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I'm working on a theoretical portfolio that includes both long and short positions. I already have the daily holding period returns, and I want to calculate both arithmetic and logarithmic returns for these positions (daily). However, I run into a problem with short positions when the return exceeds 100% (i.e., higher than 1). In such cases, I can't use the log return formula (LN(1+R)). Although these instances are rare, they do occur. Let's assume the return is 2, then for short selling it becomes LN(1-2).

I tried setting a cap, treating returns over -1 as -0.99, but this approach skews the portfolio because LN(1-0.99) equals -4.6051, which is an excessively negative value in logarithmic terms.

What would be a better solution to handle this issue?

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The easiest way is to think for a return on a short position as being the negative of the return on the long position. In that way you never get confused.

Think about the following example. On day 1 the stock price $P_1=50$, on day 2 the stock price $P_2 = 120$ and on day three the stock price $P_3 = 25$.

The arithmetic return on a long position is $\frac{P_{t+1}}{P_t} - 1$ and the log return is $\log \bigg ( \frac{P_{t+1}}{P_t} \bigg )$.

The short return is just the same as above with the opposite sign.

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    $\begingroup$ I fully agree on the note about the log return on a long positions being just the negative of the short. Another way people tend to represent returns on short positions is $(P_1-P_2)/P_1$ vs that of a long position that is the normal returns formula of $(P_2-P_1)/P_1$, which is why applying a negative sign on the log returns makes sense. $\endgroup$ Commented Aug 2, 2024 at 1:11
  • $\begingroup$ KaiSqDist: That's interesting but also possibly confusing because, in the first equation, you're using an anchor of $P_{2}$ but then dividing by $P_{1}$. My brain can't handle that !!!!!!! phdstudent's method is more like back in the day when they told you that subtraction was like adding a negative. That works better for me. $\endgroup$ Commented Aug 2, 2024 at 3:58
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    $\begingroup$ @markleeds not sure what you mean by anchor of $P_2$. The way I understand it is because $1$ comes before $2$ in terms of time, the denominator should always be $P_1$, so it makes perfect sense to me. The numerator itself is intuitive though. $\endgroup$ Commented Aug 2, 2024 at 6:31
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    $\begingroup$ Yeah, I always think of the term that is being subtracted as also being the term in the denominator so it's not intuitive to me but I see what you're doing. No problem. Different strokes for different folks. $\endgroup$ Commented Aug 2, 2024 at 13:59

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