I'm aware that one way to do significance testing on a strategy is based on the sampling distribution of its Sharpe (see, e.g., Lo, 2002 and Opdyke, 2008).
However, it appears to me that there's another very simple way to do inference directly on the average return and standard deviation implied by the Sharpe. I'm wondering (1) if this approach is valid, or whether I'm overlooking something; (2) if there's some literature on this and similar approaches.
Given a strategy with (required / targeted / IS-based) daily Sharpe ratio $SR = \frac{\mu}{\sigma}$, with returns statistics $\mu$ (mean) and $\sigma$ (standard deviation).
The question is how long of an OS period the strategy needs at minimum to establish statistically significant returns.
Assuming iid returns (big assumption), the sample mean of the daily returns in the OS period after $n$ days can be modelled with (probably more appropriate to use t distribution though) $$\hat\mu \sim \mathcal{N}\left(\mu, \frac{\sigma^2}{n}\right)$$.
We define statistical significance at level $\alpha$ as the $z_{1-\alpha}$ score (one-sided test) from zero (i.e., the null is that $\mu=0$): $$\hat\mu \geq z_{1-\alpha} \frac{\sigma}{\sqrt{n}}$$.
With the definition of the daily Sharpe $SR = \frac{\mu}{\sigma}$ and substituting $\mu=\hat\mu$ this becomes:
$$n \geq \left(\frac{z_{1-\alpha}}{SR}\right)^2$$
Intuitively this makes sense: A lower Sharpe, or higher significance level increases the minimum sampling period. For instance, at $\alpha=90\%$ significance we obtain for a strategy with an annual Sharpe of 1: $n \geq (1.28 * \sqrt{252} / 1)^2 \approx 413$ days, but at $\alpha=95\%$ already $n \geq 682$ days.
