Difference between two ways of calculating real return

Question

I've just learned the following way of calculating the real return rate $R$ of an investment:

$$R=\frac{P(1+N)-P(1+I)}{P(1+I)}$$

Where $P$ is the initial value invested, $N$ is the nominal interest rate and $I$ is the inflation rate.

However, I've seen an alternative formula for calculating the same thing, which is

$$R=N-I$$

So my question is, which one gives a better result for the effective rate of return of an invesment and why? Thanks very much in advance.

Answer 1

The second expression is an approximation which works reasonably well for small inflation $I$ but badly for high $I$

Suppose the inflation rate is $I=300\%$ (i.e. prices this year are four times what they were a year ago) and nominal interest rates were $N=100\%$ (i.e. nominally your savings doubled in a year). Then, in real terms after interest, your savings could buy half the stuff it previously could and the real interest rate was $\frac{N-I}{1+I}=-50\%$ while $N-I=-200\%$

Instead suppose the inflation rate is $I=3\%$ and nominal interest rates were $N=1\%$. Then the real interest rate was $\frac{N-I}{1+I}\approx-1.94\%$ while $N-I=-2\%$, fairly close

Answer 2

First, it's better to cancel $P$ in your first definition. Then $$ R=\frac{N-I}{1+I} $$ And the second expression is $$ R=N-I $$ Actually, it is the famous $\textit{Fisher Equation}$.

The difference that there is a $1+I$ on denominator. Of course they are different. But economists sometimes use both since they are approximately equal when $R$ and $I$ are small. If you look at the data, both real interest rate and inflation rate are just a little above 2%.

Instead of using your first definition, let's start from an alternative form $$ 1+N = (1+R)(1+I) = 1+R+I+R\cdot I $$ When $R,I$ are both small, their product can be ignored. Say $2\%\times 2\%=0.0004$. Hence, you can say $$ N= R+I \Leftrightarrow R= N-I $$

Here is the Wikipedia Link. https://en.wikipedia.org/wiki/Fisher_equation