Why Yield Assumes Reinvestment at Yield

What a Yield Does—and Does Not—Tell You

A yield summarizes a stream of dated cash flows as a rate that makes their present value equal to an initial investment. That equation alone does not promise that the investor will earn the same rate on every dollar received along the way. To interpret the yield as a rate of growth from the initial outlay to a final accumulated amount, the interim positive receipts must be reinvested at the yield until the chosen horizon.

This distinction matters whenever cash arrives before the end of an investment. A yield can be calculated from the original cash flows even if the investor spends those receipts or reinvests them at a different rate. But the terminal value corresponding to the yield equation depends on accumulating the receipts at that yield. This tutorial is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Definition: For an investment with an initial outlay $P$ at time 0 and receipts $C_t$ at the ends of periods $t=1,\ldots,n$, a yield $i$ is a rate satisfying $$ P=\sum_{t=1}^{n}\frac{C_t}{(1+i)^t}. $$ Here the periods are equally spaced, $i>-1$ is an effective rate per period, and the cash-flow amounts are treated as certain. For the usual reinvestment interpretation, the interim receipts are nonnegative and are reinvested at effective rate $i$ per period through time $n$.

The yield equation is a present-value equation: each receipt is discounted back to time 0. The corresponding terminal-value equation carries the initial outlay forward to time $n$, while each receipt is accumulated from its receipt date to time $n$. These are two forms of the same equality when the accumulation and discount rates agree.

Why the Two Equations Match

Start with the yield equation and multiply every term by $(1+i)^n$. The present value of a receipt at time $t$ is multiplied by the factor needed to move it forward from time 0 to time $n$. Its net accumulation factor is therefore $(1+i)^{n-t}$. This gives the terminal-value equation.

\begin{aligned} P&=\sum_{t=1}^{n}\frac{C_t}{(1+i)^t}\\ P(1+i)^n&=\sum_{t=1}^{n}C_t(1+i)^{n-t}. \end{aligned}

This algebra is the key justification. If each $C_t$ is received at the end of period $t$ and immediately reinvested at rate $i$, it earns interest for $n-t$ periods. Its value at time $n$ is $C_t(1+i)^{n-t}$. The initial outlay grows to $P(1+i)^n$. Thus the yield equation implies that the accumulated receipts equal the accumulated outlay under that reinvestment assumption.

Theorem: If $i$ satisfies the yield equation for the stated cash flows, then $$ \sum_{t=1}^{n} C_t(1+i)^{n-t}=P(1+i)^n. $$ When all receipts before time $n$ are reinvested at the same effective rate $i$ per period, the left side is the terminal value of those receipts. Without that reinvestment, the equality remains an algebraic identity, but its left side is not necessarily the investor’s actual terminal wealth.

For the usual conventional investment, the outlay is negative from the investor’s perspective at time 0 and the later cash flows are positive receipts. If these receipts are all at strictly positive times and at least one is positive, the present value of the receipts decreases as $i$ increases over $i>-1$. Consequently, there is at most one yield in that range. A positive yield exists when the undiscounted receipts exceed $P$; if they equal $P$, the yield is zero; and if they are less than $P$, the yield is negative. More complicated patterns with intervening negative cash flows need not have a unique yield.

Worked Examples

Worked Example: Verifying a Yield by Accumulating Receipts

An investment costs $1,000 at time 0 and pays $500 at the end of year 1 and $660 at the end of year 2. Verify that its yield is 10% per year, then find the matching terminal value at the end of year 2. Assume annual periods and, for the terminal-value interpretation, that the year-1 receipt is reinvested at an effective annual rate of 10%.

Method. First substitute the proposed yield in the present-value equation. Then accumulate each receipt to year 2 and compare the result with the initial outlay accumulated for two years.

At 10%, the present value of the first receipt is $500/1.10=\$454.54545\ldots$. The present value of the second receipt is $660/(1.10)^2=660/1.21=\$545.45454\ldots$. Their sum is $\$1,000$, so the yield equation holds.

\frac{500}{1.10}+\frac{660}{1.21} =454.54545\ldots+545.45454\ldots =1{,}000

At the end of year 2, the $500 received at year 1 has earned one year of interest: $500(1.10)=\$550$. The $660 received at year 2 has no time to earn interest before the horizon, so its accumulated value remains $660. Total terminal value is $\$550+\$660=\$1,210$. The initial outlay accumulated for two years is $1{,}000(1.10)^2=\$1,210$, as required.

Check. Both routes give $1,210. A common mistake is to accumulate the year-1 receipt for two years; it was received at the end of year 1 and earns only one year of interest before the horizon.

Worked Example: Finding a Final Receipt From the Yield Condition

A project requires $1,000 now and pays $200 at the end of year 1 and $500 at the end of year 2. It also pays an unknown amount $C_3$ at the end of year 3. If the yield is 10% per year, find $C_3$, and verify the result using terminal values. Assume annual periods and reinvestment of each interim receipt at 10% for the terminal-value interpretation.

Method. Accumulate the initial outlay to year 3 and subtract the accumulated values of the known receipts. This is the yield equation multiplied by $(1.10)^3$.

The initial outlay accumulated to year 3 is $1{,}000(1.10)^3=1{,}000(1.331)=\$1,331$. The year-1 receipt earns two years of interest, so its accumulated value is $200(1.10)^2=\$242$. The year-2 receipt earns one year, giving $500(1.10)=\$550$. The year-3 receipt earns zero years.

