Actuarial Mathematics for Modelling · Interest rates over different time periods

Force of Interest: Formula and Accumulation with Varying δ

Updated 11 October 2026 · Fact-checked

The force of interest δ(t) is the instantaneous rate of growth of an investment at time t. It equals A'(t) ÷ A(t), which is d/dt of ln A(t). To find an accumulation, integrate δ(s) from t₁ to t₂ and take the exponential. If δ is constant, e^δ = 1 + i.

Understand Force of Interest

Interest is usually quoted over a period, such as 8% a year. The force of interest goes further. It measures how fast money is growing at one instant, like a speedometer shows speed at one moment.

Think of an accumulation function A(t), the value at time t of an investment. The rate of growth at time t is A'(t). Divide by the current value A(t) to get a proportional rate. That is the force of interest: δ(t) = A'(t) ÷ A(t). It is written δ, and it is a nominal annual rate convertible continuously.

Because A'(t) ÷ A(t) is the derivative of ln A(t), you can reverse it. Integrate δ(s) from 0 to t, and you get ln A(t) − ln A(0). So A(t) = A(0) × exp(∫δ(s) ds). The integral is the total 'amount of growth' in log terms, and the exponential turns it into a growth factor.

If δ is constant, the integral is just δt, and the factor over one year is e^δ. That must equal 1 + i, so δ = ln(1 + i). This is the key link between the force of interest and the effective annual rate. If δ varies with time, you integrate instead. The same idea gives discount factors: the present value factor from time t back to 0 is exp(−∫δ(s) ds).

Key rules to remember

Definition of force of interest: δ(t) = A'(t) ÷ A(t) = d/dt [ln A(t)]; A(t) is the accumulation function. δ is an instantaneous rate, quoted per year.
Accumulation factor from t₁ to t₂: A(t₂) ÷ A(t₁) = exp( ∫ δ(s) ds from t₁ to t₂ ); Works for any δ(s), constant or varying. The limits are times, not rates.
Discount factor from t₂ back to t₁: v(t₁, t₂) = exp( −∫ δ(s) ds from t₁ to t₂ ); This is the reciprocal of the accumulation factor over the same interval.
Constant force of interest: A(t) = A(0) × e^(δt); Holds only when δ does not change with time.
Link to effective annual rate: 1 + i = e^δ, so δ = ln(1 + i); Valid when δ is constant over the year. For varying δ, use the integral over that year.
Link to discount rate and discount factor: v = e^(−δ), d = 1 − e^(−δ); v = 1 ÷ (1 + i) for a constant force.
Link to nominal rate: δ = lim of i^(m) as m → ∞; i^(m) is the nominal rate convertible m times a year. δ is its continuous limit.

How to solve Force of Interest questions

Use this method for any question on force of interest, whether δ is given, or you must find it from rates or accumulated values.

1Identify what is given: a constant δ, a function δ(t), an effective rate i, or accumulated values.
2Write down the time interval carefully. Note the start time t₁ and end time t₂, in years.
3If δ is a function, integrate it between t₁ and t₂. If δ is constant, the integral is δ × (t₂ − t₁).
4Take the exponential to get the accumulation factor. Use the negative exponential for a discount factor.
5Multiply by the amount invested (or by the amount due for a present value).
6If asked for an effective rate, set 1 + i equal to the accumulation factor over one year, then subtract 1.
7If asked for δ from accumulations, use δ = ln(1 + i), or take ln of the accumulation factor and divide by time.
8Check units and reasonableness: δ should be a bit below i for positive rates, and the answer should be in the requested form.

Quickest way: Integrate, exponentiate, then convert

When to use it: Use this when time is short and the question gives δ(t) as a simple function such as a constant, a linear function or a step function.

Write the exponent first: ∫δ(s) ds over the interval. Do not compute e^ until the integral is done.
For linear δ(s) = a + bs, the integral is a(t₂ − t₁) + b(t₂² − t₁²) ÷ 2.
For a step function, split the interval at each change and add the pieces of δ × length.
Apply exp once to the total. This is faster than multiplying several exponentials.
Convert using 1 + i = e^δ only if the question asks for an annual effective rate.

