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Two popular ways to measure returns are Arithmetic returns and Log returns. Let's define arithmetic (simple period) returns as: P(t) - P(t-1) / P(t-1). Let's define log return as Ln( P(t)/P(t-1) ) or equivalently Ln(1 + arithmetic return).

The log returns have nice properties -- logarithmic returns prevents security prices from becoming negative; we can interpret log returns as continuously compounded returns; they are approximately equal to simple returns at small values; if the security price follows Brownian motion then the log returns are normally distributed; logs convert products into sums. Jorion (2001) has a brief description of the properties and application of the log description (pg. 94): Jorion 2001

Let's say a security lost 1% (simple return) today. We would like to annualize this daily return (as it is an input to some risk model whose outputs we would like on an annualized basis). So: Ln(1 -.01) * 251 trading days = -2.52 annualized return. However, a security cannot lose more than 100% of its value. How do we explain this outcome (i.e. what is the right operation such that the lower bound on the log return is > -1)?

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You're forgetting that -2.52 is still in natural logarithm terms. So the correct answer is 2.71828183 raised to the -2.52 power which equals 0.08. Your ending portfolio value is 8% of what it was a year ago.

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  • $\begingroup$ Granted that (1 + R)^251 = .08. However, this is not a solution as it side-steps the problem by stating the problem in arithmetic returns. It is general practice to use a continuously compounded return such as Ln(1+R) for regression, to continuously compound interest, price options, etc. My point is that the log form of the return has no economic meaning when a large value of arithmetic return are involved. Even in the seemingly everyday case of annualized daily returns we seem to bump up against this < -1 issue $\endgroup$
    – Ram Ahluwalia
    Jul 14, 2011 at 5:38
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    $\begingroup$ Joshua's point was that when you convert to log returns, you also need to convert your bounds to log returns. A 100% loss corresponds to a log loss of -infinity. Requiring a lower bound on the log return of -1 is equivalent to requiring a lower bound on the arithmetic return of -63.3%. $\endgroup$
    – Frank Fingerman
    Jul 14, 2011 at 10:39
  • $\begingroup$ @Frank Fingerman - Thanks, better said than I could. $\endgroup$
    – Joshua Chance
    Jul 15, 2011 at 8:21
  • $\begingroup$ @Quant Guy - I had what I think is a similar problem: how do I compare quarterly earnings growth for stocks whose prior quarter's earnings was negative and this quarter is positive (or vice versa). My solution is ln(absolute(current.earnings))*sign(current.earnings) minus ln(absolute(past.earnings))*sign(past.earnings) --- if either quarter's earnings is zero then obviously substitute zero. That might or might not help but food for thought. $\endgroup$
    – Joshua Chance
    Jul 15, 2011 at 8:32
  • $\begingroup$ Thanks Joshua -- yes that is a very similar problem and creative solution. Thank you for sharing! $\endgroup$
    – Ram Ahluwalia
    Jul 19, 2011 at 22:32
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To get the final value, just get $e^{rt}=e^{-0.01 \cdot 252}=8.04 \%$.

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