CONTENTS

Exercise: Sharpe-Ratio Confidence Interval from the CLT

Problem

A strategy posts nn i.i.d. daily returns r1,,rnr_1, \ldots, r_n with true (unknown) mean μ\mu and standard deviation σ\sigma. The annualised Sharpe ratio is SR=252μ/σ\text{SR} = \sqrt{252}\cdot \mu/\sigma, and it is estimated by plugging in the sample mean rˉ\bar r and sample standard deviation σ^\hat\sigma:
SR^=252rˉ/σ^.\widehat{\text{SR}} = \sqrt{252}\cdot \bar r / \hat\sigma.
  1. Show that the sample Sharpe SR^\widehat{\text{SR}} is, by the CLT, approximately normal with mean SR\text{SR} and standard error (1+SR2/2)/nyears\sqrt{(1 + \text{SR}^2/2)/n_{\text{years}}}, where nyears=n/252n_{\text{years}} = n/252 — under the i.i.d. Gaussian-returns assumption (Lo, 2002).

  2. Suppose a manager reports SR^=1.0\widehat{\text{SR}} = 1.0 from a one-year backtest (nyears=1n_{\text{years}} = 1). Compute the 95% confidence interval for the true Sharpe. Does the interval exclude zero?

  3. How many years of backtest are needed to conclude at the 95% level that the true Sharpe is strictly greater than zero, given a point estimate of SR^=1.0\widehat{\text{SR}} = 1.0? Given SR^=0.5\widehat{\text{SR}} = 0.5?

  4. Conceptual. Why does the standard error grow with SR\text{SR}? Give a one-sentence intuition.

Hint

For part 1, don't re-derive the Lo (2002) result — state the standard error and use it. The key structural fact is that the sampling variance of SR^\widehat{\text{SR}} depends on the Sharpe ratio itself.

Jump to the solution when you're ready.