Interest and the Accumulation Function

Interest is the price paid for the use of money. The accumulation function $a(t)$ tracks how a single dollar invested at time 0 grows, and the amount function $A(t) = k \cdot a(t)$ scales that growth up by whatever principal k you actually put in.

By the end you'll be able to tell $a(t)$ apart from $A(t)$, and compute the effective rate of interest $i_n$ for any period n.

Predict: if you raise $i$, does the year-2 balance grow by more than twice the year-1 gain? Check with the slider below.

This fund's unit growth is $a(t) = 1 + i \cdot t + 0.03t^2$ (solid A(t) curve below is $k \cdot a(t)$ in dollars; dashed a(t) curve is the same shape per dollar of principal). Drag i and k, then click a year to see its effective rate $i_n$ bracketed in orange on the diagram.

Rate i = 5.0%

Principal k = $1,500

Highlight year n

A(2) = $1,830.00, i₂ = 12.96%

A(t) = k·a(t), with $a(t) = 1 + it + 0.03t^2$ — fund value at each integer year, and the bracketed effective rate for the chosen year

A(t), dollars a(t), per dollar of principal iₙ for the highlighted year

Every problem in this course starts from the same idea: a dollar today isn't the same as a dollar next year, and interest is the exchange rate between them.

Intuitive

Picture leaving \$1 in a growing fund and just watching it. $a(t)$ is the value of that dollar at time t — it always starts at $a(0)=1$ (no time, no growth yet), and it normally climbs from there. If you'd put in k dollars instead of 1, every value on the chart just scales up by k — that scaled-up curve is the amount function $A(t) = k \cdot a(t)$.

Formal

$a(0) = 1$ by definition, and $a(t)$ is (normally) increasing. For principal k, $A(t) = k \cdot a(t)$, so $A(0) = k$. The effective rate of interest in period n is $i_n = (A(n)-A(n-1))/A(n-1) = (a(n)-a(n-1))/a(n-1)$ — interest earned during the period, divided by the balance you started the period with. Because k cancels top and bottom, $i_n$ never depends on how much principal is riding on $a(t)$; it only depends on the shape of $a(t)$ itself.

Applied

Think of a savings account: your deposit is the principal k, the account's balance at time t is $A(t)$, and the rate printed on your statement for a given year is $i_n$ for that year — interest credited, divided by what was in the account when the year started.

Worked example

Suppose $a(t) = 1 + 0.05t + 0.01t^2$. Then $a(0)=1$, $a(1) = 1.06$, $a(2) = 1.14$. Interest earned in year 1 is $a(1)-a(0) = 0.06$, so $i_1 = 0.06/1 = $ 6%. Interest earned in year 2 is $a(2)-a(1) = 0.08$, so $i_2 = 0.08/1.06 \approx $ 7.55%. The rate changed between years because this $a(t)$ isn't a constant-rate compound function — more on that in the next lesson.

Your turn

Different fund, same idea: $a(t) = 1 + 0.03t + 0.02t^2$. Then $a(0)=1$, $a(1) = 1.05$, $a(2) = 1.14$. Interest earned in year 1 is $a(1)-a(0) = 0.05$, so $i_1 = 0.05/1 = $ 5%. Interest earned in year 2 is $a(2)-a(1) = 0.09$, so $i_2 = 0.09/1.05 = $ ____

Reveal the answer

$i_2 = 0.09/1.05 \approx $ 8.57% — divide by $a(1)$, the balance you started year 2 with, never by $a(2)$. Set i and k on the slider above and click "Year 2" to see a similarly-shaped rate bracketed on the chart.

More info — the pitfall of dividing by the wrong balance

The most common slip with $i_n = (A(n)-A(n-1))/A(n-1)$ is dividing by the ending balance $A(n)$ instead of the balance you started the period with, $A(n-1)$. The effective rate is always "interest earned ÷ what you started the period with" — never the other way around. It's also easy to assume $i_n$ is constant from period to period; it's only level if the underlying $a(t)$ is a fixed-rate compound function, which is exactly what the next lesson takes up.

Check your understanding

Question 1 of 4

Let $a(t) = 1 + 0.02t + 0.015t^2$. What is $i_1$, the effective rate of interest in year 1?

Question 2 of 4

Connecting the pieces — a(t), A(t), and iₙ together: if you double the principal k in $A(t) = k \cdot a(t)$ but leave $a(t)$ unchanged, what happens to the effective rate $i_n$ for any period n?

Question 3 of 4

A fund has $A(t) = k \cdot a(t)$ with $a(1) = 1.05$, $A(1) = 1050$, and $A(2) = 1123.50$. Find the principal k, then use it with A(2) to find $i_2$.

Question 4 of 4

By definition, what must $a(t)$ equal at $t = 0$, no matter how interest is credited?

Recap

Interest is the compensation paid for the use of money; the accumulation function $a(t)$ gives the value at time t of 1 invested at time 0, with $a(0) = 1$ always.
The amount function $A(t) = k \cdot a(t)$ rescales $a(t)$ by whatever principal k was actually invested; $A(0) = k$.
The effective rate of interest in period n is $i_n = (A(n)-A(n-1))/A(n-1) = (a(n)-a(n-1))/a(n-1)$ — interest earned in the period, over the balance you started it with. It cancels k, so it depends only on the shape of $a(t)$.

Dive deeper

AnalystPrep — Time Value of Money (SOA Exam FM Study Notes) Read the accumulation-function and effective-rate definitions with worked FM examples.
Marcel Finan — A Preparation for the Actuarial Exam FM/2 (free study guide PDF) Work the opening chapters on accumulation and amount functions.
Khan Academy — Interest and debt Build intuition for what interest is before the formal definitions.

Sources

Interest, the Accumulation Function, and Effective Rates