Force of interest varying with time

Welcome to Force of Interest Varying with Time

In your earlier studies, you likely worked with a constant rate of interest or a constant force of interest \( \delta \). That is very clean and convenient, but in the real financial world, interest rates change constantly due to inflation expectations, economic cycles, and central bank policy. That means the instantaneous rate of return—the force of interest—often depends on time, written as a function \( \delta(t) \).

Don't worry if calculus usually feels intimidating! In this chapter, we will break down step-by-step how to move from a time-varying force of interest \( \delta(t) \) to practical accumulation factors, discount factors, and equivalent annual effective interest rates.

Quick syllabus check: This chapter sits within the Theory of interest rates topic of CM1. We focus specifically on how accumulation and discounting work when the force of interest varies over time, and how to find the equivalent annual effective rate implied by that accumulation.

1. What is the Force of Interest as a Function of Time?

Let's recap the fundamental intuition. The force of interest at time \( t \), denoted by \( \delta(t) \), measures the nominal rate of growth per unit of time operating continuously over an infinitesimal moment around time \( t \).

The Differential Relationship

Suppose you invest an initial sum of money. Let \( A(t) \) denote the accumulated value of this investment at time \( t \). Over an tiny time step \( dt \), the increase in your fund is proportional to:

1. The size of the fund already accumulated, \( A(t) \)
2. The length of time elapsed, \( dt \)
3. The instantaneous rate of growth operating at that exact moment, \( \delta(t) \)

In mathematical terms:

\( dA(t) = A(t) \cdot \delta(t) \, dt \)

Rearranging this gives the formal definition of the variable force of interest:

\( \delta(t) = \frac{1}{A(t)} \frac{d}{dt}A(t) = \frac{d}{dt}\ln[A(t)] \)

Everyday Analogy: Think of \( A(t) \) as the odometer on your car showing total distance covered, and \( \delta(t) \) as your speedometer at minute \( t \). If your speed changes throughout the trip, you cannot simply multiply a single speed by the total time. Instead, you integrate your speed function over time to find out how far you've gone!

2. Accumulation Factors with a Varying Force of Interest

Let's define the accumulation factor from time \( t_1 \) to time \( t_2 \) (where \( t_1 \le t_2 \)), denoted by \( A(t_1, t_2) \). This factor tells you how much an investment of 1 unit at time \( t_1 \) grows to by time \( t_2 \).

Deriving the Formula

Starting from our differential equation:

\( \frac{d}{dt}\ln[A(t)] = \delta(t) \)

Integrating both sides with respect to \( t \) between limits \( t_1 \) and \( t_2 \):

\( \int_{t_1}^{t_2} \frac{d}{dt}\ln[A(t)] \, dt = \int_{t_1}^{t_2} \delta(t) \, dt \)

\( \ln[A(t_2)] - \ln[A(t_1)] = \int_{t_1}^{t_2} \delta(t) \, dt \)

\( \ln\left( \frac{A(t_2)}{A(t_1)} \right) = \int_{t_1}^{t_2} \delta(t) \, dt \)

Taking the exponential of both sides gives the golden rule of accumulation:

\( A(t_1, t_2) = \frac{A(t_2)}{A(t_1)} = \exp\left( \int_{t_1}^{t_2} \delta(t) \, dt \right) \)

When measuring accumulation from time 0 to time \( t \), we write:

\( A(0, t) = \exp\left( \int_0^t \delta(s) \, ds \right) \)

Key Takeaway: Whenever you want to accumulate money forward under a varying force of interest, you simply integrate the force of interest over the relevant interval and raise \( e \) to that power.

3. Discount Factors with a Varying Force of Interest

Discounting is the exact inverse of accumulation. The discount factor from time \( t_2 \) back to time \( t_1 \) (with \( t_1 \le t_2 \)), denoted by \( v(t_1, t_2) \), is the present value at time \( t_1 \) of 1 unit payable at time \( t_2 \).

Because discounting undoes accumulation:

\( v(t_1, t_2) = \frac{1}{A(t_1, t_2)} = \exp\left( -\int_{t_1}^{t_2} \delta(t) \, dt \right) \)

When discounting back to time 0 from time \( t \), the factor is often written as \( v(t) \):

\( v(t) = \exp\left( -\int_0^t \delta(s) \, ds \right) \)

Memory Aid: Accumulate goes up (positive power: \( + \int \delta \)), discount goes down (negative power: \( - \int \delta \)).

