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When a Scaled Sharpe Ratio Resembles a t-Statistic

Article Quant Q&A · Author: spacemonkey
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Summary

The document asks whether annualizing a Sharpe ratio makes it a t-statistic. It compares the annualized measure, which scales excess average returns by return volatility and the square root of the number of periods in a year, with the familiar test statistic for a sample mean. This motivates interpreting the scaled Sharpe as a measure of excess return relative to its variability.

The response asserts that the Sharpe ratio follows a Student’s t distribution and points to a statistical treatment of Sharpe ratios. However, the answer is brief and provides no derivation or assumptions. The scaling alone does not establish that a statistic has a t distribution: the relationship depends on the return model, sample size, and how volatility is estimated. The document therefore raises a useful connection but does not fully explain when the comparison is valid or how to conduct inference.

Key ideas

  • Annualized Sharpe compares average excess return with return volatility after scaling by the square root of periods per year.
  • The question is whether this scaling gives the statistic the form of a test statistic for a sample mean.
  • The response claims a Student’s t distribution but offers no supporting derivation.
  • Inference depends on assumptions about returns and volatility estimation.

Tags

StatisticsRisk management
Full text
# Is scaled Sharpe ratio a t-statistic?


# Is scaled Sharpe ratio a t-statistic?












I was just reading Quantitative Trading: How to Build Your Own Algorithmic Trading Business and it suggests annualizing Sharpe ratio in order to compare performance of strategies:

$$\text{Annualized Sharpe Ratio} = \sqrt{N_T} \frac{\bar{R_s} - R_{b}}{\sigma_{R_s}}$$

where $\bar{R_s}$ are strategy returns for a certain period, $R_b$ -- benchmark returns and $N_T$ is number of periods in a year (e.g. 12 if $R_s$ is computed monthly). This seems like a t-statistic?

$$ t_{\bar{x}} = \sqrt{n}\frac{\bar{x} - \mu}{\text{s.e.}({\bar{x}})}$$

So Sharpe ratio can be interpreted as a number of standard deviations from a benchmark returns?

## Answer by Chris (score 4)

https://quant.stackexchange.com/a/58128

Yes, Sharpe follows a student's t distribution.

https://alo.mit.edu/wp-content/uploads/2017/06/The-Statistics-of-Sharpe-Ratios.pdf

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.