The Product Identity For Chained Forward Rates

Why Chained Forward Rates Have a Product Identity

A forward rate describes accumulation over a specified future interval. When several adjacent intervals are linked, the investment accumulates through each interval in turn. The total accumulation is therefore the product of the interval accumulation factors, not the sum of the quoted rates. The product identity makes this precise: chaining the forward factors from time zero to a maturity gives the same accumulation factor as the spot rate for that endpoint.

This identity is useful as both a valuation relationship and a calculation check. It works when the interval lengths are unequal, provided each annual-effective forward rate is raised to the length of its own interval. The key is to keep rates, time units, and accumulation factors attached to the correct maturities.

Definitions and Assumptions

Definition: Let $i_t$ be the effective annual spot rate for maturity $t$ years. Its accumulation factor from time zero to time $t$ is $A(t)=(1+i_t)^t$, with $A(0)=1$. For an interval from time $s$ to time $t$, where $0\leq s<t$, let $f_{s,t}$ be the annual-effective forward rate. Its accumulation factor over that interval is $(1+f_{s,t})^{t-s}$.

Assume a single consistent term structure of rates, no fees or taxes, and reinvestment at the quoted forward rate for each stated interval. Time is measured in years, and annual-effective rates are compounded fractionally when an interval has a non-integer number of years. These are accumulation-factor relationships; they do not require the rates to be equal across intervals.

For adjacent times $0=t_0<t_1<\cdots<t_n=T$, the forward rate on interval $j$ is $f_{t_{j-1},t_j}$. The product identity is:

\prod_{j=1}^{n}(1+f_{t_{j-1},t_j})^{t_j-t_{j-1}} =A(T) =(1+i_T)^T

In words, multiplying all the forward accumulation factors from time zero to $T$ gives the endpoint spot accumulation factor. For annual intervals, each interval has length one, so its factor is simply $1+f$.

Why the Product Identity Holds

A forward factor can be expressed as the ratio of the spot accumulation factor at the end of its interval to the spot accumulation factor at the beginning. In the notation above, this means:

(1+f_{s,t})^{t-s}=\frac{A(t)}{A(s)}

This relationship follows because investing to time $t$ directly must produce the same accumulation as investing to time $s$ and then accumulating from $s$ to $t$ at the implied forward rate. Multiplying these ratios across adjacent intervals makes each intermediate spot factor cancel.

\begin{aligned} \prod_{j=1}^{n}(1+f_{t_{j-1},t_j})^{t_j-t_{j-1}} &=\prod_{j=1}^{n}\frac{A(t_j)}{A(t_{j-1})}\\ &=\frac{A(t_1)}{A(t_0)} \frac{A(t_2)}{A(t_1)} \cdots \frac{A(T)}{A(t_{n-1})}\\ &=\frac{A(T)}{A(0)}\\ &=A(T)=(1+i_T)^T. \end{aligned}

The cancellation is the essential reason for the identity: the end of one interval is the beginning of the next. Only the starting accumulation factor and the final accumulation factor remain. Since $A(0)=1$, the product is the full accumulation from time zero to maturity $T$.

The same reasoning applies if the chain starts at a later time $s$. The product of forward factors from $s$ to $T$ is $A(T)/A(s)$, the accumulation from $s$ to $T$, rather than the full accumulation from time zero.

Worked Examples

Worked Example: Three Annual Forward Rates

A hypothetical deposit of $25,000 earns an annual-effective forward rate of 2% from year 0 to year 1, 3% from year 1 to year 2, and 4% from year 2 to year 3. Find the accumulated value at year 3 and state the endpoint spot accumulation factor it implies.

Method. Multiply the three one-year accumulation factors. Each interval has length one year, so each factor is $1+f$.

The three factors are $1.02$, $1.03$, and $1.04$. First multiply the first two:

$$ 1.02(1.03)=1.0506 $$

Then multiply by the third factor:

$$ 1.0506(1.04)=1.092624 $$

Thus, the three-year accumulation factor is $1.092624$. The account value is:

\$25{,}000(1.092624)=\$27{,}315.60

The endpoint spot rate $i_3$ is defined so that its three-year accumulation factor equals this product: $(1+i_3)^3=1.092624$. This is the product identity in this example. No separate endpoint-rate calculation is needed to establish the accumulation.

Check. The final value exceeds the original deposit, as expected for three positive forward rates. A common mistake is to add the rates and calculate $\$25{,}000(1+0.02+0.03+0.04)=\$27{,}250$. That is incorrect because it omits the interest earned on earlier interest. The correct method multiplies the interval factors.

Worked Example: Forward Rates on Unequal-Length Intervals

A hypothetical investment of $12,000 earns an annual-effective forward rate of 3% from year 0 to year 2, followed by an annual-effective forward rate of 4% from year 2 to year 5. Find the value at year 5 and its endpoint spot accumulation factor.

Method. Raise each annual-effective factor to the number of years in its own interval. The first interval is two years and the second is three years.

