frac{2{,}527.2237197}{1.04} =2{,}429.061269.

Rounded to the nearest cent, the value at $t_0$ is \$2,429.06. As a check, each step moves a value at the later date back by the factor for exactly that interval. A common mistake is to apply one rate to all payments or to treat the half-year and longer intervals as if they were all one year; the given interval factors avoid both errors.

Worked Example: Payments Determined by a Changing Rule

A contract pays once at the end of each of four successive years. The first payment is \$400. The second is \$150 more than the first; the third is 80% of the second; and the fourth is \$200 more than the third. The effective annual rates over the four years are 5%, 4%, 6%, and 3%, respectively. Find the value immediately before the first year begins.

Method. First use the payment rules in chronological order to determine each $C_k$. Then value those known payments backward using the corresponding annual accumulation factors.

The payment amounts are

C_1=400,\qquad C_2=400+150=550,

C_3=0.80(550)=440,\qquad C_4=440+200=640.

The factors for the four intervals are $q_1=1.05$, $q_2=1.04$, $q_3=1.06$, and $q_4=1.03$. Start at the final payment:

V_4=0,\qquad V_3=\frac{640}{1.03}=621.3592233.

Continue backward:

V_2=\frac{440+621.3592233}{1.06} =\frac{1{,}061.3592233}{1.06} =1{,}001.2822861,

V_1=\frac{550+1{,}001.2822861}{1.04} =\frac{1{,}551.2822861}{1.04} =1{,}491.6175828,

V_0=\frac{400+1{,}491.6175828}{1.05} =\frac{1{,}891.6175828}{1.05} =1{,}801.5405551.

Rounded to the nearest cent, the value before the first year begins is \$1,801.54. The payment rules have been applied before the valuation, so every $C_k$ is known when the recursion begins. A common mistake is to discount first and then apply a percentage rule to a present value; the rule here determines the payment amount, not its value at an earlier date.

Worked Example: A Fractional-Year Interval

A stream pays \$500 at $t=0.5$, \$700 at $t=1.5$, and \$1,000 at $t=2.5$. The effective annual rate is 8% from $t=0$ to $t=0.5$, 4% from $t=0.5$ to $t=1.5$, and 6% from $t=1.5$ to $t=2.5$. Assume the 8% annual effective rate applies to the half-year interval through the factor $(1.08)^{0.5}$. Find the value at $t=0$.

Method. Convert the annual effective rate for the half-year to its half-year accumulation factor, then recurse. The interval factors are $q_1=(1.08)^{0.5}=1.0392304845$, $q_2=1.04$, and $q_3=1.06$.

Starting from the last payment,

V_2=\frac{1{,}000}{1.06}=943.3962264.

At $t=0.5$, the \$700 payment and the value of the final payment combine to give

V_1=\frac{700+943.3962264}{1.04} =\frac{1{,}643.3962264}{1.04} =1{,}580.1886792.

Discount the first payment and this later value over the half-year interval:

V_0=\frac{500+1{,}580.1886792}{(1.08)^{0.5}} =\frac{2{,}080.1886792}{1.0392304845} =2{,}001.6624898.

Rounded to the nearest cent, the value at $t=0$ is \$2,001.66. The result uses the annual effective rate over half a year, not the full one-year factor $1.08$. Applying $1.08$ to this interval would incorrectly treat six months as a full year.

Checks, Limitations, and Exam Traps

Backward recursion is especially useful when each interval or payment has its own feature. It makes the timing assumptions visible, so a candidate can match a rate to its interval rather than conceal the assumptions in a single factor. This method is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Keep the date attached to each value. $V_k$ is measured at $t_k$, after the payment there. Adding $C_k$ to $V_k$ is valid because both amounts are at $t_k$.
Use the factor for the correct interval. A factor from $t_{k-1}$ to $t_k$ is used once in that step. It is not automatically the same as the factor for any other interval.
Do not reverse the operation. To move a later value backward, divide by $q_k$. Multiplying would move a value forward and overstate its value at the earlier date when $q_k>1$.
Separate payment rules from valuation. If a payment amount is defined from earlier payments, calculate the amounts in the order specified by the rule before starting the backward valuation.
Handle the valuation-date payment separately. The recurrence as defined values payments strictly after $t_0$. A payment at $t_0$ is added without discounting if it belongs in the requested total.
Round only at the end. Retaining several digits at intermediate steps reduces small rounding errors that can compound across many intervals.

For a quick reasonableness check, if all payments are positive and all interval factors exceed 1, the value at an earlier date should be less than the sum of the payments. Also verify that the final step has the requested valuation date as its endpoint; a correct numerical value assigned to the wrong date is not a correct valuation.

Check Your Understanding

For each problem, identify the interval factors, work from the final payment backward, and round money to the nearest cent.

A stream pays \$300 at the end of year 1 and \$500 at the end of year 2. The effective annual rates are 5% in year 1 and 4% in year 2. Find the value at time 0. (Answer: \$743.59)
Three annual payments are \$200, \$300, and \$450. The effective annual rates over the successive years are 3%, 5%, and 4%. Find the value immediately before the first year. (Answer: \$871.65)
A stream pays \$400 at $t=0.5$ and \$900 at $t=1.5$. The effective annual rate is 4% over the first half-year and 6% over the following full year. Use $(1.04)^{0.5}$ for the first interval. Find the value at $t=0$. (Answer: \$1,224.80)
Annual payments follow this rule: $C_1=\$250$, $C_2=C_1+\$50$, and $C_3=0.5C_2+\$100$. The effective annual rates for the three years are 4%, 3%, and 5%. Find the value immediately before year 1. (Answer: \$742.71)

What You'll Learn