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This one took me a minute to set up properly.
Start by writing the Sharpe ratio formula and expressing the filtered Sharpe in terms of the full-sample mean and standard deviation, noting that removing zeros changes both. Then analyze the ratio of the two Sharpe ratios to determine when the filtered Sharpe is higher or lower, considering the sign of the mean. Finally, generalize by considering arbitrary removal of n observations and discuss the impact on the Sharpe ratio.
Pro tip: Emphasize that the Sharpe ratio is not invariant to removing zero returns; it depends on the signal-to-noise ratio. A common mistake is to assume that removing zeros always improves the Sharpe ratio, but it can actually decrease it if the mean return is negative.
Let the full sample have 2n observations with n zeros. Compute the full-sample Sharpe ratio S_full = μ_full / σ_full, where μ_full is the mean and σ_full is the standard deviation. Compute the filtered Sharpe ratio S_filt = μ_filt / σ_filt using only the n non-zero observations.
Relate μ_filt and σ_filt to μ_full and σ_full. Since half the observations are zero, μ_full = μ_filt / 2. Also, the sum of squares of the full sample equals the sum of squares of the non-zero observations, so σ_full^2 = (1/(2n)) Σ x_i^2 - (μ_filt/2)^2, while σ_filt^2 = (1/n) Σ x_i^2 - μ_filt^2. Derive the relationship between σ_full and σ_filt.
Form the ratio S_filt / S_full = (μ_filt / σ_filt) / (μ_full / σ_full) = 2 * (σ_full / σ_filt). Determine whether this ratio is greater than 1, which depends on the sign of μ_filt and the relative variability. Show that S_filt > S_full if and only if μ_filt > 0 (assuming non-zero variance).
Consider removing any n observations, not necessarily zeros. The filtered Sharpe ratio can be higher or lower depending on which observations are removed. Discuss that removing low-volatility or negative-mean observations can increase the Sharpe ratio, while removing high-mean observations can decrease it. The conclusion depends on the properties of the removed subset.
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