Uniqueness of Yield for a Fixed Callable Cash-Flow Scenario

Why Uniqueness Matters for a Fixed Scenario

A callable bond can have more than one possible redemption date. Once a particular redemption scenario is selected, however, its future coupons and redemption payment form a fixed set of cash flows. The yield for that scenario is a rate that makes the present value of those specified payments equal to the observed price. A key question is whether two different admissible yields could both do that.

For a fixed scenario with positive future payments, the answer is no: its present value strictly decreases with the annual effective yield. This gives an important distinction. There can be at most one yield for each fixed scenario, but this result by itself does not say which redemption scenario applies or compare yields across different scenarios.

Definition: A fixed callable cash-flow scenario specifies the assumed redemption date, every coupon paid before or on that date, and the redemption amount. Write the resulting positive cash flows as $C_1,\ldots,C_n$, paid at times $t_1,\ldots,t_n$ years after the valuation date, where $C_j>0$ and $t_j>0$. The scenario’s present value at annual effective yield $y$ is $$ V(y)=\sum_{j=1}^{n} C_j(1+y)^{-t_j}, \qquad y>-1. $$

The time $t_j$ is measured from the valuation date, not from the previous coupon date. If valuation occurs between coupon dates, a payment may be due after a fractional number of years; the formula uses that fractional time. It assumes an annual effective yield with discounting by $(1+y)^{-t_j}$. A different yield quotation convention must be translated into its corresponding discount factors before using the formula.

In many problems, the admissible yields are further restricted, for example to $y\geq 0$. We will call the lower bound $b$, with $b>-1$, and consider yields in $y\geq b$. The boundary is part of the problem’s assumptions: a conclusion about nonnegative yields need not hold if negative yields are permitted.

Why Present Value Strictly Decreases

Each positive payment is discounted more heavily when the yield rises. The following derivative makes that observation precise, including when payment times are fractional.

V'(y)=\sum_{j=1}^{n} -t_j C_j(1+y)^{-t_j-1}.

For every $y>-1$, each factor $t_j$, $C_j$, and $(1+y)^{-t_j-1}$ is positive. Thus every term in the derivative is negative, so $V'(y)<0$. Therefore $V(y)$ is strictly decreasing throughout its domain.

Theorem: For a fixed scenario with at least one positive payment at a strictly positive time, the present value is strictly decreasing in the annual effective yield on $y>-1$. Consequently, for any specified price, there is at most one yield in any admissible interval that makes the present value equal to that price.

Strict decrease proves “at most one”; it does not automatically prove that a yield exists. For the common case $y\geq b>-1$, with all payments positive and at positive times, $V(b)$ is the largest value on the permitted range, and $V(y)$ approaches zero as $y$ increases without bound. Since $V$ is continuous, a positive target price $P$ has exactly one solution in $y\geq b$ if $0<P\leq V(b)$. If the restriction is instead $y>b$, the condition is $0<P<V(b)$: the boundary value cannot be attained.

\lim_{y\to\infty}V(y)=0, \qquad \begin{cases} \text{one solution for } y\geq b, & 0<P\leq V(b),\\ \text{no solution for } y\geq b, & P>V(b). \end{cases}

These existence conditions use the assumptions that every payment is positive, is due after the valuation date, and the permitted range includes its lower boundary $b$. A payment at time zero would not be discounted and would change the limiting value as $y\to\infty$. Negative cash flows would also invalidate the term-by-term argument for strict decrease.

Worked Examples

Worked Example: Checking a Price and Solving for the Unique Yield

A fixed scenario pays $5 at the end of year 1 and $105 at the end of year 2. The observed price is $100, and only nonnegative annual effective yields are admissible. Determine whether a yield exists and find it.

Method. First check the value at the boundary and at a trial yield. Then use strict decrease for uniqueness and solve the value equation. The common mistake to avoid is miscomputing the year-1 discounted payment.

At the boundary $y=0$, the present value is $$ V(0)=5+105=110. $$

At $y=10\%$, the year-1 payment is $5/1.10=4.54545\ldots$, not $90.9091\ldots$. The year-2 payment has discount factor $1.10^2=1.21$, so $$ V(0.10)=\frac{5}{1.10}+\frac{105}{1.21} =4.54545\ldots+86.77686\ldots =91.32231\ldots. $$

The price $100 lies between $V(0)=110$ and $V(0.10)\approx91.32231$. Continuity gives a yield between 0% and 10%, and strict decrease shows that this yield is unique.

To find it, set $x=1+y$. Then $$ 100=\frac{5}{x}+\frac{105}{x^2} \quad\Longrightarrow\quad 100x^2-5x-105=0. $$

The discriminant is $(-5)^2-4(100)(-105)=42{,}025=205^2$. The positive root is $$ x=\frac{5+205}{200}=1.05, \qquad y=x-1=0.05. $$

Check. At 5%, $5/1.05=4.76190\ldots$ and $105/1.1025=95.23809\ldots$, giving a present value of $100.00. The unique admissible yield is 5.00%.

Worked Example: A Fixed Call-Date Scenario

A hypothetical bond is valued at $100. Under one specified call scenario, it pays a $6 coupon at the end of year 1, then a $6 final coupon and a $100 call price at the end of year 2. Assume an annual effective yield $y\geq0$. Find the yield for this scenario and explain why there cannot be another nonnegative yield.

