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Actuarial Mathematics for Modelling · Annuity and accumulation functions

Accumulation and Discount Functions: a(t), A(t) and v(t) Explained

Updated 11 October 2026 · Fact-checked

The accumulation function a(t) gives the value at time t of 1 invested at time 0, with a(0) = 1. The amount function is A(t) = A(0) × a(t). The discount function is v(t) = 1 ÷ a(t). To find a(t) from the force of interest δ(s), use a(t) = exp(∫ δ(s) ds from 0 to t).

Understand Accumulation and Discount Functions

Start with a simple idea. You invest money today. Later it grows because it earns interest. The accumulation function a(t) tells you how much 1 invested at time 0 is worth at time t. So a(0) = 1, because no time has passed.

The amount function A(t) is the same idea for any starting sum. If you invest A(0) at time 0, then A(t) = A(0) × a(t). So a(t) = A(t) ÷ A(0). It is the growth factor, and A(t) is the money value.

Under standard assumptions, a(t) is increasing in t, because interest is positive. It is continuous when there are no sudden jumps in the fund value. Under compound interest at effective annual rate i, a(t) = (1 + i)^t. Under simple interest, a(t) = 1 + it. Do not assume a(t) has either form unless the question says so.

The discount function reverses accumulation. It gives the present value at time 0 of 1 payable at time t. It equals v(t) = 1 ÷ a(t). Under compound interest this is v^t, where v = 1 ÷ (1 + i). A payment of X at time t has present value X × v(t).

The force of interest δ(t) is the instantaneous rate of growth: δ(t) = a′(t) ÷ a(t) = d/dt ln a(t). Integrate this and you recover a(t). This is the route you use whenever the question gives you δ(t) rather than a rate.

Key rules to remember

Amount function
A(t) = A(0) × a(t)
a(t) = A(t) ÷ A(0). Also a(0) = 1.
Growth between two times
Value at t2 = Value at t1 × a(t2) ÷ a(t1)
Needs a(t) defined from time 0. For t1 < t2, the accumulation factor is a(t2) ÷ a(t1).
Discount function
v(t) = 1 ÷ a(t)
Present value at time 0 of 1 due at time t.
Force of interest
δ(t) = a′(t) ÷ a(t) = d/dt [ln a(t)]
Valid where a(t) is differentiable.
Accumulation from force of interest
a(t) = exp( ∫ from 0 to t of δ(s) ds )
Uses a(0) = 1.
Accumulation between two times
exp( ∫ from t1 to t2 of δ(s) ds ) = a(t2) ÷ a(t1)
Use for a payment moved from t1 to t2.
Discount from force of interest
v(t) = exp( − ∫ from 0 to t of δ(s) ds )
Reverse of accumulation.
Compound interest
a(t) = (1 + i)^t and v(t) = (1 + i)^(−t)
Holds for a constant effective annual rate i, which is equivalent to constant δ = ln(1 + i).
Simple interest
a(t) = 1 + it
Not equal to compound interest for fractional t. Check the question.

How to solve Accumulation and Discount Functions questions

Use this method for any question on a(t), A(t), v(t) or δ(t).

  1. 1Write down what you are given: a(t), A(t), a rate, or δ(t). Note the time units.
  2. 2Check that a(0) = 1. If A(t) is given, find a(t) = A(t) ÷ A(0).
  3. 3If δ(t) is given, integrate from the start time to the end time. Use exp of the integral to get the accumulation factor.
  4. 4If you need a present value, divide by the accumulation factor. That is, multiply by v = 1 ÷ a.
  5. 5If the two times are not 0 and t, use the ratio a(t2) ÷ a(t1), or integrate δ(s) from t1 to t2.
  6. 6Write down the working with the correct notation, then compute.
  7. 7Check that the answer is sensible: accumulation factors should exceed 1 for positive interest, and discount factors should be below 1.

Quickest way: Integrate δ, then exponentiate

When to use it: Use when you are given a force of interest, either constant or a simple function of time, and need a value or a factor.

  1. Write the limits: from the starting time to the end time.
  2. Integrate δ(s) first. Do not exponentiate until the integral is finished.
  3. Take exp of the result for accumulation, or exp of its negative for discounting.
  4. For a constant δ, the factor is e^(δ × length of time).
  5. Quickly check the sign and size before you move on.