C_3=1{,}331-242-550=\$539

To check the yield equation directly, discount each receipt:

\begin{aligned} \frac{200}{1.10}+\frac{500}{1.21}+\frac{539}{1.331} &=181.81818\ldots+413.22314\ldots+404.95868\ldots\\ &=1{,}000. \end{aligned}

The terminal receipts are $\$242+\$550+\$539=\$1,331$, equal to the accumulated initial outlay. Check. The final receipt is $539. A common mistake is to give the year-1 cash flow three years of reinvestment; it is received at the end of year 1, so it compounds for only two years.

Worked Example: A Different Reinvestment Rate Changes Terminal Wealth

An investment costs $2,000 and pays $500 at the end of year 1, $700 at the end of year 2, and $1,287 at the end of year 3. Its yield is 10% per year. If the year-1 and year-2 receipts can instead be reinvested at an effective annual rate of 4%, find the actual terminal value at year 3 and compare it with the terminal value implied by the yield. Assume annual periods, immediate reinvestment of each interim receipt, and no taxes or transaction costs.

Method. First verify the yield and its implied terminal value at 10%. Then accumulate the interim receipts at the actual reinvestment rate of 4%, without changing the timing of any receipt.

At 10%, the present values are $500/1.10=\$454.54545\ldots$, $700/1.21=\$578.51240\ldots$, and $1{,}287/1.331=\$966.94215\ldots$. Their sum is $2,000, confirming the 10% yield. The terminal value implied by that yield is $2{,}000(1.10)^3=\$2,662$. Equivalently, the accumulated receipts are $500(1.10)^2=\$605$, $700(1.10)=\$770$, and $1,287, totaling $2,662.

At 4%, the year-1 receipt accumulates for two years to $500(1.04)^2=500(1.0816)=\$540.80$. The year-2 receipt accumulates for one year to $700(1.04)=\$728$. The year-3 receipt remains $1,287. Therefore the actual terminal value is $\$540.80+\$728+\$1,287=\$2,555.80$.

The shortfall relative to the yield-implied terminal value is $\$2,662-\$2,555.80=\$106.20$. Check. The lower reinvestment rate produces a lower terminal value, even though the original cash flows still have a 10% yield. A common mistake is to call $2,662 guaranteed terminal wealth; it is the terminal value only under the stated 10% reinvestment assumption.

Assumptions Behind the Reinvestment Interpretation

The terminal-value reading is useful, but it depends on the cash-flow model. Before using it, make the timing and reinvestment assumptions explicit:

Period timing is consistent. The yield is an effective rate per period, and receipts occur at the period ends specified in the equation. The exponent $n-t$ counts the remaining periods from receipt to the horizon.
Interim receipts are available to reinvest. The usual interpretation assumes positive receipts can be reinvested from their receipt dates through the selected horizon.
The reinvestment rate matches the yield. Each receipt earns the same effective rate per period used in the yield equation. A different rate generally gives a different terminal value.
The receipts and initial outlay are treated consistently. The basic equation assumes the stated amounts are the relevant cash flows, without omitted taxes, fees, or additional contributions.

A yield remains a property of the stated cash-flow pattern even when these conditions are not met in practice. If receipts are spent, there is no accumulated terminal value from those receipts. If they are reinvested at changing rates, their terminal values must use those rates instead. And if an intermediate cash flow is negative, it represents a payment or funding need, not a positive receipt available to reinvest. The algebraic equation can still be multiplied through by $(1+i)^n$, but its interpretation as growth of reinvested proceeds requires care.

Common Mistakes

Confusing yield with a guaranteed realized rate. A yield solves a cash-flow equation; it does not guarantee that interim receipts can earn that same rate.
Using the wrong number of periods. A receipt at time $t$ accumulates for $n-t$ periods to a horizon at time $n$, not for $n$ periods.
Changing the cash-flow dates while accumulating. The terminal-value equation preserves the original dates. It does not move a receipt to the initial date or the horizon before calculating its value.
Ignoring multiple sign changes. When intermediate outflows occur, a single yield may not exist or may not be unique; the standard single-outlay, positive-receipts interpretation may fail.

Key takeaway: Multiplying the yield equation by $(1+i)^n$ gives a terminal-value equality. It describes actual terminal wealth only when interim positive cash flows are reinvested at the yield for the periods remaining to the horizon.

Check Your Understanding

For each question, use the stated annual periods and distinguish the yield-implied value from any value based on another reinvestment rate.

An investment costs $800 and pays $300 at the end of year 1 and $638 at the end of year 2. Verify a 10% yield by comparing the accumulated receipts at year 2 with the accumulated outlay. (Answer: $968 on each side.)
A project costs $1,500 and has a 8% yield. It pays $400 at the end of year 1, $600 at the end of year 2, and $C_3$ at the end of year 3. Find $C_3$. (Answer: $775.008.)
An investment costs $2,000 and has a 10% yield, with receipts of $500 at year 1, $700 at year 2, and $1,287 at year 3. If the first two receipts are reinvested at 4% instead, find the year-3 terminal value and the shortfall from the value implied by the yield. (Answer: Terminal value $2,555.80; shortfall $106.20.)
Explain why multiplying a yield equation by $(1+i)^n$ does not by itself prove that an investor will actually have that terminal amount. (Answer: It is an algebraic identity; actual terminal wealth matches it only if interim receipts are reinvested at the yield for the remaining periods.)
A cash-flow pattern has an initial outlay, a positive receipt at year 1, and a required payment at year 2. Why is it not generally valid to describe every term in the accumulated equation as reinvested proceeds? (Answer: The year-2 negative cash flow is a funding requirement, not a positive receipt available for reinvestment.)

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