Common mistakes in Force of Interest

Using δ = i or treating δ as the effective rate.
Both are annual percentages, so they look interchangeable.
Fix: Remember δ = ln(1 + i). For positive rates, δ is smaller than i. Convert before you use either in an effective-rate formula.
Integrating from 0 to t when the question asks for the interval t₁ to t₂.
Students memorise the formula for A(t) from time 0 and apply it automatically.
Fix: Always write the limits of the integral from the question. Circle the start and end times before you integrate.
Forgetting the minus sign for a discount factor.
The accumulation formula is learned first, and the discount version is treated as an afterthought.
Fix: Present value uses exp(−∫δ). Sense check: a present value should be smaller than the future amount when δ is positive.
Multiplying δ by a time in months or an incorrect number of years.
Questions often mix months and years.
Fix: Convert all times to years before using δ, since δ is an annual rate. For example, 6 months is 0.5 years.
Integrating δ(t) without taking the exponential, giving ∫δ as the accumulation factor.
Students stop once the integral is calculated.
Fix: The integral gives ln of the factor. Always apply exp, then check the result is around 1 or above for accumulation.
Applying a constant-δ formula to a varying force.
e^(δt) is simple and the time dependence is overlooked.
Fix: Check whether δ depends on t. If it does, integrate. You can only write δt when δ is constant on that interval.

Worked examples

Example 1

The force of interest is δ(t) = 0.02 + 0.004t per year, for t ≥ 0. Calculate the accumulated value at time 5 of ₹1,00,000 invested at time 0. Give your answer to the nearest rupee.

Show the solution

The accumulation factor is exp(∫δ(s) ds) from 0 to 5.
Integral = ∫(0.02 + 0.004s) ds = 0.02s + 0.002s².
At s = 5: 0.02 × 5 + 0.002 × 25 = 0.10 + 0.05 = 0.15.
At s = 0: the value is 0, so the integral equals 0.15.
Accumulation factor = e^0.15 = 1.161834.
Accumulated value = 1,00,000 × 1.161834 = ₹1,16,183.

Answer: ₹1,16,183 (approximately).

Example 2

An investment earns a constant force of interest of 6% per year. Find (a) the effective annual rate of interest and (b) the present value of ₹2,50,000 due in 3 years.

Show the solution

(a) 1 + i = e^0.06.
e^0.06 = 1.061837, so i = 0.061837, or 6.1837% per year.
(b) Discount factor over 3 years = exp(−0.06 × 3) = e^(−0.18).
e^(−0.18) = 0.835270.
Present value = 2,50,000 × 0.835270 = ₹2,08,818 (to the nearest rupee).
Check: using v = 1 ÷ 1.061837 = 0.941765, v³ = 0.835270. This agrees.

Answer: (a) i ≈ 6.18% per year. (b) Present value ≈ ₹2,08,818.

Exam tips

Write δ(t) and your integral limits at the top of the answer. Examiners award method marks for a correct set-up even if the arithmetic slips.
Keep full calculator precision for e^x values. Rounding δ or the exponent early can change the final rupee amount.
For written answers, state that the integral gives ln of the accumulation factor. This shows you understand the derivation, not just the recipe.
In Paper B (computer-based) questions, use the built-in integration or exp functions, and show the formula in a cell or code comment so the working can be followed.
Watch for questions that switch between δ, i, d and i^(m). Convert everything to δ first, then convert back at the end.

Practice questions from Interest rates over different time periods

Force of Interest: frequently asked questions

What is the formula for force of interest in actuarial mathematics?

The force of interest is δ(t) = A'(t) ÷ A(t), which equals d/dt of ln A(t). It is the instantaneous proportional rate of growth at time t. If the accumulation is e^(δt), δ is constant.

How do I calculate accumulation with a varying force of interest?

Integrate δ(s) from the start time to the end time, then take the exponential. Multiply the result by the amount invested. For a step function, add the δ × length for each interval before taking the exponential.

What is the relation between force of interest and effective rate?

When δ is constant, 1 + i = e^δ, so δ = ln(1 + i). If δ varies, the effective rate for a year is exp(∫δ(s) ds over that year) − 1. For positive rates, δ is lower than i.

Is the force of interest the same as a nominal rate?

It is the limiting case of a nominal rate i^(m) as m tends to infinity. It is a nominal annual rate convertible continuously. It is not the same as an effective rate.