The first interval factor is:

$$ (1.03)^2=1.0609 $$

The second interval factor is:

$$ (1.04)^3=1.124864 $$

Multiply the two factors:

$$ 1.0609(1.124864)=1.1933682176 $$

Therefore, the account value at year 5 is:

\$12{,}000(1.1933682176)=\$14{,}320.4186112

Rounded to the nearest cent, the accumulated value is $14,320.42. The product $1.1933682176$ is also the five-year endpoint spot accumulation factor, so $(1+i_5)^5=1.1933682176$.

Check. The balance after the first two years is $\$12{,}000(1.0609)=\$12{,}730.80$. Accumulating that balance for the remaining three years gives $\$12{,}730.80(1.124864)=\$14{,}320.4186112$, the same value. A common mistake is to use $1.03$ and $1.04$ just once each. Those are annual-effective rates, and the intervals last two and three years, respectively.

Worked Example: Chaining Continuous Forward Forces

A hypothetical fund of $5,000 accumulates continuously at force of interest 2% from time 0 to year 1, 4% from year 1 to year 3, and 3% from year 3 to year 4. Find the year-4 value and the equivalent endpoint force over the full four years.

Method. For a constant force $\delta$ over an interval of length $t$, the accumulation factor is $e^{\delta t}$. The interval lengths are 1, 2, and 1 years.

The factors are $e^{0.02}$, $e^{0.08}$, and $e^{0.03}$. Their product is:

e^{0.02}e^{0.08}e^{0.03} =e^{0.02+0.08+0.03} =e^{0.13} \approx 1.138828

The final value, rounded to the nearest cent, is:

\$5{,}000e^{0.13} \approx \$5{,}694.14

Let $\bar{\delta}$ be the constant force that produces the same accumulation over four years. It must satisfy $e^{4\bar{\delta}}=e^{0.13}$, so $4\bar{\delta}=0.13$ and $\bar{\delta}=0.0325$, or 3.25% per year.

Check. The interval forces contribute $0.02(1)+0.04(2)+0.03(1)=0.13$ to the exponent. The endpoint force gives $0.0325(4)=0.13$, so both descriptions produce the same accumulation factor. This continuous form uses the same chaining principle: multiply exponential factors, which adds their exponents.

Annual-Effective and Continuous Forms

The product identity depends on the accumulation factors, not on adding the quoted rates. For annual-effective forwards over intervals of different lengths, the factors have the form $(1+f)^{\text{length}}$. For continuous forward forces, the factors have the form $e^{\delta\,\text{length}}$. In either case, the factors across adjacent intervals combine into the factor over the full horizon.

Theorem: For adjacent intervals covering time zero to $T$, the product of the annual-effective forward accumulation factors equals $(1+i_T)^T$, the endpoint spot accumulation factor. Under continuous accumulation, the product of the interval factors $e^{\delta_j(t_j-t_{j-1})}$ equals the endpoint factor $e^{\delta_T T}$, where $\delta_T$ is the constant force producing that same total accumulation over $T$.

This is an equality of accumulation factors. The spot rate and the sequence of forward rates are different ways to describe accumulation over the same horizon; the identity does not say that their quoted rates should be added, or that all interval rates must match.

Common Errors and Practical Checks

When applying the identity, first mark the interval boundaries and write the factor for each interval. Then check that the product covers the entire horizon with no gap or overlap. For annual-effective rates, confirm that the exponent on each factor equals the interval length in years. For continuous forces, confirm that each force is multiplied by its interval length in the exponent.

Adding the rates. Chained accumulation compounds, so the factors are multiplied. Earlier interest remains invested and can earn interest in later intervals.
Ignoring interval length. A 4% annual-effective forward rate over three years contributes $(1.04)^3$, not $1.04$.
Mixing conventions. Do not treat an annual-effective rate as a continuous force. Convert conventions before comparing accumulation factors.
Stopping at an intermediate date. The product equals the spot accumulation factor at the final endpoint only when the intervals form a complete adjacent chain from time zero to that endpoint.

A reliable final check is to compare the product of the interval factors with the endpoint spot factor on the same time basis. If an initial principal is given, multiplying either factor by that principal must produce the same accumulated value.

Check Your Understanding

For each problem, form the interval accumulation factors first. Round dollar answers to the nearest cent.

A hypothetical account starts with $8,000 and earns annual-effective forward rates of 1.5%, 2.5%, and 3% over three consecutive one-year intervals. What is the year-3 value? (Answer: $8,572.69)
A hypothetical investment of $10,000 earns 2% per year from year 0 to year 2 and 4% per year from year 2 to year 5, with annual-effective forward rates. Find the year-5 value. (Answer: $11,703.09)
A hypothetical principal of $20,000 earns annual-effective forward rates of 1%, -2%, and 3% over three consecutive one-year intervals. Find the final accumulation factor and account value. (Answer: factor $1.019494$; value $20,389.88)
A hypothetical $10,000 fund earns continuously at force 1% from time 0 to year 2, 2.5% from year 2 to year 3, and 2% from year 3 to year 6. Find the year-6 value. (Answer: $11,107.11)
Three consecutive interval accumulation factors are 1.06, 1.08, and 0.97. A hypothetical $5,000 is invested at time zero. Find the endpoint accumulation factor and accumulated value. (Answer: factor $1.110456$; value $5,552.28)

What You'll Learn