How Log Returns Relate to Simple Returns and Geometric Averages

Summary

The document explains the conversion between simple returns and log returns using consecutive asset prices. A simple return is the price ratio minus one, while a log return is the logarithm of that ratio. Taking the exponential of a log return and subtracting one recovers the simple return; taking the logarithm of one plus a simple return gives the log return.

This identity clarifies why log prices are useful for representing price changes as percentage movements. The response does not derive a broader result connecting averages over multiple periods, or explain when arithmetic versus geometric averaging is appropriate. It offers no empirical evidence or trading method, so its value is a concise mathematical clarification rather than a complete treatment of return aggregation.

Key ideas

• A log return is the logarithm of the ratio between consecutive prices.
• A simple return is the consecutive price ratio minus one.
• The two return measures convert through the exponential and logarithm functions.
• The explanation does not establish a full comparison of arithmetic and geometric averages across multiple periods.

Tags

# What is the relationship between arithmetic versus geometric averages and simple versus logarithmic prices?


# What is the relationship between arithmetic versus geometric averages and simple versus logarithmic prices?












I know that the geometric mean is used in order to make percentage returns across time comparable. Similarly, I know that log prices make percentage returns comparable for example when prices are charted.

What is the mathematical basis for this relationship? Put differently, in what way do log returns "correspond" to geometric means and in what way to simple returns "correspond" to arithmetic means?

## Answer by Richi Wa (score 4, accepted)

https://quant.stackexchange.com/a/17194

If you wanted to see the following (price $S_t$, log return $r$, simple return $R$) then $$ r = \log(S_{t+1}) - \log(S_t) = \log(S_{t+1}/S_{t}), $$ and $$ R = S_{t+1}/S_{t}-1, $$ thus $$ R = \exp(r)-1 $$ and $$r = \log(1+R).$$ Was this the question?

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