Continuous Forward Rates From Spot Rates

Why Spot Rates Determine an Interval Forward Force

A spot rate describes accumulation from time 0 to one particular maturity. A forward force describes accumulation over an interval that begins after time 0. To connect the two, compare the accumulation factors to the interval’s start and end dates. The ratio removes the accumulation before the start date and leaves only the factor for the interval.

This tutorial uses continuously compounded spot rates. Time is measured in years, and each spot force is annualized over the time from 0 to its maturity. Rates and maturities must use consistent time units. The central task is not to subtract the two spot rates: it is to subtract their accumulated exponents and divide by the interval length.

Definition: Let $\delta_t$ be the continuously compounded spot force for maturity $t>0$. A unit invested at time 0 accumulates to $e^{t\delta_t}$ at time $t$. For $0\le s<t$, the continuously compounded forward force $f_{s,t}$ is the constant annual force over the interval from $s$ to $t$ whose accumulation factor equals the ratio of the spot accumulation factors: $$ e^{(t-s)f_{s,t}}=\frac{e^{t\delta_t}}{e^{s\delta_s}}. $$ For $s=0$, the starting accumulation factor is 1, so the same relation applies with $e^{s\delta_s}=1$.

The definition assumes positive accumulation factors, as required for the logarithm used to solve for the force. Exponential accumulation factors are positive for any finite real force, including negative forces. The spot forces need not be positive.

Deriving the Forward-Force Formula

Take natural logarithms of both sides of the defining relation. The logarithm turns each continuously compounded accumulation factor into its accumulated exponent. Dividing by the elapsed time then gives the annualized force for the interval.

\begin{aligned} e^{(t-s)f_{s,t}}&=\frac{e^{t\delta_t}}{e^{s\delta_s}}\\ e^{(t-s)f_{s,t}}&=e^{t\delta_t-s\delta_s}\\ (t-s)f_{s,t}&=t\delta_t-s\delta_s\\ f_{s,t}&=\frac{t\delta_t-s\delta_s}{t-s}. \end{aligned}

The units provide a useful check. The products $t\delta_t$ and $s\delta_s$ are accumulated exponents over their respective maturities. Their difference is the exponent for the interval, and dividing by $t-s$, measured in years, gives an annual force.

Theorem: Given continuously compounded spot forces $\delta_s$ and $\delta_t$ at maturities $s$ and $t$, with $0\le s<t$, the continuously compounded forward force over $[s,t]$ is $$ f_{s,t}=\frac{t\delta_t-s\delta_s}{t-s}. $$ Equivalently, the interval accumulation factor is $$ e^{(t-s)f_{s,t}}=e^{t\delta_t-s\delta_s}. $$

If $s=0$, interpret the accumulated exponent at time 0 as zero. The formula becomes $f_{0,t}=\delta_t$: over an interval starting immediately, the forward force is the spot force for that maturity. If $s>0$, both endpoint spot forces matter. The interval force is generally not equal to either endpoint spot force, nor to their simple difference.

Calculating a Forward Force from Two Spot Forces

Worked Example: A Forward Force from Years 1 and 4

A hypothetical term structure has continuously compounded spot forces $\delta_1=0.021$ and $\delta_4=0.033$. Find the implied continuously compounded forward force from year 1 to year 4, and check it using the interval accumulation factor. State the rate as a percentage per year.

Method. Use the accumulated spot exponents at both endpoints, subtract the exponent at year 1 from the exponent at year 4, and divide by the three-year interval.

At year 1, the accumulated exponent is $1(0.021)=0.021$. At year 4, it is $4(0.033)=0.132$. The interval exponent is therefore $0.132-0.021=0.111$. The interval length is $4-1=3$ years, so

f_{1,4}=\frac{4(0.033)-1(0.021)}{4-1} =\frac{0.132-0.021}{3} =\frac{0.111}{3} =0.037.

The forward force is $0.037$, or $3.70\%$ per year. Its three-year accumulation factor is $e^{3(0.037)}=e^{0.111}$. The ratio of the spot accumulation factors is $e^{4(0.033)}/e^{1(0.021)}=e^{0.132-0.021}=e^{0.111}$, so the check agrees.

Common mistake avoided. Subtracting the spot forces gives $0.033-0.021=0.012$, or $1.20\%$, which is not the interval force. The spot forces apply to different full maturities; their accumulated exponents, not the rates alone, must be compared.

Unequal and Fractional Maturities

The same calculation applies when maturities are fractional or the interval is not a whole number of years. Keep all time values in years, use the stated annual spot forces, and divide by the exact elapsed time. Do not round the interval length or either accumulated exponent early.

Worked Example: A Forward Force Over a Fractional Interval

A hypothetical curve has continuously compounded spot forces $\delta_{1.5}=0.024$ and $\delta_4=0.031$. Find the continuously compounded forward force from year 1.5 to year 4. Check the resulting interval factor against the spot-factor ratio.

Method. First calculate each maturity’s accumulated exponent. The exponents are $1.5(0.024)=0.036$ and $4(0.031)=0.124$. Thus the exponent over the interval is $0.124-0.036=0.088$. The interval lasts $4-1.5=2.5$ years, so

f_{1.5,4} =\frac{4(0.031)-1.5(0.024)}{4-1.5} =\frac{0.124-0.036}{2.5} =\frac{0.088}{2.5} =0.0352.

