<p>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 $\textit{Fisher Equation}$.</p>

<p>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%.</p>

<p>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
$$
N= R+I \Leftrightarrow R= N-I
$$</p>

<p>Here is the Wikipedia Link.
<a href="https://en.wikipedia.org/wiki/Fisher_equation" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Fisher_equation</a></p>

CC BY-SA 3.0
2017-11-23T04:34:27.740Z
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
CC BY-SA 3.0
2017-11-23T04:40:33.510Z
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
CC BY-SA 3.0
2017-11-23T03:01:13.020Z
2017-11-23T03:01:13.020Z
Andy Xu
<p>The second expression is an approximation which works reasonably well for small inflation $I$ but badly for high $I$ </p>

<p>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 $N-I=-200\%$</p>

<p>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</p>

CC BY-SA 3.0
2017-11-23T12:02:55.950Z
2017-11-23T12:02:55.950Z
Henry
<p>I've just learned the following way of calculating the real return rate $R$ of an investment:</p>

<blockquote>
  <p>$$R=\frac{P(1+N)-P(1+I)}{P(1+I)}$$</p>
</blockquote>

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

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

<blockquote>
  <p>$$R=N-I$$</p>
</blockquote>

<p>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.</p>

<finance>
Difference between two ways of calculating real return
Sasaki
# Difference between two ways of calculating real return

(Asked by: username_0 on 2017-11-22)

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.

**Tags:** finance


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## 2 answers

### Answer by username_2 on 2017-11-23. Score: 3 | Accepted answer

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


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### Answer by username_1 on 2017-11-23. Score: 3

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>

> **Comments:**
> **username_0**: 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.  
> **username_1**: 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: $(1+N)/(1+I)-1$.  

---

