Comparing Two Sharpe Ratios: Jobson-Korkie-Memmel

Strategy A has a backtested Sharpe ratio of 1.4, Strategy B has 1.1. Is A genuinely better, or is that 0.3 gap just the kind of thing you'd expect to see by chance even if the two strategies were equally good? A plain eyeball comparison of two numbers can't answer this — you need a test that accounts for how noisy Sharpe ratio estimates are, and, importantly, that two strategies tested over the same time period usually have correlated returns, which changes how much the two estimates can diverge by chance alone.

Why you can't just run two separate tests

Testing whether A's Sharpe differs from zero, and separately whether B's does, still doesn't tell you whether A differs from B — that's a different question. Treating the two estimates as independent and running a standard two-sample test overstates how surprising a gap is, because both strategies are typically driven by the same market and move together to some degree. Ignoring that shared variation makes the test too eager to declare a difference significant.

The Jobson-Korkie test (later corrected by Memmel, whose version is the one actually used) builds a test statistic for the difference in Sharpe ratios that explicitly incorporates the covariance between the two strategies' returns, not just each one's own variance. Given two return series with means $\mu_A, \mu_B$, standard deviations $\sigma_A, \sigma_B$, and covariance $\sigma_{AB}$, the test statistic (using Memmel's small-sample correction, with $n$ observations) is approximately

$$z = \frac{\hat{S}_A - \hat{S}_B}{\sqrt{\dfrac{1}{n}\Big(2 - 2\rho_{AB} + \tfrac{1}{2}\hat{S}_A^2 + \tfrac{1}{2}\hat{S}_B^2 - \rho_{AB}^2\hat{S}_A\hat{S}_B\Big)}} ,$$

where $\hat{S}_A, \hat{S}_B$ are the two estimated Sharpe ratios and $\rho_{AB}$ is the correlation between the strategies' returns. In plain English: the denominator is the standard error of the difference between two Sharpe estimates, and it shrinks as the strategies become more correlated — because when two strategies move together, part of what drives one Sharpe estimate also drives the other, so a gap between them is less likely to be pure noise. Compare $z$ to a standard normal distribution to get a p-value.

Worked example

Two strategies over 3 years of monthly returns ($n=36$): $\hat{S}_A = 1.4$, $\hat{S}_B = 1.1$ (monthly-scale estimates), correlation between their monthly returns $\rho_{AB} = 0.6$. Plugging in: the bracket term is $2 - 2(0.6) + 0.5(1.4^2) + 0.5(1.1^2) - (0.6)^2(1.4)(1.1) \approx 2 - 1.2 + 0.98 + 0.605 - 0.554 \approx 1.83$. Dividing by $n=36$ and taking the square root gives a standard error of about $\sqrt{1.83/36} \approx 0.225$. The test statistic is $z = (1.4-1.1)/0.225 \approx 1.33$, which is not significant at the usual 5% threshold (would need roughly $z > 1.96$) — despite the two Sharpe ratios looking meaningfully different at first glance, the data doesn't yet rule out that this is noise.