Force of Interest
The force of interest \(\delta_t\) is what a nominal rate becomes when you compound infinitely often. Integrate it and you recover the whole accumulation history — even when the rate itself is changing.
By the end you'll be able to recover the accumulation function \(a(t)\) from a constant or time-varying force via \(a(t) = \exp\left(\int_0^t \delta_s\, ds\right)\), and convert a constant force to an effective annual rate \(i\) with \(e^{\delta} = 1 + i\).
Predict: if you double a constant force \(\delta\) (set b = 0, drag a), does the 1-year accumulation factor double too? Drag the slider and check the readout.
The force is \(\delta_t = a + bt\) — a straight line you shape with the two sliders. Set b = 0 for a constant force. The shaded area under \(\delta_t\) from 0 to t is exactly \(\int_0^t \delta_s\, ds\), and the right-hand chart plots \(a(t) = \exp\) of that shaded area — drag t and watch both panels move together.
δ_t = 0.050 + 0.000·t (constant δ = 0.050) · a(3.0) = exp(0.150) = 1.162 · i = e^δ−1 = 5.13%
The force of interest packs the whole shape of an accumulation function \(a(t)\) into a single instantaneous rate \(\delta_t\), and integrating that rate back up recovers \(a(t)\) exactly — for a constant rate or a rate that drifts over time.
Take a nominal rate and compound it more and more often — quarterly, monthly, daily, every second. As the number of compounding periods grows without bound, the rate you're compounding settles down to a fixed number: the force of interest \(\delta\). It's what "continuous compounding" means in practice — interest credited every instant, not just at the end of each period.
\(\delta_t\) is defined as the fund's growth rate per unit of current balance: \(\delta_t = a'(t)/a(t) = \dfrac{d}{dt}\ln a(t)\). Integrating this differential relationship from 0 to t recovers the master formula \(a(t) = \exp\left(\int_0^t \delta_s\, ds\right)\) — it works for any pattern of interest, because \(\delta_t\) is defined pointwise at every instant. When \(\delta_t = \delta\) is constant, the integral collapses to \(\delta t\), so \(a(t) = e^{\delta t}\); setting \(t=1\) gives the equivalence \(e^{\delta} = 1+i\), i.e. \(\delta = \ln(1+i)\) and \(i = e^{\delta}-1\). Discounting runs the same integral with a minus sign: the present-value factor over \([0,t]\) is \(\exp(-\int_0^t \delta_s\, ds)\).
A fund manager quoting "5% continuously compounded" is quoting a force of interest, not an effective rate — before comparing it to a competitor's 5.1% effective annual rate, you have to convert: \(i = e^{0.05}-1 \approx 5.127\%\), which beats the 5.1% quote. When a rate drifts over the year (a central bank easing policy, say), you can't shortcut the conversion — you integrate the whole path \(\delta_t\) to get the true accumulation.
Constant force \(\delta = 0.05\). Then \(i = e^{0.05}-1 \approx\) 5.127%, and 1000 grows in 3 years to \(1000\cdot e^{0.05 \cdot 3} = 1000\cdot e^{0.15} \approx\) 1161.83.
Same idea, a different shape: \(\delta_t = 0.02 + 0.005t\). First find the accumulated value at \(t=2\): \(\int_0^2 (0.02+0.005t)\,dt = 0.02(2) + 0.005\cdot 2^2/2 = 0.04 + 0.01 = 0.05\), so \(a(2) = e^{0.05} \approx 1.0513\). Now finish it for \(t=6\): \(\int_0^6 (0.02+0.005t)\,dt = 0.02(6) + 0.005\cdot 6^2/2 = 0.12 + 0.09 = \) ____, so \(a(6) = e^{(\text{that sum})} \approx\) ____
Reveal the answer
\(0.12 + 0.09 = 0.21\), so \(a(6) = e^{0.21} \approx\) 1.2337. Set a = 0.02 and b = 0.005 on the sliders above, drag t to 6, and check the right-hand chart's readout against this value.
More info — why you can't skip the integral when δ_t varies
A common slip is plugging a single value of t into \(e^{\delta_t \cdot t}\) instead of integrating \(\int_0^t \delta_s\, ds\). That shortcut only works because a constant force gives \(\int_0^t \delta\, ds = \delta t\) — the integral of a constant really is "rate times time." The moment \(\delta_s\) itself depends on s, the area under the curve is no longer a rectangle; it's whatever shape \(\delta_s\) traces out, and you have to add up (integrate) every instant's contribution separately. See the AnalystPrep link in Dive deeper for more practice recovering \(a(t)\) this way.
Check your understanding
Constant force of interest \(\delta = 0.08\). What is the equivalent annual effective rate \(i\)?
With a time-varying force \(\delta_t = 0.03 + 0.02t\), what is \(a(4) = \exp\left(\int_0^4 \delta_s\, ds\right)\)?
As the compounding frequency \(m\) in a nominal rate \(i^{(m)}\) grows without bound, what does \(i^{(m)}\) converge to?
You double a constant force of interest \(\delta\). Does the 1-year accumulation factor \(a(1) = e^{\delta}\) double too?
Recap
- \(\delta_t = a'(t)/a(t) = \dfrac{d}{dt}\ln a(t)\), and \(a(t) = \exp\left(\int_0^t \delta_s\, ds\right)\) recovers the accumulation from the force — constant or time-varying.
- Constant force: \(a(t)=e^{\delta t}\), with equivalence \(e^{\delta}=1+i\), so \(\delta=\ln(1+i)\) and \(i=e^{\delta}-1\). The discount factor is \(v=e^{-\delta}\).
- \(\delta\) is the limit of a nominal rate \(i^{(m)}\) as \(m\to\infty\) — continuous compounding.
- With a time-varying \(\delta_t\), you must integrate the whole path — plugging a single t into \(\delta_t\) only works when the force is constant.
Dive deeper
- Wikibooks — Financial Math FM/Formulas Confirm δ_t=d/dt ln a(t) and δ=ln(1+i).
- AnalystPrep — Time Value of Money (SOA Exam FM Study Notes) Practice recovering a(t) from a variable force via a(t)=exp(∫₀ᵗ δ_s ds).
- Introduction to Computational Finance (bookdown) — The Time Value of Money See continuous compounding derived as the m→∞ limit FV=V·e^(Rn).
Sources
- The Force of Interest