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Assume a given stock's log returns are normally distributed, its average annual log return = 100% and annual standard deviation (or volatility) = 200%. Given a trading year of 250 days, what are $\mu$ and $\sigma$ (average and sd) of daily log returns? I'm attempting this using R.

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  • $\begingroup$ What is a log return? Is it the logarithm of returns, and if so and the native data is normal, why are you using it, that would make it log-normal? $\endgroup$
    – Carl
    Sep 30, 2016 at 23:12
  • $\begingroup$ Yes, logarithm of returns. I am intending to run simulations to calculate net returns given the above parameters for daily returns. However I might be overthinking this. Would mu just be ln(1+1)/250 and sigma be 2/sqrt(250) or am I way off? Thanks $\endgroup$
    – James
    Sep 30, 2016 at 23:33
  • $\begingroup$ Oh, sorry, read too quickly. Yes, you are overthinking it. The daily values are just examples of the more stable estimators, as the number of days increases, the standard error of the mean decreases, and the standard deviations (sd) become more reliable as well. Indeed, to be exact, even as crude estimates, the sd for $n<10$ should be small number corrected. $\endgroup$
    – Carl
    Sep 30, 2016 at 23:55
  • $\begingroup$ Thanks for your help Carl. Just to be clear, I have the correct values above? Or should sigma be ln(2)/sqrt(250). My original question should read "annual st. deviation of log returns = 200%" sorry for the confusion. $\endgroup$
    – James
    Oct 1, 2016 at 0:12

2 Answers 2

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Analysis

We can model the return on any given day $t$ as a random variable $X_t.$ This permits the log return after $n$ trading days to be expressed as

$$S_n = X_1 + X_2 + \cdots + X_n.$$

For simplicity--and as the foundation for more detailed considerations if we need them later--let's further suppose all these variables are equally distributed with common means of $\alpha$ and variances of $\sigma^2.$ Assuming the returns are not appreciably correlated (the "Random Walk Down Wall Street" hypothesis), these assumptions imply

$$E[S_n] = n\alpha$$

(expectation is linear) and

$$\operatorname{Var}(S_n) = n\sigma^2$$

(variance is bilinear).

In these terms the question supposes $n\alpha = 100\% = 1$ and $\sqrt{n\sigma^2} = 200\% = 2.$ With $n=250,$ we find

$$\alpha = \frac{100\%}{n} = \frac{1}{250} = 0.4\%$$

and

$$\sigma = \sqrt{\frac{(200\%)^2}{n}} = \sqrt{\frac{2^2}{250}} = \frac{2}{\sqrt{250}} \approx 12\%.$$


Commentary

1. The Rule of 16.

These are useful general rules. For instance, whenever I hear on the news that "the markets changed by $x\%$ today," I attribute most of that change to variance--after all, a random fluctuation averaging $12\%$ per day is usually far greater than a constant change of $0.4\%$ per day--and think to myself "that daily change translates to an annual change of around $\pm x\times\sqrt{250} \approx \pm16 x\%.$" Given a long history of annual returns in the -40% to 40% range, this number provides a helpful perspective on the magnitude of the daily change. For instance, it suggests daily changes of around $0\%$ to $3\%$ or so are going to be commonplace, so only substantially larger changes (in either direction) are really noteworthy.

Call this the "Rule of 16," if you like.

2. The meaning of "100%" log return

Incidentally, if by "annual log return of 100%" you really mean a doubling after one year, then the log return is $\log(1 + 1) = \log 2 = 69.3\%.$ The only change you need to make to the foregoing is that now the daily value of $\alpha$ is around $69\%/250 \approx 0.3\%,$ down from the original value of $0.4\%.$

3. Estimates or not?

Bear in mind that this Rule of 16 merely expresses mathematical relationships among model parameters. Individual returns $X_t$ are, as stated at the outset, random, and will almost always differ from $\alpha$ or $\sigma.$ They are not, however, necessarily estimates as might be supposed by some. When you apply the rule by taking a year's return and computing the daily parameters, you are just finding a different--and maybe insightful--way of re-expressing a given fact about things that have already occurred. When you apply the rule in reverse, as I often do, you might be misunderstood as forecasting or predicting future returns based on past returns. As everyone knows and will gleefully tell you, past returns are no guarantee of future market behavior. Of course you know that and will understand such calculations not as predictions but just ways to understand and interpret the information you have.

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The individual day values should not be corrected, in the long term, they have the same mean and SD as the 250 day data because there is no trend in you example problem. Please do not divide the mean of SD by anything, the daily mean, on the average, is very close to the yearly mean, and the actual daily SD, you cannot really judge by just one day's data, unless you have minute by minute data for that day. Then it gets a bit more complicated because the SD you have is the day by day SD, not the year to year SD, if you got that SD by using all 250 trading days in the SD formula.

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  • $\begingroup$ This reads like a list of things not to do, but what should one do? What is the answer? $\endgroup$
    – whuber
    May 1, 2022 at 15:15
  • $\begingroup$ @whuber Without trending information within the trading calendar, there is no unique answer anymore that one can provide an answer for which numbers were added together to create a sum. You want an answer, provide enough information to get one. $\endgroup$
    – Carl
    May 2, 2022 at 18:16
  • $\begingroup$ When there's not enough information to answer, don't answer. Explain, in comments, what information would be needed. $\endgroup$
    – whuber
    May 2, 2022 at 18:17
  • $\begingroup$ @whuber OK, will do in the future. For now, I just explained in a comment why there is no answer, and my answer of no answer is still an answer of sorts, and if you delete it, you will delete the explanatory comments as well. $\endgroup$
    – Carl
    May 2, 2022 at 18:22
  • $\begingroup$ I have no intention of deleting your post! But it very well might serve better as a comment because it doesn't answer the question. $\endgroup$
    – whuber
    May 2, 2022 at 18:47

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