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How can we interpret the difference between two log points? Is it correct to interpret this difference in percentage points?

Thanks.

Marko

Marko
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2 Answers2

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$$\log(x)-\log(y) = \log(x/y)$$

If $x$ differs from $y$ by a factor of $1+\epsilon$ then we have,

$$\log(x)-\log(y) = \log(x/y) = \log(1+\epsilon) \approx \epsilon, $$

The last approximation is only valid for small values of $\epsilon$. Since expsilon can be interpreted as a percent difference I would say that the comparison of the difference of logarithms to a percentage is is only sensible if the difference is small.

I think a more reasonable interpretation is just to think back to the definition of logarithms are. The difference of $\log(x)$ and $\log(y)$ just represents the power of $e$ by which $x$ must be multiplied to equal $y$.

Spencer
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Just to elaborate some on what Spencer wrote:

$\text{log}(x)-\text{log}(y)=\text{log}(x/y)=\text{log}(\frac{x-y}{y}+1)=\text{log}(\frac{\%\Delta}{100}+1) $

where $\%\Delta$ is the percentage change. A first order Taylor approximation around $\%\Delta=0$ gives

$\text{log}(x)-\text{log}(y)=\text{log}(\frac{\%\Delta}{100}+1) \approx \frac{\%\Delta}{100}$

The exact percentage change is of course given by

$100(\text{e}^{\text{log}(x)-\text{log}(y)}-1)$

PaulB
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