# Difference between two ways of calculating real return

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.

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

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

• So, I understand now that the first equation is the more accurate way of calculating the real return, and that the second one is just a good approximation when $R$ and $I$ are small. However, $N-I$ makes perfect intuitive sense to me, and I really can't see why we have to divide it by $1+I$ to get the actual rate of return. Could you provide some intuition on this? Thanks. – Sasaki Nov 23 '17 at 4:34
• The most accurate way is to write the equation as $1+N=(1+R)(1+I)$, which, mathematically, gives you the first defnition. It implies that the norminal income $P(1+N)$ is inflated by $(1+I)$, so your real income is $P(1+N)/(1+I)$ and the real return rate is $(1+N)/(1+I)-1$. It's why there is always an $I$ on the denominator. I'm not sure if it is the intuition you need. But for me, if I use the first definition, I prefer using the expression: $(1+N)/(1+I)-1$. – Andy Xu Nov 23 '17 at 4:40