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Why Log Returns Equal the Log of One Plus Arithmetic Returns

Article Quant Q&A · Author: Austin
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Summary

The document explains why a log return is calculated from the ratio of ending value to starting value, rather than by taking the logarithm of the arithmetic return itself. Arithmetic return is the change in value divided by the initial value, which is also the ending-to-starting value ratio minus one. Adding one to arithmetic return therefore recovers the gross value ratio; taking its natural logarithm gives the log return.

The accepted answer presents the conversion directly: log return equals the natural logarithm of one plus arithmetic return. This identity clarifies that the two return measures use different inputs and are not obtained by applying a logarithm to the arithmetic return alone. The explanation is algebraic and contains no market data, empirical comparison, or discussion of when either convention is preferable. It provides a basic quantitative-finance definition rather than a trading method, and it assumes positive starting and ending values so the logarithm is defined.

Key ideas

  • Arithmetic return is the ending value divided by the starting value, minus one.
  • Adding one to arithmetic return recovers the gross value ratio.
  • Log return is the natural logarithm of that gross ratio, or equivalently the logarithm of one plus arithmetic return.
  • The logarithmic definition requires a positive value ratio.

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# Returns vs log returns formula


# Returns vs log returns formula












Probably something very simple I'm missing, but if returns is:

$R = \frac{V_f}{V_i} -1$

Then why is log returns $R = log(\frac{V_f}{V_i})$ instead of $R = log(\frac{V_f}{V_i} -1)$?

## Answer by alexbougias (score 6, accepted)

https://quant.stackexchange.com/a/41821

Let $R$ denote the arithmetic return and $r$ the log returns.

$$R=\frac{V_f-V_i}{V_i} \textrm { and } r=\ln\left(\frac{V_f}{V_i}\right)$$

Arithmetic and log returns are connected as:

$$R=\frac{V_f-V_i}{V_i} =\frac{V_f}{V_i}-1$$

Hence, $R+1=\frac{V_f}{V_i}$. Taking log on both sides.

$$\ln\left(\frac{V_f}{V_i}\right)=\ln(R+1) \textrm{ and } r=\ln(R+1)$$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.