Normality assumption in Sharpe ratio

Question

I have read that the Sharpe ratio imposes a normality assumption, but I fail to see how. Standard deviation is statistic for any type of distribution. Anyone have any ideas?

Answer 1

You are correct that you can compute Sharpe ratios on portfolios with any return distribution. The issue is comparing Sharpe ratio's of non-normally distributed portfolios (which in reality is almost any portfolio). To take an extreme example. Consider two portfolios, with returns in excess of benchmark.

1. 50% chance of 10% return, 50% chance of a 20% return
2. 50% chance of 10% return, 50% chance of a 100% return

The Sharpe ratios are $$ 1. \frac{0.5 \cdot 0.1 + 0.5 \cdot 0.2}{\sqrt{0.5 (0.1 - 0.15)^2 + 0.5 (0.2 - 0.15)^2}} = 3 \\ 2. \frac{0.5 \cdot 0.1 + 0.5 \cdot 1}{\sqrt{0.5 (0.1 - 0.55)^2 + 0.5 (1 - 0.55)^2}} \approx 1.22 $$

Portfolio 2 clearly dominates portfolio 1, but its Sharpe ratio is much lower.

Answer 2

The Sharpe ratio is just a transformation of Student's t-test and it is a special case of the t-test so that all requirements to use a t-test apply to any use of the Sharpe ratio.

A statistic is any function that uses data from a sample. The Sharpe ratio is a statistic. Although it does not assume normality explicitly, it does assume the existence of a second moment and it is a poor statistic in small samples if the data generating function has a second moment but is not near to the normal distribution and the sample size is small.

It has been argued since Mandelbrot in 1963 that the distribution of returns lacks a mean and hence a variance, or standard deviation. I have written a proof that no standard deviation exists for stocks that I am about to submit for publication. Not all statistical distributions have a standard deviation.

The gist of my argument is that returns are a future value divided by a present value minus one. Under the Markowitz assumptions, of many buyers and sellers and markets in equilibrium the logical market behavior is for actors to bid their expectations. By the central limit theorem, as the number of buyers and sellers become large, the distribution of a set of expectations must converge to normality. This implies that the distribution is the ratio of two normal distributions, which would be $$\frac{1}{\pi}\frac{\gamma}{\gamma^2+(r-\mu)^2}.$$

If you take expectations over the distribution you will find that neither a mean nor a variance exists. This distribution is the Cauchy distribution. It doesn't really explain returns because of the limitation of liability, liquidity constraints, bankruptcy and mergers however it explains the overwhelming amount of the uncertainty. You have to move away from the Markowitz assumptions to get realistic distributions. NIST describes the Cauchy distribution thusly.

The Cauchy distribution is important as an example of a pathological case. Cauchy distributions look similar to a normal distribution. However, they have much heavier tails. When studying hypothesis tests that assume normality, seeing how the tests perform on data from a Cauchy distribution is a good indicator of how sensitive the tests are to heavy-tail departures from normality. Likewise, it is a good check for robust techniques that are designed to work well under a wide variety of distributional assumptions. The mean and standard deviation of the Cauchy distribution are undefined. The practical meaning of this is that collecting 1,000 data points gives no more accurate an estimate of the mean and standard deviation than does a single point.

One practical issue though is that the Sharpe ratio cannot exist. It is not difficult to show that a t-test would be perfectly inefficient as the sample size went to infinity.

So to answer your question exactly, a standard deviation does not exist for all distributions, and while there isn't a strict assumption of normality it will only be efficient for small samples if normality holds.