Converting Sharpe-Based Log Returns to Geometric Returns

Summary

The document examines how to infer an annual geometric return from an annual Sharpe ratio, a risk-free yield, and annualized volatility. Its example uses a Sharpe ratio of 0.5, a three-month Treasury bill yield of 5% as a risk-free proxy, and volatility of 10%. Multiplying the Sharpe ratio by volatility gives a 5% excess return under the stated assumptions; the question is whether this can be combined directly with the quoted yield.

The proposed calculation converts the yield to a continuously compounded rate, adds the excess log return, then exponentiates to express the result geometrically. It arrives at 10.383% and asks whether that is correct. The document presents this as a tentative calculation, not a validated answer. It does not establish that Sharpe inputs use log returns, or specify the return conventions, compounding assumptions, and annualization details needed to confirm the conversion.

Key ideas

• The example combines a Sharpe ratio, annualized volatility, and a Treasury bill yield proxy.
• Multiplying the Sharpe ratio by volatility gives the stated 5% excess return.
• The proposed approach converts the risk-free yield to log form before combining returns.
• Exponentiation converts the combined log return back to a geometric return.
• The calculation is posed for review and does not resolve the assumptions required for a definitive result.

Tags

# Determine expected geometric return from Sharpe ratio


# Determine expected geometric return from Sharpe ratio












I'm trying to calculate the expected annual geometric return, given that I'm provided with an annual Sharpe ratio (0.5), the yield on a 3-month T-Bill (5%) (using this yield as a proxy for the risk-free return), and the annualized volatility (10%).

At first blush, I think that the expected annual return is 10% by simply calculating $(0.5)(10\%) + 5\% = 10\%$, but I think I'm neglecting the fact that I'm blending log returns and geometric returns.

I know that my expected excess return is 5%, though I know also that Sharpe is calculated with log returns, so this is a 5% log excess return. If the 3-month yield is 5%, this is a geometric yield, so I need to turn it into a log return: $\ln{(1+5\%)} = 0.04879 = 4.879\%$. Summing these together I get the expected annual log return of $5\% + 4.879\% = 9.879\%$, which I can then convert back into the geometric return: $\exp{(9.879\%)}-1 = 10.383\%$.

Therefore, I think that the expected annual geometric return is 10.383%. Am I doing something wrong?

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.