4. Two Common Exam Formats for \( \delta(t) \)

In CM1 exam questions, \( \delta(t) \) typically appears in one of two ways:

1. As a piecewise constant function (step function)
2. As a smooth continuous mathematical function of \( t \) (such as linear, polynomial, or exponential functions)

Case A: Piecewise Constant (Step Function)

Suppose the force of interest is constant across specific time intervals:

\( \delta(t) = \delta_1 \) for \( 0 \le t \le t_1 \)
\( \delta(t) = \delta_2 \) for \( t_1 < t \le t_2 \)

The accumulation from \( 0 \) to \( t_2 \) splits neatly into separate product terms:

\( A(0, t_2) = \exp\left( \int_0^{t_1} \delta_1 \, dt + \int_{t_1}^{t_2} \delta_2 \, dt \right) = \exp(\delta_1 t_1) \cdot \exp(\delta_2 (t_2 - t_1)) \)

Case B: Smooth Continuous Function of \( t \)

Here, \( \delta(t) \) is given by a formula such as \( \delta(t) = a + b t \) or \( \delta(t) = a + b t^2 \). To solve these:

Step 1: Set up the definite integral \( \int_{t_1}^{t_2} \delta(t) \, dt \).
Step 2: Evaluate the integral carefully using standard calculus rules.
Step 3: Substitute the result into \( \exp(\dots) \) for accumulation, or \( \exp(-\dots) \) for discounting.

5. Step-by-Step Worked Examples

Worked Example 1: Continuous Function of Time

Question: The force of interest at time \( t \) (in years) is given by:

\( \delta(t) = 0.03 + 0.008 t \quad \text{for } t \ge 0 \)

Calculate:
(a) The accumulated value at time \( t = 5 \) of an investment of £2,000 made at time \( t = 0 \).
(b) The present value at time \( t = 2 \) of a payment of £5,000 due at time \( t = 6 \).

Solution:

Part (a): Accumulating £2,000 from \( t = 0 \) to \( t = 5 \).
First, evaluate the integral of \( \delta(t) \) from \( 0 \) to \( 5 \):

\( \int_0^5 (0.03 + 0.008t) \, dt = \left[ 0.03t + 0.004t^2 \right]_0^5 \)

\( = (0.03 \times 5 + 0.004 \times 5^2) - 0 = 0.15 + 0.10 = 0.25 \)

Now, calculate the accumulation factor:

\( A(0, 5) = \exp(0.25) \approx 1.284025 \)

The accumulated value of £2,000 is:

\( 2{,}000 \times A(0, 5) = 2{,}000 \times 1.284025 = £2{,}568.05 \)

Part (b): Discounting £5,000 from \( t = 6 \) back to \( t = 2 \).
First, integrate \( \delta(t) \) between limits \( 2 \) and \( 6 \):

\( \int_2^6 (0.03 + 0.008t) \, dt = \left[ 0.03t + 0.004t^2 \right]_2^6 \)

\( = (0.03 \times 6 + 0.004 \times 36) - (0.03 \times 2 + 0.004 \times 4) \)

\( = (0.18 + 0.144) - (0.06 + 0.016) = 0.324 - 0.076 = 0.248 \)

Since we are finding the present value, we apply the discount factor:

\( v(2, 6) = \exp(-0.248) \approx 0.780362 \)

The present value at time \( t = 2 \) is:

\( 5{,}000 \times v(2, 6) = 5{,}000 \times 0.780362 = £3{,}901.81 \)

Worked Example 2: Piecewise-Defined Force of Interest

Question: An investor places £10,000 into a fund at time \( t = 0 \). The force of interest is defined as:

\( \delta(t) = \begin{cases} 0.04 & 0 \le t \le 3 \\ 0.01 + 0.01t & 3 < t \le 7 \end{cases} \)

Find the value of the investment at time \( t = 7 \).

Solution:
We need the accumulation factor \( A(0, 7) \). Because the definition of \( \delta(t) \) changes at \( t = 3 \), we split the integral into two parts:

\( \int_0^7 \delta(t) \, dt = \int_0^3 0.04 \, dt + \int_3^7 (0.01 + 0.01t) \, dt \)

Evaluating the first interval:

\( \int_0^3 0.04 \, dt = [0.04t]_0^3 = 0.12 \)

Evaluating the second interval:

\( \int_3^7 (0.01 + 0.01t) \, dt = \left[ 0.01t + 0.005t^2 \right]_3^7 \)

\( = (0.01 \times 7 + 0.005 \times 49) - (0.01 \times 3 + 0.005 \times 9) \)

\( = (0.07 + 0.245) - (0.03 + 0.045) = 0.315 - 0.075 = 0.240 \)

Summing the two integrals:

\( \int_0^7 \delta(t) \, dt = 0.12 + 0.24 = 0.36 \)

Thus, the overall accumulation factor is:

\( A(0, 7) = \exp(0.36) \approx 1.433329 \)

The accumulated value of £10,000 at time 7 is:

\( 10{,}000 \times 1.433329 = £14{,}333.29 \)

6. The Equivalent Annual Effective Rate of Interest

In Objective 1.1 of the CM1 syllabus, candidates are explicitly required to calculate:

"the equivalent annual rate implied by accumulation over a period where the force of interest is a function of time."

What Does This Mean?

Suppose an investment accumulates from time \( t_1 \) to time \( t_2 \) (a total duration of \( n = t_2 - t_1 \) years) under a variable force of interest \( \delta(t) \).
The equivalent constant annual effective rate of interest, denoted by \( i \), is the single, constant effective annual rate that would yield exactly the same overall accumulation over that exact \( n \)-year term.

