4
$\begingroup$

I just finished reading "Fooled by Randomness" by Nassim Taleb. He, inter alia, gives the following example to prove one of his points:

A 15% return with 10% volatility per annum translates into a 93% probability of making money in any given year. However, when looking at a quarter, this probability decreases to 77%. The respective probability in a given month/day/hour/minute/second is 67%, 54%, 51.3%, 50.17% and 50.02%, respectively.

Although probably trivial, this was an astonishing realizition for me. Given one is equally happy/sad by making gains/losses, a person that uses a smaller timeframe to assess his performance will be much sadder, than one that uses a larger timeframe (this was the point he was arguing for). He does not show the maths behind the aforementioned probability decrease conditional on narrowing down the timeframe, though. Could someone clarify this intuition mathematically? Thanks in advance.

$\endgroup$

1 Answer 1

4
$\begingroup$

This is a consequence of means scaling linearly and standard deviations scaling in a square-root way.

Here is the Wikipedia definition of volatility:

"annualized" volatility $\sigma_{\text{annually}}$ is the standard deviation of an instrument's yearly logarithmic returns.

Thus, an annual return of 15% with a volatility of 10% is a log-normally distributed random variable with log-mean $0.15$ and log-standard deviation $0.10$:

$$ X_{\text{annual}}\sim LN(0.15,0.1). $$

We have a positive return if $X>1$. We can calculate the probability of this in R:

> plnorm(1,0.15,0.1,lower.tail=FALSE)
[1] 0.9331928

This matches Taleb's claim of a 93% chance of being in the money over a year.

What is the corresponding return over a quarter? For a quarterly view, the log-mean is divided by $4$ - but the log-SD is divided by $\sqrt{4}$ (see the Wikipedia page again):

$$ X_{\text{quarterly}}\sim LN\big(\frac{0.15}{4},\frac{0.1}{\sqrt{4}}\big) $$

with a chance for a quarterly positive return of

> plnorm(1,0.15/4,0.1/sqrt(4),lower.tail=FALSE)
[1] 0.7733726

And on a monthly basis, we get

$$ X_{\text{monthly}}\sim LN\big(\frac{0.15}{12},\frac{0.1}{\sqrt{12}}\big) $$

> plnorm(1,0.15/12,0.1/sqrt(12),lower.tail=FALSE)
[1] 0.6674972

And so on.


In any case, one typically sees this argument turned around. Here, we are going from a more stable long-term view to a less stable short-term one. Usually, one first looks at the more unstable short-term situation (which we of course observe sooner) and then calms down when one considers that in the longer term, the chances of a positive return are much better.

$\endgroup$
7
  • $\begingroup$ Good answer - thanks. I actually knew about the "square root of t"-rule. But for some reason I did not make the connection to it. $\endgroup$
    – shenflow
    Jan 26, 2021 at 22:32
  • $\begingroup$ Why has the log-normal distribution been used to make the example and how might the time scaling differ under another distribution fitted to returns $\endgroup$
    – develarist
    Jan 26, 2021 at 22:36
  • 1
    $\begingroup$ @develarist: the log-normal is a convention. It has similarly easy properties for a multiplicative variable as a normally distributed one has for an additive variable - which we make use of here. For instance, multiplying independent lognormals again yields a lognormal, with log-means and log-variances simply adding up, which is exactly analogous to addition of normally distributed variables, and we use this exactly here. For other distributions, things are likely different, but there are few distributions that make our life so easy. $\endgroup$ Jan 26, 2021 at 22:43
  • 1
    $\begingroup$ ... that said, Taleb is famous for arguing that the assumption of (log-)normality is mathematically easy, but does not capture reality well enough. Instead, one should use fatter tails. Apparently, his bets along these lines have paid off handsomely. $\endgroup$ Jan 26, 2021 at 22:44
  • 1
    $\begingroup$ Unfortunately, I'm not familiar with that distribution. If it works out similarly, please post an answer here! $\endgroup$ Jan 26, 2021 at 22:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.