Common mistakes in Accumulation and Discount Functions

  • Forgetting a(0) = 1 when finding a constant in a(t).

    Students write a(t) = at² + b and leave b unknown.

    Fix: Use a(0) = 1 first. It fixes the constant immediately.

  • Exponentiating each term of δ(t) separately, or exponentiating before integrating.

    Students rush and apply exp to the integrand.

    Fix: Integrate the full δ(s) first. Then take exp of the whole integral.

  • Using a(t) instead of a(t2) ÷ a(t1) when the payment starts at a later time.

    Students treat every accumulation as starting from time 0.

    Fix: Divide a(t2) by a(t1), or integrate δ from t1 to t2.

  • Mixing up accumulation and discounting by multiplying instead of dividing.

    The two factors are reciprocals, so it is easy to swap them.

    Fix: Ask whether you are moving money forward or backward in time. Forward means multiply by a. Backward means multiply by v = 1 ÷ a.

  • Assuming compound interest when only simple interest is given, or the reverse.

    Compound interest is the default in most textbook problems.

    Fix: Read the wording. Use 1 + it only when told simple interest applies.

Worked examples

Example 1

The force of interest is δ(t) = 0.02 + 0.004t for t ≥ 0, where t is in years. Find (a) a(5), (b) the present value at time 0 of ₹1,00,000 payable at time 5.

Show the solution
  1. Integrate δ(s) from 0 to 5: ∫(0.02 + 0.004s) ds = 0.02s + 0.002s².
  2. At s = 5: 0.02 × 5 + 0.002 × 25 = 0.10 + 0.05 = 0.15.
  3. So a(5) = exp(0.15) = 1.16183.
  4. The discount function is v(5) = 1 ÷ a(5) = exp(−0.15) = 0.86071.
  5. Present value = 1,00,000 × 0.86071 = ₹86,071.

Answer: (a) a(5) = e^0.15 ≈ 1.1618. (b) Present value ≈ ₹86,071.

Example 2

An accumulation function is a(t) = 1 + 0.05t + 0.01t² for t ≥ 0. Find (a) the amount at time 4 of ₹2,00,000 invested at time 0, (b) the force of interest at time 4.

Show the solution
  1. Check a(0) = 1 + 0 + 0 = 1, so a(t) is valid.
  2. Compute a(4) = 1 + 0.05 × 4 + 0.01 × 16 = 1 + 0.20 + 0.16 = 1.36.
  3. Amount A(4) = 2,00,000 × 1.36 = ₹2,72,000.
  4. Differentiate: a′(t) = 0.05 + 0.02t, so a′(4) = 0.05 + 0.08 = 0.13.
  5. Force of interest δ(4) = a′(4) ÷ a(4) = 0.13 ÷ 1.36 = 0.09559.

Answer: (a) ₹2,72,000. (b) δ(4) ≈ 0.0956, or 9.56% per year.

Exam tips

  • In MCQs, the quickest check is a(0) = 1. Use it to rule out options that fail this test.
  • When δ(t) is given, set out the integral clearly. Marks are given for the integral limits and the exp step.
  • State your units and the time convention (years from time 0) in written answers.
  • If a question asks for a value between two times t1 and t2, look for a(t2) ÷ a(t1) and show it explicitly.
  • In Paper B (R or Excel), use a clear function for a(t) and check that it returns 1 at t = 0.

Practice questions from Annuity and accumulation functions

Accumulation and Discount Functions: frequently asked questions

What is the difference between a(t) and A(t)?

a(t) is the value at time t of 1 invested at time 0, so a(0) = 1. A(t) is the value of an actual investment, with A(t) = A(0) × a(t). You use a(t) as a growth factor and A(t) as a money amount.

How do I find a(t) from the force of interest?

Integrate δ(s) from 0 to t, then take the exponential. That is a(t) = exp(∫ from 0 to t of δ(s) ds). For a constant δ, this gives a(t) = e^(δt).

How is the discount function related to a(t)?

It is the reciprocal: v(t) = 1 ÷ a(t). It gives the present value of 1 payable at time t. Under compound interest, v(t) = (1 + i)^(−t).

Must a(t) always be increasing?

In standard theory with non-negative interest, yes. If the force of interest becomes negative at some point, a(t) can fall over that period, so check δ(t) before you assume it.