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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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  • $\begingroup$ How do you know that the expected return used in the question is a log return? $\endgroup$
    – KaiSqDist
    Commented Nov 14, 2023 at 14:35
  • $\begingroup$ Because Sharpe ratios are calculated with log returns, which I understand to be the case since Sharpe ratios can be scaled with the square root of time. Being able to scale the Sharpe ratio in this way is only possible if we use log returns in its calculation. $\endgroup$
    – Ringleader
    Commented Nov 14, 2023 at 15:03

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