A stronger signal needs surprisingly little data

Question

A high-conviction signal has a per-trade signal-to-noise ratio of $0.2$ . Its z-statistic grows as $z = 0.2\sqrt{n}$ with the number of trades $n$ .

About how many trades are needed for this edge to reach two-sided significance at 5%, and what does comparing it to fainter signals show?

Accepted Answer

The computation. Two-sided significance needs z \geq 1.96. Solving 0.2\sqrt{n} = 1.96: \sqrt{n} = \frac{1.96}{0.2} = 9.8, \qquad n = 9.8^2 \approx 96 \approx 97 	ext{ trades}. The comparison. A per-trade edge of 0.2 needs only about 97 trades, versus roughly 385 for an edge of 0.1 and about 1537 for an edge of 0.05. Doubling the effect size cuts the required sample by a factor of four, because the sample size scales with the inverse square of the effect. Strong signals reveal themselves quickly; weak ones hide in the noise for a long time. The interpretation warning. The flip side is that a …

A stronger signal needs surprisingly little data

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