The Relationship Between Sharpe Ratios and Mean-Return t-Statistics
Summary
The document asks how signal returns, a regression slope, the Sharpe ratio, and a test statistic for mean return relate. Its answer compares the standardized mean return with the t-statistic for testing a zero mean, which scales that standardized value by the square root of the sample size. It then relates periodic Sharpe ratios to annualized Sharpe ratios under an independent, identically distributed return assumption.
The central intuition is that a periodic Sharpe resembles a standardized score, while annualization introduces a square-root-of-periods scaling similar to a t-statistic. The answer also interprets Sharpe values through a normal-return win-probability analogy. These comparisons are approximate and depend on assumptions such as iid returns, normality, and a zero benchmark return; they do not fully resolve how to interpret the regression slope or establish that a signal-regression t-stat always equals a Sharpe-based statistic.
Key ideas
- A t-statistic for a zero-mean return scales the standardized mean by the square root of the sample size.
- Annualized Sharpe ratios use a similar square-root scaling under an iid assumption.
- A periodic Sharpe can be viewed as a standardized return measure when the benchmark is zero.
- The normal-distribution win-rate intuition depends on assumptions and does not establish equivalence for every signal regression.
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Full text
# How does one show that the Sharpe Ratio is closely related to the t-statistic of the mean differential return?
# How does one show that the Sharpe Ratio is closely related to the t-statistic of the mean differential return?
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.
## Answer by Newquant (score 4, accepted)
https://quant.stackexchange.com/a/76059
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).Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.