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 $R = log(\frac{V_f}{V_i} -1)$?


closed as off-topic by phdstudent, skoestlmeier, LocalVolatility, Helin, Matthew Gunn Sep 19 '18 at 18:13

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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.

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

  • $\begingroup$ Ahh didn't realize the log return was ln(R+1) instead of just ln(R), thanks. $\endgroup$ – Austin Sep 19 '18 at 17:28
  • 1
    $\begingroup$ @Austin Another fun fact is that for arithmetic return $R$ near $0$, $R \approx r$. The linear approximation (first order Taylor expansion) of $\log(1+x)$ near $x=0$ is $x$. Example: $\log (1.02) = .0198$ $\endgroup$ – Matthew Gunn Sep 19 '18 at 18:18

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