The forward force is $0.0352$, or $3.52\%$ per year. The interval accumulation factor is $e^{2.5(0.0352)}=e^{0.088}$. The spot-factor ratio is $e^{0.124}/e^{0.036}=e^{0.088}$, confirming the result.

Common mistake avoided. Dividing the exponent difference by 4 would annualize over the full maturity, not over the interval from year 1.5 to year 4. The denominator must be $2.5$, the elapsed interval length.

Negative Forward Forces and Accumulation

A forward force can be negative even when the spot forces used to calculate it are positive. The sign depends on the difference between the endpoint accumulated exponents, $t\delta_t-s\delta_s$, not on whether each spot force is individually above or below zero. A negative force means that the interval accumulation factor is less than 1. It does not mean the factor is negative.

Worked Example: A Negative Forward Force and an Investment Value

A hypothetical curve has continuously compounded spot forces $\delta_2=0.040$ and $\delta_5=0.010$. Find the forward force from year 2 to year 5. Then calculate the value at year 5 of $\$10{,}000$ invested at year 2 under that implied interval force. Round the value to the nearest cent.

Method. Calculate the accumulated spot exponents at years 2 and 5: $2(0.040)=0.080$ and $5(0.010)=0.050$. The interval exponent is $0.050-0.080=-0.030$. The interval lasts $5-2=3$ years, giving

f_{2,5} =\frac{5(0.010)-2(0.040)}{5-2} =\frac{0.050-0.080}{3} =\frac{-0.030}{3} =-0.010.

The forward force is $-0.010$, or $-1.00\%$ per year. Over three years the accumulation factor is $e^{3(-0.010)}=e^{-0.030}$. Therefore, the value of the investment is

\$10{,}000e^{-0.030} =\$10{,}000(0.9704455\ldots) =\$9{,}704.46.

The check from spot factors gives $e^{5(0.010)}/e^{2(0.040)}=e^{0.050-0.080}=e^{-0.030}$, the same interval factor. The value is rounded only at the final step, to the nearest cent.

Common mistake avoided. A negative force does not produce a negative account value. The exponential factor $e^{-0.030}$ is positive and less than 1, so the investment decreases over the interval.

Common Calculation Traps

Forward-force questions often provide enough information to make the calculation look like a simple comparison of rates. The safe approach is to write the endpoint exponents and the interval length before substituting numbers.

Subtracting spot forces directly. The expression $\delta_t-\delta_s$ ignores that each force applies over a different maturity. Use $t\delta_t-s\delta_s$ in the numerator.
Using the wrong denominator. Divide by $t-s$, not by $t$, unless the interval actually begins at time 0.
Reversing the ratio. The interval factor is the accumulation to $t$ divided by the accumulation to $s$. Reversing it changes the sign of the exponent.
Mixing time units. If maturities are expressed in years and forces are annual, keep the calculation in years. Convert months to years before using the formula.
Rounding before the final step. Retain the full calculated exponent or force through the accumulation-factor check, then round the requested final result.

Write the endpoint data.
Record $s$, $t$, $\delta_s$, and $\delta_t$, with $s<t$ and time measured consistently.

Calculate the accumulated exponents.
Compute $s\delta_s$ and $t\delta_t$; use zero for the starting exponent if $s=0$.

Find and annualize the interval exponent.
Subtract the starting exponent from the ending exponent, then divide by $t-s$.

Check the factor.
Verify that $e^{(t-s)f_{s,t}}=e^{t\delta_t}/e^{s\delta_s}$, preserving the order of the ratio.

Why the Logarithmic Method Is Useful

Continuously compounded accumulation is exponential: a constant force $f$ over $t-s$ years gives the factor $e^{(t-s)f}$. Taking logarithms makes the accumulation exponents additive across time and lets a quotient of factors become a difference of exponents. That is why the forward-force formula uses maturity-weighted spot forces. It also explains why the method works for noninteger maturities and for negative forces without requiring a separate formula.

For exam calculations, keep the distinction clear: $\delta_s$ and $\delta_t$ are spot forces measured from time 0, while $f_{s,t}$ is the annualized force only over the interval from $s$ to $t$. This distinction is central to questions aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, use continuously compounded annual spot forces and time in years. Keep full precision until the requested final result.

A hypothetical curve has $\delta_1=0.020$ and $\delta_3=0.030$. Find $f_{1,3}$, as a decimal and a percentage per year. (Answer: $0.035$, or $3.50\%$ per year)
A hypothetical curve has $\delta_2=0.025$ and $\delta_6=0.030$. Find the forward force from year 2 to year 6. (Answer: $0.0325$, or $3.25\%$ per year)
A hypothetical curve has $\delta_{1.5}=0.040$ and $\delta_{3.5}=0.025$. Find the forward force from year 1.5 to year 3.5. (Answer: $0.01375$, or $1.375\%$ per year)
A hypothetical curve has $\delta_2=0.030$ and $\delta_5=0.018$. Find the forward force from year 2 to year 5 and the year-5 value of $\$5{,}000$ invested at year 2. Round the value to the nearest cent. (Answer: $0.010$, or $1.00\%$ per year; $\$5{,}152.27$)

What You'll Learn