Exam FMMeasurement of interestFree to read

Effective interest and discount rates

The effective rate i measures interest paid at the end of a period and the effective discount rate d measures it at the start; v = 1 − d = 1/(1+i) converts between them.

The formulas

Accumulation

$$a(t) = (1+i)^t$$

value at time t of 1 invested now

Discount factor

$$v = \dfrac{1}{1+i} = 1 - d$$

Discount rate

$$d = \dfrac{i}{1+i} = iv$$

Interest from discount

$$i = \dfrac{d}{1-d}$$

The identity worth memorising

$$i - d = id$$

Where it comes from

1. Interest of $i$ on 1 is paid at the END of the period, so 1 grows to $1+i$.
2. The same interest measured at the START of the period is $d$, and 1 at the end is worth $1-d$ now, so $v = 1-d$.
3. Setting the two views of the same growth equal: $v = \frac{1}{1+i}$, hence $d = 1 - \frac{1}{1+i} = \frac{i}{1+i}$.
4. Multiplying $d = \frac{i}{1+i}$ through by $1+i$ gives $d + di = i$, that is $i - d = id$.

Worked example

This answer is recomputed from the site’s own interest-theory and probability functions every time the test suite runs, so the page and the mathematics cannot drift apart.

Memory hooks

• Discount is interest paid in advance, so d is always slightly SMALLER than i, and i − d = id is the gap.
• v = 1 − d and v = 1/(1+i) are the same statement - whichever you remember gives you the other rate.

Traps

• Treating d as i and discounting one period too few.
• Using i = d(1+d) instead of i = d/(1 − d).

Related

• Nominal rates convertible m-thly
• Force of interest
• Equations of value and the time line

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