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I'm doing some analysis on log returns and I notice that returns can exceed 100%. For example, if a security's close price \$1 today and \$10 yesterday, the log return is $ln(1) - ln(10) = -230\%$! Under arithmetic computation, returns for a long position cannot exceed 100% (i.e. the initial investment). So how would i interpret a log return of -230%?

Thanks,

Alex

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closed as off-topic by Quantopik, olaker♦ Jan 17 '16 at 7:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Basic financial questions are off-topic as they are assumed to be common knowledge for those studying or working in the field of quantitative finance." – Quantopik, olaker
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The change in the log price is not a percentage, it's a value (pure number). To compute the percentage change, you take e^(change in log value). So, a change of -2.3 in the log price would equate to e^-2.3 which is 0.1 or 10%, so the price change would be a decrease of 90% (which, of course, is what a change from 10 to 1 actually is). $\endgroup$ – barrycarter Jan 16 '16 at 15:58
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Large? ?

The relationship between normal and log returns is $$(normal return) = exp(log return)-1$$ Therefore log-returns can be from $-\infty$ to $+\infty$ while normal ones can only be between $-1$ and $+\infty$.

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The result is:

$ e^{(-230%)} - 1 = -89% $

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If the value is $1 today and was $10 yesterday

return = today/yesterday - 1 = 1/10 - 1 = -0.9 = -90%

log return  = ln(1 - 0.9) = -2.302585

check : A = P e^rt = $10 * e^-2.302585 = $1 (i.e. today's value)

Since investments can end up in the red it's interesting to consider a return that exceeds -100%, for instance if today's value is -$1

return = today/yesterday - 1 = -1/10 - 1 = -1.1 = -110%

Now the log return method breaks down since the natural log cannot be taken for a negative number (unless imaginary numbers are used)

log return  = ln(1 - 1.1) = -2.30259 + 3.14159 i

The curve for ln(x) is only in the positive domain of x.

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image source

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