Normality assumption in Sharpe ratio

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?

• This is a basic statistics question. The Sharpe ratio is calculated using the mean and variance of a distribution, therefore it is a less descriptive measure the more the mean and variance do not completely describe the distribution. Jan 16 '14 at 6:32
• @JoshuaUlrich, I disagree with that notion. SR calculations for any return distributions are 100% accurate. SR of a bond portfolio can be fairly compared with an emerging market stock portfolio. After all the returns are scaled by their own volatility measure. The problem arises due to a problem with the definition of SR itself, it penalizes out-sized returns to the upside, which are desirable yet not properly accounted for in SR computations.
– Matt
Jan 17 '14 at 2:32
• @MattWolf: so you would argue that there's no normality assumption? Even if you account for upside/downside deviation, the Sharpe Ratio still has issues if the distributions are non-normal. Jan 17 '14 at 15:03
• @JoshuaUlrich, that is not what I said, of course are the moments different for different distributions. My point was, and sorry if that was not clear, that it is impossible to attach an accurate distributional assumption to any financial asset returns, hence the most often used normal and log-normal assumptions for most asset returns, which validates the usage of mean and stdev under normal distributional assumptions. In fact most all financial asset returns exhibit dynamic distributional features.
– Matt
Jan 18 '14 at 11:07
• @Matt Wolf, I get what you're saying, but is it because the standard deviation is a unitless measure? Apr 4 '17 at 22:13

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.

• Very good example!
– Ric
Jan 16 '14 at 8:18
• Many thanks to both David Nehme and Joshua Ulrich for answering my question. Jan 16 '14 at 18:27
• @user6997, you should mark this as correct answer if you think it properly addresses your question
– Matt
Jan 17 '14 at 2:25
• @David, I would argue that this particular observation is not really because of different distributional assumptions but because of the intrinsic flaws of SR itself which is that it penalizes large upside returns because of an increase in volatility such upside returns cause. It is absolutely fair to compare SRs of portfolios of entirely different asset classes with intrinsically different distributional assumptions.
– Matt
Jan 17 '14 at 2:34

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.

• Looking forward to reading your paper. It seems like you have an outsider mindset. Why is the convergence of two normally distributed sets of expectations the ratio of two normal distributions? Sees, if anything, if you be bimodal. And how is this related to the Cauchy? Apr 24 '17 at 4:53
• Actually, I am an old-time insider. I just am an outsider on this issue because I am a very strict empiricist. As to the expectations question, the sampling distribution of an arbitrarily large set of expectations will tend toward normality. If you add the assumption of independence and equilibrium and work in the error space, then you end up with a Cauchy. If the errors are not independent and/or are systematically away from the equilibrium then you get a very ugly transform of the Cauchy distribution. Apr 24 '17 at 5:52
• If you do other assumptions, such as using a single price auction, you end up with yet another type of distribution as the ratio of two Gumbel distributions. And, of course, if you alter the assumptions yet again, you can even end up with a log-normal distribution for a set of single period discount bonds with simultaneous maturities. Apr 24 '17 at 5:54
• Likewise, if you collapse liquidity as in 2008, or play games with the macro economy as the Fed did in August of 1929 you get yet again another set of ugly distributions once the system goes into free-fall. Liquidity should always be a real world component of your distribution of returns since you cannot sell your stock if nobody can buy it. Apr 24 '17 at 5:58
• Finally, the limitation on liability really matters a lot in the macro economy because it truncates losses at -100% and this appears to matter for economic growth, surprisingly. It appears that abolishing slavery matters quite a bit more than appears in standard literature because it forces behavioral changes on the part of creditors. Apr 24 '17 at 6:00