The Formula

Under a constant annual effective rate \( i \), 1 unit invested for \( n \) years grows to \( (1 + i)^n \).
Under the varying force of interest, 1 unit grows to \( \exp\left( \int_{t_1}^{t_2} \delta(t) \, dt \right) \).

Equating the two expressions:

\( (1 + i)^n = \exp\left( \int_{t_1}^{t_2} \delta(t) \, dt \right) \)

Solving for \( i \):

\( 1 + i = \left[ \exp\left( \int_{t_1}^{t_2} \delta(t) \, dt \right) \right]^{\frac{1}{n}} = \exp\left( \frac{1}{n} \int_{t_1}^{t_2} \delta(t) \, dt \right) \)

\( i = \exp\left( \frac{1}{n} \int_{t_1}^{t_2} \delta(t) \, dt \right) - 1 \)

Did You Notice? The quantity \( \frac{1}{n} \int_{t_1}^{t_2} \delta(t) \, dt \) is simply the average force of interest over the period, often denoted by \( \bar{\delta} \). Therefore:

\( 1 + i = \exp(\bar{\delta}) \implies i = e^{\bar{\delta}} - 1 \)

This provides an intuitive shortcut: find the average force of interest across the time window, and convert that single average force to an effective rate!

Worked Example 3: Equivalent Annual Effective Rate

Question: Using the force of interest from Worked Example 1, where \( \delta(t) = 0.03 + 0.008t \):
Find the equivalent annual effective rate of interest over the 5-year period from \( t = 0 \) to \( t = 5 \).

Solution:
From Worked Example 1, we know the accumulation over the 5-year period is:

\( A(0, 5) = \exp\left( \int_0^5 \delta(t) \, dt \right) = \exp(0.25) \approx 1.284025 \)

Here, the term is \( n = 5 \) years. We set up the equation of equivalence:

\( (1 + i)^5 = 1.284025 \)

Taking the 5th root on both sides:

\( 1 + i = (1.284025)^{1/5} = (e^{0.25})^{0.2} = e^{0.05} \approx 1.051271 \)

\( i = 1.051271 - 1 = 0.051271 \text{ or } 5.127\% \)

Sanity check: The average force of interest over the period is \( \bar{\delta} = \frac{0.25}{5} = 0.05 \). The equivalent effective rate is \( e^{0.05} - 1 = 5.127\% \). Since effective rates are always slightly higher than their continuous force counterparts for positive rates, \( 5.127\% > 5\% \) makes total sense!

6. Evaluating General Cashflows Under a Varying Force

Once you can calculate \( v(t) \), finding the present value of cashflows under a varying force is a direct application of standard discounting principles:

• Discrete cashflows: For discrete payments \( C_k \) at times \( t_k \), the present value at time 0 is:

\( \text{PV} = \sum_k C_k \cdot v(t_k) = \sum_k C_k \cdot \exp\left( -\int_0^{t_k} \delta(t) \, dt \right) \)

• Continuous cashflows: For a continuous payment stream at rate \( \rho(t) \) from time \( a \) to \( b \), the present value is:

\( \text{PV} = \int_a^b \rho(t) \cdot v(t) \, dt = \int_a^b \rho(t) \cdot \exp\left( -\int_0^t \delta(s) \, ds \right) dt \)

(Detailed integration of variable continuous payment streams is explored further in the Cashflow Streams chapter.)

7. Common Exam Pitfalls to Avoid

Examiners regularly test this topic in both Paper A and Paper B. Watch out for these frequent mistakes:

• Forgetting the limits when integrating: When finding accumulation between \( t_1 \) and \( t_2 \), remember to integrate from \( t_1 \) to \( t_2 \), NOT from 0 to \( t_2 - t_1 \), unless \( \delta(t) \) is strictly a function of the time elapsed rather than calendar time.

• Sign errors in the exponent: Remember that discounting requires a negative power: \( \exp\left(-\int \delta \, dt\right) \). A positive sign accumulates forward.

• Mixing up annual rate \( i \) and force \( \delta \): \( 1 + i = \exp(\bar{\delta}) \), which means \( i = e^{\bar{\delta}} - 1 \). Do not simply state \( i = \bar{\delta} \).

• Calculus integration mistakes: When integrating powers of \( t \), don't forget to divide by the new power: \( \int b t \, dt = \frac{b t^2}{2} \). A surprising number of marks are lost to simple arithmetic errors under exam pressure!

8. Summary of Key Formulas

Keep these core formulas front and centre during your revision:

• Definition of force: \( \delta(t) = \frac{1}{A(t)} \frac{d}{dt}A(t) = \frac{d}{dt}\ln[A(t)] \)

• Accumulation factor: \( A(t_1, t_2) = \exp\left( \int_{t_1}^{t_2} \delta(t) \, dt \right) \)

• Discount factor: \( v(t_1, t_2) = \exp\left( -\int_{t_1}^{t_2} \delta(t) \, dt \right) \)

• Equivalent annual effective rate over \( n \) years: \( i = \exp\left( \frac{1}{n} \int_{t_1}^{t_2} \delta(t) \, dt \right) - 1 = e^{\bar{\delta}} - 1 \)