I see it being mentioned in many places, such as here, and even here.
How should I interpret it?
Suppose I have an array of signals, I, and returns of those signals, R
Then my regression is
R = a + BI
And my t-test on B will thus be B-0/std(B), and as such, B is the mean differential returns? Am I missing something or does this seem right?
TLDR: Want to understand how signals, returns of those signals fit into a regression to show the t-test of that regression vs sharpe is scaled.
It makes sense. Intuitively, I consider the Sharpe ratio as a proxy for win-rate. Let's assume normal returns.
The Z score is defined as: $$ \frac{x - \mu}{\sigma} $$
The T-stat for zero-mean is defined as: $$ \frac{\mu}{\sigma} * \sqrt{N}$$
The Sharpe ratio of periodic returns is defined as: $$ \frac{x - r}{\sigma} $$
The annualised Sharpe (assuming iid) is given by: $$ \frac{(x - r)*T}{\sigma * \sqrt{T}} = \frac{(x - r)}{\sigma} * \sqrt{T}$$
You can see the form of all these equations is very similar. The annualised Sharpe ratio, which takes periodic returns like daily or weekly, then multiplies by the $\sqrt{N}$ where N is the number of periods in a year (52 weeks, 252 days, etc). For 0 r, this tells you over the course of a year the probability estimate for that return being positive. An annualised sharpe ratio of 1 implies a win rate at the yearly level of ~84%. This is equivalent to a daily Sharpe ratio (or equivalently a z-score) of ~ 0.06 (52% win rate).
So in conclusion, the Sharpe ratio with 0 r evaluated at periodic returns is analogous to a z-score, and the annualised Sharpe ratio is analogous to a t-stat, as both are scaled by $\sqrt{N}$. For returns taken over short periods, it is generally acceptable to use the Sharpe and/or t-stat values to generate p-values from a normal distribution for success probabilities (in the Sharpe's case).