Method. Combine payments due at the same time, then solve the present-value equation. Treat the selected call date and call price as fixed. The common mistake to avoid is either omitting the final coupon or treating the call price as the entire year-2 cash flow.

At year 2 the total payment is $6+100=106$, so the scenario’s equation is $$ 100=\frac{6}{1+y}+\frac{106}{(1+y)^2}. $$

Let $x=1+y$. Multiplying through by $x^2$ gives $$ 100x^2-6x-106=0. $$

The discriminant is $36+4(100)(106)=42{,}436=206^2$. The positive root is $x=(6+206)/200=1.06$, hence $y=0.06$.

Check. At 6%, the present value is $$ \frac{6}{1.06}+\frac{106}{1.1236} =5.660377\ldots+94.339622\ldots =100.00. $$

Both cash flows are positive and due after valuation, so the present value is strictly decreasing in $y$. The 6% yield is therefore the only yield for this fixed call scenario, not merely the only root found by the quadratic calculation.

Worked Example: Fractional Payment Times and a Yield Boundary

At a valuation date between coupon dates, a specified redemption scenario has payments of $11 in 0.5 years and $133.10 in 1.5 years. The observed price is $110. Use an annual effective yield and restrict attention to $y\geq0$. Show that there is a unique yield, and verify it.

Method. Use the actual times from the valuation date, including the fractional years. Check the boundary value, then verify a candidate yield. The common mistake to avoid is rounding the half-year timing to a whole year.

At the zero-yield boundary, no discounting occurs: $$ V(0)=11+133.10=144.10. $$

At $y=21\%$, $1+y=1.21$, so $(1.21)^{0.5}=1.1$ and $(1.21)^{1.5}=1.331$. Thus $$ V(0.21)=\frac{11}{1.1}+\frac{133.10}{1.331} =10+100 =110. $$

The price is below $V(0)$, so the boundary and limit conditions ensure existence; strict decrease ensures uniqueness. The fractional-time calculation verifies that the unique yield is 21.00%.

What the Result Does and Does Not Establish

The proof applies after the cash-flow scenario has been fixed. For a callable bond, selecting a different redemption date generally changes the final coupon, redemption payment, or both. That produces a different present-value function. Each such function can have at most one admissible yield under the stated positivity assumptions, but uniqueness for one scenario does not imply that all scenarios have the same yield.

The boundary check is also essential. In the first example, the price was below the zero-yield value, so a nonnegative yield existed. If a price exceeds the value at the lowest permitted yield, increasing the yield can only lower present value further. There is then no solution in the permitted range, even if an equation-solving method returns a negative rate outside it.

Fix the scenario.
List every coupon and redemption payment included, and measure each payment time from the valuation date.

State the yield convention and permitted range.
For annual effective yield, use $C_j(1+y)^{-t_j}$; record any lower bound such as $y\geq0$.

Check the boundary value.
Compare the target price with the present value at the smallest permitted yield to determine whether a solution can exist.

Use monotonicity, then solve.
Positive payments at positive times make present value strictly decreasing, so any admissible solution is unique. Use algebra or numerical methods to locate it.

A common exam trap is to confuse “at most one” with “exactly one.” Strict decrease rules out two roots but does not guarantee any root. Another is to use whole-year exponents for payments between coupon dates. Finally, do not carry a yield equation from one assumed call date into another without rebuilding the cash flows: the uniqueness proof is scenario-specific.

Key takeaway: For a fixed scenario of positive future cash flows, annual-effective-yield present value is continuous and strictly decreasing. Check the target price against the value at the allowed yield boundary to decide whether the unique possible yield actually exists.

Check Your Understanding

For each problem, identify the admissible yield conclusion from the given fixed scenario. Unless stated otherwise, use annual effective yield and report rates to two decimal places where a rate is requested.

A scenario pays $20 at year 1 and $120 at year 2. Its price is $125, and $y\geq0$. Does an admissible yield exist, and how many? (Answer: Exactly one; $V(0)=\$140>\$125$, and $V(y)$ tends to zero.)
A fixed call scenario pays $8 at year 1 and $108 at year 2. Its price is $100, with $y\geq0$. Find the yield and state why it is unique. (Answer: 8.00%; it is unique because the scenario’s present value strictly decreases.)
A scenario pays $15 in 0.5 years and $90 in 1.5 years. Its price is $90, with $y\geq0$. Is there a unique yield, and can you bracket it between 0% and 21%? (Answer: Yes; $V(0)=\$105$, and $V(0.21)=15/1.1+90/1.331\approx\$81.25$, so the unique yield is between 0% and 21%.)
A fixed scenario pays $10 at year 1 and $130 at year 3. Its price is $140, with $y\geq0$. How many admissible yields are there? (Answer: Exactly one, at 0.00%, because $V(0)=\$140$ and strict decrease makes every positive yield produce a lower value.)
A scenario pays $20 at year 1 and $120 at year 2. Its price is $150, with $y\geq0$. Could a numerical solver find an admissible yield? (Answer: No; $V(0)=\$140<\$150$, and present value decreases for $y\geq0$, so no admissible solution exists.)

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