Years to prove a Sharpe of 2.0 beats a benchmark of 1.0

Question

Your strategy has an observed annualized Sharpe of 2.0. You want to show, with 95% confidence, that its true Sharpe exceeds a benchmark Sharpe of 1.0. Treat returns as roughly independent and normal.

Roughly how many years of track record do you need? (Use z = 1.645.)

Accepted Answer

Minimum track record length. Solving (SR - SR^)/\operatorname{SE} \geq z with \operatorname{SE}(SR) \approx \sqrt{(1 + 	frac{1}{2}SR^2)/T}: T \geq \left(1 + 	frac{1}{2}SR^2ight)\left(\frac{z}{SR - SR^}ight)^2. Plug in the numbers. With SR = 2.0, SR^ = 1.0, z = 1.645: T \geq (1 + 2)\left(\frac{1.645}{2.0 - 1.0}ight)^2 = 3 	imes (1.645)^2 = 3 	imes 2.706 \approx 8.1 	ext{ years}. Interpretation. You need about 8 years (round up to 9). Two forces trade off here: the wide gap over the benchmark (a full 1.0) works in your favor and pulls the required years down, while the high Sharpe (2…

Years to prove a Sharpe of 2.0 beats a benchmark of 1.0

Your answer

More Statistics questions