Sample size for a one-sided strategy comparison

Question

You want to test whether strategy B has a higher mean daily PnL than strategy A, running both live and splitting capital. Each strategy's daily PnL has a standard deviation of about $\sigma = 3$ units. You want to detect a true mean advantage of $\Delta = 1$ unit, using a one-sided test at $\alpha = 0.05$ with $80\%$ power.

Roughly how many days per strategy do you need?

Accepted Answer

The formula, one-sided version. n \approx \frac{2\,\sigma^2 \left(z{\alpha} + z\betaight)^2}{\Delta^2}. The only change from the two-sided case is z\alpha = 1.645 in place of z{\alpha/2} = 1.96. Constants. (z\alpha + z\beta)^2 = (1.645 + 0.84)^2 = (2.485)^2 = 6.18. Plug in. \sigma^2 = 9, \Delta^2 = 1: n \approx \frac{2 	imes 9 	imes 6.18}{1} = 111.3 \approx 112. So about 112 days per strategy. The signal here is small (\Delta is only one third of a standard deviation), which is why the count is over a hundred even with the one-sided discount. Takeaway. A one-sided test swaps 1.96 for 1.64…

Sample size for a one-sided strategy comparison

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